Chemical kinetics

Chemical kinetics
Chemical kinetics, also known as reaction kinetics, is the study of rates of chemical processes. Chemical kinetics includes investigations of how different experimental conditions can influence the speed of a chemical reaction and yield information about the reaction s mechanism and transition states, as well as the construction of mathematical models that can describe the characteristics of a chemical reaction.

Factors affecting reaction rate
Depending upon what substances are reacting, the reaction rate varies. Acid/base reactions, the formation of salts, and ion exchange are fast reactions. When covalent bond formation takes place between the molecules and when large molecules are formed, the reactions tend to be very slow. Nature and strength of bonds in reactant molecules greatly influence the rate of its transformation into products.
Physical state
The physical state (solid, liquid, or gas) of a reactant is also an important factor of the rate of change. When reactants are in the same phase, as in aqueous solution, thermal motion brings them into contact. However, when they are in different phases, the reaction is limited to the interface between the reactants. Reaction can occur only at their area of contact; in the case of a liquid and a gas, at the surface of the liquid. Vigorous shaking and stirring may be needed to bring the reaction to completion. This means that the more finely divided a solid or liquid reactant the greater its surface area per unit volume and the more contact it with the other reactant, thus the faster the reaction. To make an analogy, for example, when one starts a fire, one uses wood chips and small branches — one does not start with large logs right away. In organic chemistry, on water reactions are the exception to the rule that homogeneous reactions take place faster than heterogeneous reactions.
Surface area of solids
In a solid, only those particles that are at the surface can be involved in a reaction. Crushing a solid into smaller parts means that more particles are present at the surface, and the frequency of collisions between these and reactant particles increases, and so reaction occurs more rapidly. For example, Sherbet (powder) is a mixture of very fine powder of malic acid (a weak organic acid) and sodium hydrogen carbonate. On contact with the saliva in the mouth, these chemicals quickly dissolve and react, releasing carbon dioxide and providing for the fizzy sensation. Also, fireworks manufacturers modify the surface area of solid reactants to control the rate at which the fuels in fireworks are oxidised, using this to create different effects. For example, finely divided aluminium confined in a shell explodes violently. If larger pieces of aluminium are used, the reaction is slower and sparks are seen as pieces of burning metal are ejected.
Concentration
The reactions are due to collisions of reactant species. The frequency with which the molecules or ions collide depends upon their concentrations. The more crowded the molecules are, the more likely they are to collide and react with one another. Thus, an increase in the concentrations of the reactants will usually result in the corresponding increase in the reaction rate, while a decrease in the concentrations will usually have a reverse effect. For example, combustion will occur more rapidly in pure oxygen than in air (21% oxygen).
The rate equation shows the detailed dependence of the reaction rate on the concentrations of reactants and other species present. Different mathematical forms are possible depending on the reaction mechanism. The actual rate equation for a given reaction is determined experimentally and provides information about the reaction mechanism. The mathematical expression of the rate equation is often given by
rate = k [A]x[B]y
Here x and y are constants for each reactant, while [A] and [B] are molar concentrations of reactants. Also k is the reaction rate constant which can only be determined experimentally.
Temperature
Temperature usually has a major effect on the rate of a chemical reaction. Molecules at a higher temperature have more thermal energy. Although collision frequency is greater at higher temperatures, this alone contributes only a very small proportion to the increase in rate of reaction. Much more important is the fact that the proportion of reactant molecules with sufficient energy to react (energy greater than activation energy: E > Ea) is significantly higher and is explained in detail by the Maxwell–Boltzmann distribution of molecular energies.
The rule of thumb that the rate of chemical reactions doubles for every 10 °C temperature rise is a common misconception. This may have been generalized from the special case of biological systems, where the ? (temperature coefficient) is often between 1.5 and 2.5.
A reaction s kinetics can also be studied with a temperature jump approach. This involves using a sharp rise in temperature and observing the relaxation time of the return to equilibrium. A particularly useful form of temperature jump apparatus is a shock tube, which can rapidly jump a gas s temperature by more than 1000 degrees.
Catalysts
Catalysis

Generic potential energy diagram showing the effect of a catalyst in a hypothetical endothermic chemical reaction. The presence of the catalyst opens a different reaction pathway (shown in red) with a lower activation energy. The final result and the overall thermodynamics are the same.
A catalyst is a substance that alters the rate of a chemical reaction but remains chemically unchanged afterwards. The catalyst increases the rate of the reaction by providing a different reaction mechanism to occur with a lower activation energy. In autocatalysis a reaction product is itself a catalyst for that reaction leading to positive feedback. Proteins that act as catalysts in biochemical reactions are called enzymes. Michaelis–Menten kinetics describe the rate of enzyme mediated reactions. A catalyst does not affect the position of the equilibrium, as the catalyst speeds up the backward and forward reactions equally.
In certain organic molecules, specific substituents can have an influence on reaction rate in neighbouring group participation.[citation needed]
Pressure
Increasing the pressure in a gaseous reaction will increase the number of collisions between reactants, increasing the rate of reaction. This is because the activity of a gas is directly proportional to the partial pressure of the gas. This is similar to the effect of increasing the concentration of a solution.
In addition to this straightforward mass-action effect, the rate coefficients themselves can change due to pressure. The rate coefficients and products of many high-temperature gas-phase reactions change if an inert gas is added to the mixture; variations on this effect are called fall-off and chemical activation. These phenomena are due to exothermic or endothermic reactions occurring faster than heat transfer, causing the reacting molecules to have non-thermal energy distributions (non-Boltzmann distribution). Increasing the pressure increases the heat transfer rate between the reacting molecules and the rest of the system, reducing this effect.
Condensed-phase rate coefficients can also be affected by (very high) pressure; this is a completely different effect than fall-off or chemical-activation. It is often studied using diamond anvils.
A reaction s kinetics can also be studied with a pressure jump approach. This involves making fast changes in pressure and observing the relaxation time of the return to equilibrium.
Experimental methods
The experimental determination of reaction rates involves measuring how the concentrations of reactants or products change over time. For example, the concentration of a reactant can be measured by spectrophotometry at a wavelength where no other reactant or product in the system absorbs light.
For reactions which take at least several minutes, it is possible to start the observations after the reactants have been mixed at the temperature of interest.
Fast reactions
For faster reactions, the time required to mix the reactants and bring them to a specified temperature may be comparable or longer than the half-life of the reaction.[7] Special methods to start fast reactions without slow mixing step include
Stopped flow methods, which can reduce the mixing time to the order of a millisecond[7][8][9]Chemical relaxation methods such as temperature jump and pressure jump, in which a pre-mixed system initially at equilibrium is perturbed by rapid heating or depressurization so that it is no longer at equilibrium, and the relaxation back to equilibrium is observed.[7][10][11][12] For example, this method has been used to study the neutralization H3O+ + OH? with a half-life of 1 ?s or less under ordinary conditions.[7][12]. Flash photolysis, in which a laser pulse produces highly excited species such as free radicals, whose reactions are then studied.[9][13][14][15]
Equilibrium
While chemical kinetics is concerned with the rate of a chemical reaction, thermodynamics determines the extent to which reactions occur. In a reversible reaction, chemical equilibrium is reached when the rates of the forward and reverse reactions are equal [the principle of dynamic equilibrium ] and the concentrations of the reactants and Products no longer change. This is demonstrated by, for example, the Haber–Bosch process for combining nitrogen and hydrogen to produce ammonia. Chemical clock reactions such as the Belousov–Zhabotinsky reaction demonstrate that component concentrations can oscillate for a long time before finally attaining the equilibrium.
Free energy
In general terms, the free energy change (?G) of a reaction determines whether a chemical change will take place, but kinetics describes how fast the reaction is. A reaction can be very exothermic and have a very positive entropy change but will not happen in practice if the reaction is too slow. If a reactant can produce two different products, the thermodynamically most stable one will in general form, except in special circumstances when the reaction is said to be under kinetic reaction control. The Curtin–Hammett principle applies when determining the product ratio for two reactants interconverting rapidly, each going to a different product. It is possible to make predictions about reaction rate constants for a reaction from free-energy relationships.
The kinetic isotope effect is the difference in the rate of a chemical reaction when an atom in one of the reactants is replaced by one of its isotopes.
Chemical kinetics provides information on residence time and heat transfer in a chemical reactor in chemical engineering and the molar mass distribution in polymer chemistry.

Order and molecularity of reaction
Molecularity in chemistry is the number of molecules that come together to react in an elementary (single-step) reaction[1] and is equal to the sum of stoichiometric coefficients of reactants in this elementary reaction.[2] Depending on how many molecules come together, a reaction can be unimolecular, bimolecular or trimolecular.
The kinetic order of any elementary reaction or reaction step is equal to its molecularity, and the rate equation of an elementary reaction can therefore be determined by inspection, from the molecularity.[1]
The kinetic order of a complex (multistep) reaction, however, cannot be equated to molecularity since molecularity only describes elementary reactions or steps.
Unimolecular reactions
In a unimolecular reaction, a single molecule rearranges atoms forming different molecules.[1] This is illustrated by the equationand is described by the first order rate law where [A] is the concentration of species A, t is time, and kr is the reaction rate constant.
As can be deduced from the rate law equation, the number of A molecules that decay is proportional to the number of A molecules available. An example of a unimolecular reaction, is the isomerization of cyclopropane to propene:

Unimolecular reactions can be explained by the Lindemann-Hinshelwood mechanism.
Bimolecular reactions
In a bimolecular reaction, two molecules collide and exchange energy, atoms or groups of atoms.[1]This can be described by the equation

which corresponds to the second order rate law: Here, the rate of the reaction is proportional to the rate at which the reactants come together. An example of a bimolecular reaction is the SN2-type nucleophilic substitution of methyl bromide by hydroxide ion:[3]

Trimolecular reactions
A termolecular[4][5] (or trimolecular)[6] reaction in solutions or gas mixtures involves three reactant molecules simultaneously colliding.[4] However the term termolecular is also used to refer to three body association reactions of the type Where the M over the arrow denotes that to conserve energy and momentum a second reaction with a third body is required. After the initial bimolecular collision of A and B an energetically excited reaction intermediate is formed, then, it collides with a M body, in a second bimolecular reaction, transferring the excess energy to it.[7].The reaction can be explained as two consecutive reactions:
These reactions frequently have a pressure and temperature dependence region of transition between second and third order kinetics.[8]Catalytic reactions are often three-component, but in practice a complex of the starting materials is first formed and the rate-determining step is the reaction of this complex into products, not an adventitious collision between the two species and the catalyst. For example, in hydrogenation with a metal catalyst, molecular dihydrogen first dissociates onto the metal surface into hydrogen atoms bound to the surface, and it is these monatomic hydrogens that react with the starting material, also previously adsorbed onto the surface.
Reactions of higher molecularity are not observed due to very small probability of simultaneous interaction between 4 or more molecules.

Difference between molecularity and order of reaction
It is important to distinguish molecularity from order of reaction. The order of reaction is an empirical quantity determined by experiment from the rate law of the reaction. It is the sum of the exponents in the rate law equation.[10] Molecularity, on the other hand, is deduced from the mechanism of an elementary reaction, and is used only in context of an elementary reaction. It is the number of molecules taking part in this reaction. This difference can be illustrated on the reaction between nitric oxide and hydrogen: The observed rate law is , so that the reaction is third order. Since the order does not equal the sum of reactant stoechiometric coefficients, the reaction must involve more than one step. The proposed two-step mechanism[11] has a rate-limiting first step whose molecularity corresponds to the overall order of 3:
On the other hand, the molecularity of this reaction is undefined, because it involves a mechanism of more than one step. However, we can consider the molecularity of the individual elementary reactions that make up this mechanism: the first step is termolecular because it involves three reactant molecules, while the second step is bimolecular because it involves two reactant molecules.
Pseudo reactions
The rate law or rate equation for a chemical reaction is an equation that links the reaction rate with the concentrations or pressures of the reactants and constant parameters (normally rate coefficients and partial reaction orders).[1] For many reactions the rate is given by a power law such as where [A] and [B] express the concentration of the species A and B (usually in moles per liter (molarity, M)). The exponents x and y are the partial orders of reaction for A and B and the overall reaction order is the sum of the exponents. These are often positive integers, but they may also be zero, fractional, or negative. The constant k is the reaction rate constant or rate coefficient of the reaction and has units of 1/time. Its value may depend on conditions such as temperature, ionic strength, surface area of an adsorbent, or light irradiation.
Elementary (single-step) reactions have reaction orders equal to the stoichiometric coefficients for each reactant. The overall reaction order, i.e. the sum of stoichiometric coefficients of reactants, is always equal to the molecularity of the elementary reaction. Complex (multi-step) reactions may or may not have reaction orders equal to their stoichiometric coefficients.
The rate equation of a reaction with an assumed multi-step mechanism can often be derived theoretically using quasi-steady state assumptions from the underlying elementary reactions, and compared with the experimental rate equation as a test of the assumed mechanism. The equation may involve a fractional order, and may depend on the concentration of an intermediate species.
A reaction can also have an undefined reaction order with respect to a reactant if the rate is not simply proportional to some power of the concentration of that reactant; for example, one cannot talk about reaction order in the rate equation for a bimolecular reaction between adsorbed molecules.

Rate equation
Chemical reaction is an equation that links the reaction rate with the concentrations or pressures of the reactants and constant parameters (normally rate coefficients and partial reaction orders).[1] For many reactions the rate is given by a power law such as where [A] and [B] express the concentration of the species A and B (usually in moles per liter (molarity, M)). The exponents x and y are the partial orders of reaction for A and B and the overall reaction order is the sum of the exponents. These are often positive integers, but they may also be zero, fractional, or negative. The constant k is the reaction rate constant or rate coefficient of the reaction and has units of 1/time. Its value may depend on conditions such as temperature, ionic strength, surface area of an adsorbent, or light irradiation.
Elementary (single-step) reactions have reaction orders equal to the stoichiometric coefficients for each reactant. The overall reaction order, i.e. the sum of stoichiometric coefficients of reactants, is always equal to the molecularity of the elementary reaction. Complex (multi-step) reactions may or may not have reaction orders equal to their stoichiometric coefficients. The rate equation of a reaction with an assumed multi-step mechanism can often be derived theoretically using quasi-steady state assumptions from the underlying elementary reactions, and compared with the experimental rate equation as a test of the assumed mechanism. The equation may involve a fractional order, and may depend on the concentration of an intermediate species. A reaction can also have an undefined reaction order with respect to a reactant if the rate is not simply proportional to some power of the concentration of that reactant; for example, one cannot talk about reaction order in the rate equation for a bimolecular reaction between adsorbed molecules:
Derivation of order of reaction
Zero order reaction:

Zero order
For zero-order reactions, the reaction rate is independent of the concentration of a reactant, so that changing its concentration has no effect on the speed of the reaction. Thus, the concentration changes linearly with time. This may occur when there is a bottleneck which limits the number of reactant molecules that can react at the same time, for example if the reaction requires contact with an enzyme or a catalytic surface.[7] Many enzyme-catalyzed reactions are zero order, provided that the reactant concentration is much greater than the enzyme concentration which controls the rate, so that the enzyme is saturated. For example, the biological oxidation of ethanol to acetaldehyde by the enzyme liver alcohol dehydrogenase (LADH) is zero order in ethanol.[8] Similarly reactions with heterogeneous catalysis can be zero order if the catalytic surface is saturated. For example, the decomposition of phosphine (PH3) on a hot tungsten surface at high pressure is zero order in phosphine which decomposes at a constant rate.[7] In homogeneous catalysis zero order behavior can come about from reversible inhibition. For example, ring-opening metathesis polymerization using third-generation Grubbs catalyst exhibits zero order behavior in catalyst due to the reversible inhibition that is occur between the pyridine and the ruthenium center.[9]
In some reactions, the rate is apparently independent on the reactant concentration. The rates of these zero-order reactions do not vary with increasing nor decreasing reactants concentrations. This means that the rate of the reaction is equal to the rate constant(k). Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions. Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.
There are two general conditions that can give rise to zero-order rates:
1. Only a small fraction of the reactant molecules are in a location or state in which they are able to react, and this fraction is continually replenished from the larger pool.
2. When two or more reactants are involved, the concentrations of some are much greater than those of others
This situation commonly occurs when a reaction is catalyzed by attachment to a solid surface (heterogeneous catalysis) or to an enzyme.
Example 1: Decomposition of Nitrous Oxide
Important example of this type of reaction is the decomposition of ammonia gas over tungsten filament is a good example of zero order reaction:

Nitrous oxide will decompose exothermically into nitrogen and oxygen, at a temperature of approximately 575 °C on the surface of nickel metal:

This reaction in the presence of a hot platinum wire (which acts as a catalyst) is zero-order, but it follows more conventional second order kinetics when carried out entirely in the gas phase.


In this case, the N2O molecules that react are limited to those that have attached themselves to the surface of the solid catalyst. Once all of the sites on the limited surface of the catalyst have been occupied, additional gas-phase molecules must wait until the decomposition of one of the adsorbed molecules frees up a surface site.
Enzyme-catalyzed reactions in organisms begin with the attachment of the substrate to the active site on the enzyme, leading to the formation of an enzyme-substrate complex. If the number of enzyme molecules is limited in relation to substrate molecules, then the reaction may appear to be zero-order.
This is most often seen when two or more reactants are involved. Thus if the reaction
A+B?products (1)
is first-order in both reactants so that
rate=k?[A][B] (2)
If B is present in great excess, then the reaction will appear to be zero order in B (and first order overall). This commonly happens when B is also the solvent that the reaction occurs in
Differential Form of the Zeroth Order Rate Law


From last equation it can be found that there is a direct proportionality between time and increase in concentration of product(x). the rate constant will be:
Ko= x/t and its unit will be concentration unit over time
(i.e. mol/liter/minute).
Rate constant can be calculated graphically by plotting concentration of product versus time of reaction and the slope then will be equal to (k).









Figure xx: calculating of rate constant for zero order reaction using graphical method


Figure 1: Rate vs. time (A) and Concentration vs. time for a zero order reaction[
Integrated form of zeroth order rate law:
Rate=?d[A]/dt =k[A] 0 =k=constant (3)


Figure 1: Rate vs. time (A) and Concentration vs. time for a zero order reaction.
Integrated Form of the Zeroth Order Rate Law
Integration of the differential rate law yields the concentration as a function of time. Start with the general rate law equations
Rate=k[A] n (4)
First, write the differential form of the rate law with n=0 n=0
[A]=[A] 0 ?kt (8)
The integrated form of the rate law allows us to find the population of reactant at any time after the start of the reaction.
Graphing Zero-order Reactions
[A]=?kt+[A] 0 (9)
is in the form y = mx+b where slope = m = -k and the y- intercept = b = [A] 0 [A]0
Zero-order reactions are only applicable for a very narrow region of time. Therefore, the linear graph shown below (Figure 2) is only realistic over a limited time range. If we were to extrapolate the line of this graph downward to represent all values of time for a given reaction, it would tell us that as time progresses, the concentration of our reactant becomes negative. We know that concentrations can never be negative, which is why zero-order reaction kinetics is applicable for describing a reaction for only brief window and must eventually transition into kinetics of a different order.

Figure 2: (left) Concentration vs. time of a zero-order reaction. (Right) Concentration vs. time of a zero-order catalyzed reaction.
To understand where the above graph comes from, let us consider a catalyzed reaction. At the beginning of the reaction, and for small values of time, the rate of the reaction is constant; this is indicated by the blue line in Figures 2; right. This situation typically happens when a catalyst is saturated with reactants. With respect to Michaelis-Menton kinetics, this point of catalyst saturation is related to the V max Vmax . As a reaction progresses through time, however, it is possible that less and less substrate will bind to the catalyst. As this occurs, the reaction slows and we see a tailing off of the graph (Figure 2; right). This portion of the reaction is represented by the dashed black line. In looking at this particular reaction, we can see that reactions are not zero-order under all conditions. They are only zero-order for a limited amount of time.
If we plot rate as a function of time, we obtain the graph below (Figure 3). Again, this only describes a narrow region of time. The slope of the graph is equal to k, the rate constant. Therefore, k is constant with time. In addition, we can see that the reaction rate is completely independent of how much reactant you put in.

Figure 3: Rate vs. time of a zero-order reaction.
Relationship Between Half-life and Zero-order Reactions
The half-life. T=1/2, t1/2, is a timescale in which each half-life represents the reduction of the initial population to 50% of its original state. We can represent the relationship by the following equation.
[A]=1/2 of [A]o (10)
Using the integrated form of the rate law, we can develop a relationship between zero-order reactions and the half-life.
[A]=[A]o?kt (11)
Substitute

Determination of half time for zero order reaction
Half time (t1/2) is the time that is required to consume a half concentration of the reactant (i.e. x=1/2 a):
x=ko ×t 1/2a= ko × t1/2 t1/2= a/2ko


Questions
1. Using the integrated form of the rate law, determine the rate constant k of a zero-order reaction if the initial concentration of substance A is 1.5 M and after 120 seconds the concentration of substance A is 0.75 M.
2. Using the substance from the previous problem, what is the half-life of substance A if its original concentration is 1.2 M?
3. If the original concentration is reduced to 1.0 M in the previous problem, does the half-life decrease, increase, or stay the same? If the half-life changes what is the new half-life?
4. Given are the rate constants k of three different reactions:
• Reaction A: k = 2.3 M-1s-1
• Reaction B: k = 1.8 Ms-1
• Reaction C: k = 0.75 s-1
Which reaction represents a zero-order reaction?
5. True/False: If the rate of a zero-order reaction is plotted as a function of time, the graph is a strait line where \( rate = k\ ).
Answers
1. The rate constant k is 0.00624 M/s
2. The half-life is 96 seconds.
3. Since this is a zero-order reaction, the half-life is dependent on the concentration. In this instance, the half-life is decreased when the original concentration is reduced to 1.0 M. The new half-life is 80 seconds.
4. Reaction B represents a zero-order reaction because the units are in M/s. Zero-order reactions always have rate constants that are represented by molars per unit of time. Higher order reactions, however, require the rate constant to be represented in different units.
5. True. When using the rate function rate=k[A] n rate=k[A]n with n equal to zero in zero-order reactions. Therefore, rate is equal to the rate constant k.

First-Order Reactions
A first order reaction depends on the concentration of only one reactant (a unimolecular reaction). Other reactants can be present, but each will be zero order. The rate law for such a reaction is The half-life is independent of the starting concentration and is given by Examples of such reactions are:
• In organic chemistry, the class of SN1 (nucleophilic substitution unimolecular) reactions consists of first-order reactions. For example, in the reaction of aryldiazonium ions with nucleophiles in aqueous solution ArN2+ + X? ? ArX + N2, the rate equation is r = k[ArN2+], where Ar indicates an aryl group.[10]
Pseudo-first order
If the concentration of a reactant remains constant (because it is a catalyst, or because it is in great excess with respect to the other reactants), its concentration can be included in the rate constant, obtaining a pseudo–first-order (or occasionally pseudo–second-order) rate equation. For a typical second-order reaction with rate equation r = k[A][B], if the concentration of reactant B is constant then r = k[A][B] = k [A], where the pseudo–first-order rate constant k = k[B]. The second-order rate equation has been reduced to a pseudo–first-order rate equation, which makes the treatment to obtain an integrated rate equation much easier.
One way to obtain a pseudo-first order reaction is to use a large excess of one reactant (say, [B]?[A]) so that, as the reaction progresses, only a small fraction of the reactant in excess (B) is consumed, and its concentration can be considered to stay constant. For example, the hydrolysis of esters by dilute mineral acids follows pseudo-first order kinetics where the concentration of water is present in large excess:
CH3COOCH3 + H2O ? CH3COOH + CH3OH
The hydrolysis of sucrose in acid solution is often cited as a first-order reaction with rate r = k[sucrose]. The true rate equation is third-order, r = k[sucrose][H+][H2O]; however, the concentrations of both the catalyst H+ and the solvent H2O are normally constant, so that the reaction is pseudo–first-order.
A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.

The Differential Representation
Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements. The differential equation describing first-order kinetics is given below:
-
From above equation unit of rate constant will be inverted time unit (i.e. sec-1, min.- -1, hour-1 etc……) .
Plotting Ln a/(a-x) versus time will enable to calculate k1 graphically:







Figure xx: evaluating of k1 using graphitic method
The "rate" is the reaction rate (in units of molar/time) and k is the reaction rate coefficient (in units of 1/time). However, the units of k vary for non-first-order reactions. These differential equations are separable, which simplifies the solutions as demonstrated below.
The Integral Representation
First, write the differential form of the rate law.
Rate=?d[A]dt =k[A] (1.2)
Rearrange to give:
d[A]/[A]=?kdt (1.3)
Second, integrate both sides of the equation.
? d[A]/[A]= ??kdt (1.4)

Recall from calculus that: Upon integration,
ln[A]?ln[A]o=?kt (1.7)
Rearrange to solve for [A] to obtain one form of the rate law:
ln[A]=ln[A]o=?kt (1.8)
This can be rearranged to:
ln[A]=?kt+ln[A]o (1.9)
This can further be arranged into y=mx +b form:
ln[A]=?kt+ln[A]o (1.10)
The equation is a straight line with slope m:
mx=?kt (1.11)
and y-intercept b:
b=ln[A]o (1.12)
Now, recall from the laws of logarithms that
ln([A]t[A]o =?kt (1.13)
where [A] is the concentration at time tt and [A]o is the concentration at time 0, and k k is the first-order rate constant.

Figure xx: Decay profiles for first-order reactions with large and small rate constants.
Because the logarithms of numbers do not have any units, the product ?kt also lacks units. This concludes that unit of k in a first order of reaction must be time-1. Examples of time-1 include s-1 or min-1. Thus, the equation of a straight line is applicable:
ln[A]=?kt+ln[A]o . (1.14)
To test if it the reaction is a first-order reaction, plot the natural logarithm of a reactant concentration versus time and see whether the graph is linear. If the graph is linear and has a negative slope, the reaction must be a first-order reaction.
To create another form of the rate law, raise each side of the previous equation to the exponent, e:
ln[A]=ln[A]o?kt (1.15)
Simplifying gives the second form of the rate law:
[A]=[A]o e?kt (1.16)
The integrated forms of the rate law can be used to find the population of reactant at any time after the start of the reaction. Plotting ln[A] with respect to time for a first-order reaction gives a straight line with the slope of the line equal to -k. More information can be found in the article on rate laws.
This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law. Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms. The reason that the exponential function y=exy=ex so efficiently describes such changes is that dy/dx = ex; that is, ex is its own derivative, making the rate of change of y y identical to its value at any point.
Graphing First-order Reactions
The following graphs represents concentration of reactants versus time for a first-order reaction.

Plotting ln[A] ln?[A] with respect to time for a first-order reaction gives a straight line with the slope of the line equal to ?k ?k .

Half-time of first order reactions
The half-life (t1/2) is a timescale on which the initial concentration is decreased by half of its original value, represented by the following equation.

From above equation it can be seen that half time of reaction(t1/2) does not depend on the concentration. Important example of first order reaction is the decomposition of diazonium benzene salt:

From measuring the volume of nitrogen gas that evolved at any time(Vt) and that evolved at the end of reaction (V?) , rate constant k1 can be calculated as follows:

Another example of first order reaction is the acidic hydrolysis of methyl acetate :

Rate constant can be calculated after knowing the volumes of NaOH , Vt and Vf and Vi that are consumed by titration with acid as follows:

In this context, the k1 can be calculated using variation in concentration, rotation angle, absorbance, electrical conductance.
Example: For a desired reaction, the change in the UV- visible absorption (A) as a function of time was varied as shown below and this reaction was followed first order reaction find e value of k1 for this reaction?
Absorbance 0 0.005 0.036 0.12 0.162 0.1999 0.249
Time (sec.) 0 300 750 1500 2400 3600 ?


From these results, it is clear that the reactants don’t absorb the light, while the product absorbs light at the applied wavelength :

K1= 2.303/t Log( A?- Ao)/ (A?- At)

(A?- Ao)/ (A?- At) 0.244 0.213 0.129 0.087 0.05
Log(A?- Ao)/ (A?- At) 1.369 1.328 1.243 1.111 1.084
Time (sec.) 0 750 1500 2400 3600







By plotting Log(A?- Ao)/ (A?- At) on Y-axis against time in second on X-axis , k1 can be calculated from the slope as follows:
Slope= k1/2.303 , k1= slope x 2.303= 2.303x 1.92 x 10-4= 4.42 x 10-4 sec-1










Example: chloroform reacts with presence of sodium methoxide in methanol solution, concentration of product was estimated by titration with silver nitrate (0.01 N) and the obtained results are shown below, find k1 for this reaction?





Time (min.) 0 4 9 15 22 30 41 50 ?
Volume of AgNO3(ml) 1.17 3.03 4.49 5.97 7.39 8.87 10.48 11.7 15.98


K1= 2.303/t Log (V?- Vo)/ (V?- Vt), at t=4, V?= 15.98, Vt= 3.03, Vo= 1.17:
K1= 2.303/4Log ( 15.98- 1.71)/ (15.98-3.03),
K1= 0.0243 min.-1 ,
By repeating this process for each case and then k1 is equal to the mean values of k













Figure 2: Half lives graphically demonstrated for first-order reaction. Notice the the half-life is independent of initial concentration. This is not the case with other reaction orders.

Example 1: Estimated Rate Constants
The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant?
SOLUTION
Use Equation 20 that relates half life to rate constant for first order reactions:
k=0.693 600s =0.00115s ?1 (1.21)
As a check, dimensional analysis can be used to confirm that this calculation generates the correct units of inverse time.
Notice that, for first-order reactions, the half-life is independent of the initial concentration of reactant, which is a unique aspect to first-order reactions. The practical implication of this is that it takes as much time for [A] to decrease from 1 M to 0.5 M as it takes for [A] to decrease from 0.1 M to 0.05 M. In addition, the rate constant and the half life of a first-order process are inversely related.

Example 2: Determining Half life
If 3.0 g of substance A A decomposes for 36 minutes the mass of unreacted A remaining is found to be 0.375 g. What is the half life of this reaction if it follows first-order kinetics?
SOLUTION
There are two ways to approach this problem: The :simple inspection approach" and the "brute force approach"
Approach #1: "The simple Inspection Approach"
This approach is used when one can recognize that the final concentration of A A is 1 8 18 of the initial concentration and hence three half lives (1 2 ×1 2 ×1 2 ) (12×12×12) have elapsed during this reaction. Then use equation 18:
t 1/2 =36min 3 =12min (1.22)
This approach works only when the final concentration is (1 2 ) n (12)n that of the initial concentration, then n n is the number of half lives that have elapsed. If this is not the case, then approach #2 can be used.
Approach #2: "The brute force approach"
This approach involves solving for k k from the integral rate law equation (Eq. 12 or 17) and then relating k k to the t 1/2 t1/2 via Equation 20.
[A]t/[A]o=e?kt (1.23)
k=?ln[A]t[A]ot=?ln0.375g 3g 36min =0.0578min ?1 (1.24)
t1/2=ln2k ?0.693 0.0578min ?1 ?12min (1.25)
The first approach is considerably faster (if the number of half lives evolved is apparent).
Practice Problems
Calculate the half-life of the reactions below:
1. If 4.00 g A are allowed to decompose for 40 min, the mass of A remaining undecomposed is found to be 0.80 g.
2. If 8.00 g A are allowed to decompose for 34 min, the mass of A remaining undecomposed is found to be 0.70 g.
3. If 9.00 g A are allowed to decompose for 24 min, the mass of A remaining undecomposed is found to be 0.50 g.
Determine the percent H2O2 that decomposes in the time using k=6.40×10 ?5 s ?1 k=6.40×10?5s?1
4. The time for the concentration to decompose is 600.0 s after the reaction begins. Use the value of k above.
5. The time for the concentration to decompose is 450 s after the reaction begins. Use the value of k above.
Solutions
Use the half life reaction that contains initial concentration and final concentration. Plug in the appropriate variables and solve to obtain:
1. 17.2 min
2. 9.67 min
3. 5.75 min
4. Rearranging Eq. 17 to solve for the [H 2 O 2 ] t /[H 2 O 2 ] 0 [H2O2]t/[H2O2]0 ratio
[H 2 O 2 ] t [H 2 O 2 ] 0 =e ?kt (1.26) (1.26)[H2O2]t[H2O2]0=e?kt
This is a simple plug and play application once you have identified this equation.
[H 2 O 2 ] t=600s [H 2 O 2 ] 0 =e ?(6.40×10 ?5 s ?1 )(600s) (1.27) (1.27)[H2O2]t=600s[H2O2]0=e?(6.40×10?5s?1)(600s)

[H 2 O 2 ] t=600s [H 2 O 2 ] 0 =0.9629 (1.28) (1.28)[H2O2]t=600s[H2O2]0=0.9629
So 100-96.3=3.71% of the hydrogen peroxide has decayed by 600 s.
5. Rearranging Eq. 17 to solve for the [H 2 O 2 ] t /[H 2 O 2 ] 0 [H2O2]t/[H2O2]0 ratio
[H 2 O 2 ] t [H 2 O 2 ] 0 =e ?kt (1.29) (1.29)[H2O2]t[H2O2]0=e?kt
This is a simple plug and play application once you have identified this equation.
[H 2 O 2 ] t=450s [H 2 O 2 ] 0 =e ?(6.40×10 ?5 s ?1 )(450s) (1.30) (1.30)[H2O2]t=450s[H2O2]0=e?(6.40×10?5s?1)(450s)

[H 2 O 2 ] t=450s [H 2 O 2 ] 0 =0.9720 (1.31) (1.31)[H2O2]t=450s[H2O2]0=0.9720
So 100-96.3=2.8% of the hydrogen peroxide has decayed by 450 s.

Second-Order Reactions
Many important biological reactions, such as the formation of double-stranded DNA from two complementary strands, can be described using second order kinetics. In a second-order reaction, the sum of the exponents in the rate law is equal to two. The two most common forms of second-order reactions will be discussed in detail in this section.
Reaction Rate: Integration of the second-order rate law
d[A]/dt=?k[A]2 (1.1)
(1.2)
which is easily rearranged into a form of the equation for a straight line and yields plots similar to the one shown on the left below.
(1.3)
Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to first-order reactions. For this reason, the concept of half-life for a second-order reaction is far less useful. Reaction rates are discussed in more detail here. Reaction orders are defined here. Here are explanations of zero and first order reactions.
Case 1: Identical Reactants
Two of the same reactant (A) combine in a single elementary step.
A+A?P (1.4)
2A?P (1.5)
The reaction rate for this step can be written as
Rate= ?1/2 d[A]/dt = +d[P]/dt (1.6)
and the rate of loss of reactant A:
dA/dt= ?k[A][A]= ?k[A]2 -(1.7)
where k is a second order rate constant with units of M-1 min-1 or M-1 s-1. Therefore, doubling the concentration of reactant A will quadruple the rate of the reaction. In this particular case, another reactant (B ) could be present with A ; however, its concentration does not affect the rate of the reaction, i.e., the reaction order with respect
to B is zero, and we can express the rate law as rate =k[A]2[B]0.

Case 2: Different Reactants
Two different reactants (A and B) combine in a single elementary step:
A+B?P (1.8)
The reaction rate for this step can be written as:
Rate=?d[A]/dt =?d[B]/dt=+d[P]/dt (1.9)
and the rate of loss of reactant A :
-d[A]/dt = ?k[A][B] (1.10)
where the reaction order with respect to each reactant is 1. This means that when the concentration of reactant A is doubled, the rate of the reaction will double, and quadrupling the concentration of reactant in a separate experiment will quadruple the rate. If we double the concentration of A and quadruple the concentration of B at the same time, then the reaction rate is increased by a factor of 8. This relationship holds true for any varying concentrations of A or B.
Derivative and Integral Forms
To describe how the rate of a second-order reaction changes with concentration of reactants or products, the differential (derivative) rate equation is used as well as the integrated rate equation. The differential rate law can show us how the rate of the reaction changes in time, while the integrated rate equation shows how the concentration of species changes over time. The latter form, when graphed, yields a linear function and is, therefore, more convenient to look at. Nonetheless, both of these equations can be derived from the above expression for the reaction rate. Plotting these equations can also help us determine whether or not a certain reaction is second-order.
Case 1: A + A ? P (Second Order Reaction with Single Reactant)
The rate at which A decreases can be expressed using the differential rate equation.
?d[A]/dt =k[A]2 (1.11)
The equation can then be rearranged:
d[A]/[A]2 = ?k dt (1.12)
Since we are interested in the change in concentration of A over a period of time, we integrate between t=0 t=0 and t the time of interest.


To solve this, we use the following rule of integration (power rule):

We then obtain the integrated rate equation.

Upon rearrangement of the integrated rate equation, we obtain an equation of the line:


The crucial part of this process is not understanding precisely how to derive the integrated rate law equation, rather it is important to understand how the equation directly relates to the graph which provides a linear relationship. In this case, and for all second order reactions, the linear plot of 1/[A]t versus time will yield the graph below:

This graph is useful in a variety of ways. If we only know the concentrations at specific times for a reaction, we can attempt to create a graph similar to the one above. If the graph yields a straight line, then the reaction in question must be second order. In addition, with this graph we can find the slope of the line and this slope is k k , the reaction constant. The slope can be found be finding the "rise" and then dividing it by the "run" of the line. For an example of how to find the slope, please see the example section below. There are alternative graphs that could be drawn.
The plot of [A]t versus time would result in a straight line if the reaction were zeroth order. It does, however, yield less information for a second order graph. This is because both the graphs of a first or second order reaction would look like exponential decays. The only obvious difference, as seen in the graph below, is that the concentration of reactants approaches zero more slowly in a second-order, compared to that in a first order reaction.

Case 2: A + B ? P (Second Order Reaction with multiple reactants)
As before, the rate at which A decreases can be expressed using the differential rate equation:
d[A]/dt= ?k[A][B] (1.17)
Two situations can be identified.

Situation 2a: [A]0 ?[B]0
Situation 2a is the situation that the initial concentration of the two reactants are not equal. Let x be the concentration of each species reacted at time t
Let [A]0=a and [B]0=b, then [A]= (a?x) ;[B]=b?x The expression of rate law becomes:
?dx/dt=?k(a?x) ([b?x) (1.18)
which can be rearranged to:
dx/(a?x)( b?x)= -kdt (1.19)
We integrate between t=0 (when x=0) and tt , the time of interest.
? dx/(a?x)(b?x)= k?dt (1.20)
To solve this integral, we use the method of partial fractions.

Evaluating the integral gives us:


Then for a?b k2 equal to:

Plotting Ln b(a-x)/a(b-x) versus time gives linear relationship as follows


This graph can be used in the same manner as the graph in the section above or written in the other way:
Situation 2b: [A]0=[B]0
Because A+B?P
Since A and B react with a 1 to 1 stoichiometry, [A]=[A]0?x and [B]=[B]0?x
at any time t , [A]=[B][A]=[B] and the rate law will be,
rate=k[A][B]=k[A][A]=k[A]2. (1.27)
Thus, it is assumed as the first case:

Upon intigration we have:


Utilizing inigration method for this equation we obtain the following integrated form for second order reaction for case (a=b):

Or by rearrange equation 1-32


Example 1
The following chemical equation reaction represents the thermal decomposition of gas E into K and G at 200° C?
5E(g )? 4K(g)+G(g) (1.28)
This reaction follows a second order rate law with regards to E . What is the initial rate of decomposition of E . For this reaction suppose that the rate constant at 200° C is equivalent to 4.0×10?2M?1s?1 and the initial concentration is 0.050M0.050M ?
SOLUTION
Start by defining the reaction rate in terms of the loss of reactants
Rate (initial)=?1/5d[E]/dt (1.29)
and then use the rate law to define the rate of loss of E
d[E]/dt= ?k[A]2 (1.30)
We already know k and [A]i but we need to figure out x . To do this look at the units of k and one sees it is M-1s-1 which means the overall reaction is a second order reaction with x=2 .
Initial rate=(4.0×10?2 M?1 s?1) (0.050M)2= 1×10?4 M -1s?1 (1.31)
Half-Life
Another characteristic used to determine the order of a reaction from experimental data is the half-life (t 1/2 t1/2 ). By definition, the half life of any reaction is the amount of time it takes to consume half of the starting material. For a second-order reaction, the half-life is inversely related to the initial concentration of the reactant (A). For a second-order reaction each half-life is twice as long as the life span of the one before.
Consider the reaction 2A?P 2A?P :
We can find an expression for the half-life of a second order reaction by using the previously derived integrated rate equation.
1/[A]t – 1/[A]o= kt (1.32)
Since,
[A]t 1/2=1/2/[A]o (1.33)
When,
t=t1/2 (1.34)
Our integrated rate equation becomes:
1/1/2[A]o?1/[A]o= kt1/2 (1.35)
After a series of algebraic steps,
2/[A]o?1/[A]o=kt1/2 (1.36)
1/[A]o=kt1/2 (1.37)
We obtain the equation for the half-life of a second order reaction:
t1/2=1/k[A]o (1.38)
This inverse relationship suggests that as the initial concentration of reactant is increased, there is a higher probability of the two reactant molecules interacting to form product. Consequently, the reactant will be consumed in a shorter amount of time, i.e. the reaction will have a shorter half-life. This equation also implies that since the half-life is longer when the concentrations are low, species decaying according to second-order kinetics may exist for a longer amount of time if their initial concentrations are small.
Note that for the second scenario in which A+B? P , the half-life of the reaction cannot be determined. As stated earlier, [A]o[A]o cannot be equal to [B]o. Hence, the time it takes to consume one-half of A is not the same as the time it takes to consume one-half of B. Because of this, we cannot define a general equation for the half-life of this type of second-order reaction.
Example 1.1
If the only reactant is the initial concentration of A A , and it is equivalent to [A] 0 =4.50×10 ?5 M [A]0=4.50×10?5M and the reaction is a second order with a rate constant k=0.89M ?1 s ?1 k=0.89M?1s?1 what is the rate of the reaction?
SOLUTION
1 k[A]0=1/(4.50x10?5 M)(0.89M?1s?1)= 2.50×104s (1.39)

The graph below is the graph that tests if a reaction is second order. The reaction is second order if the graph has a straight line, as is in the example below.

Practice Problems
1. Given the following information, determine the order of the reaction and the value of k, the reaction constant.
Concentration (M) Time (s)
1.0 10
0.50 20
0.33 30

*Hint: Begin by graphing

2. Using the following information, determine the half life of this reaction, assuming there is a single reactant.
Concentration (M) Time (s)
2.0 0
1.3 10
0.9633 20

3. Given the information from the previous problem, what is the concentration after 5 minutes?
Solutions
1
a. Make graphs of concentration vs. time (zeroth order), natural log of concentration vs. time (first order), and one over concentration vs. time (second order).
b. Determine which graph results in a straight line. This graph reflects the order of the reaction. For this problem, the straight line should be in the 3rd graph, meaning the reaction is second order.
c. The numbers should have are:
1/Concentration(M-1) Time (s)
1 10
2 20
3 30
The slope can be found by taking the "rise" over the "run". This means taking two points, (10,1) and (20,2). The "rise" is the vertical distance between the points (2-1=1) and the "run" is the horizontal distance (20-10=10). Therefore the slope is 1/10=0.1. The value of k, therefore, is 0.1 M-2s-1.

2
a. Determine the order of the reaction and the reaction constant, k, for the reaction using the tactics described in the previous problem. The order of the reaction is second, and the value of k is 0.0269 M-2s-1.
b. Since the reaction order is second, the formula for t1/2 = k-1[A]o-1. This means that the half life of the reaction is 0.0259 seconds.
3
a. Convert the time (5 minutes) to seconds. This means the time is 300 seconds.
b. Use the integrated rate law to find the final concentration. The final concentration is .1167 M.

Example:
A solution of ethyl acetate 0.019 N at 293 K, this solution was saponificated with NaOH (0.002 N), within 23 minute 10% of the solution was saponificated. This reaction is a second order reaction how the time changes when reducing the concentrations of reactants by ten times?
Solution:

a= 0.019 N, b= 0.002 N, and x= 0.019 N,



When reduction of concentrations at the same temperature, rate constant(k) doesn t change and the time can be evaluated as follows:


Example 2: Reaction between thiosulphate and n-bropyl bromide was studied kinetically at 310 K,

Concentration of thiosulphate was estimated at different times via titration with iodine (0.0257) N for for each 10.02 mL of thio solution and the initial concentration of thio was (0.1)N. The obtained results are listed in table below, find rate constant (k2) for this reaction?
Time (sec.) 0 1110 2010 3192 5052 7 7380 11232 78840
V(ml) 37.63 35.20 33.63 31.90 29.86 28.04 26.01 22.24
Concentration of thio can be calculated as follows:
Thio= iodine
NxV= Nx V
Nx 10.02= 0.02572 x 37.63
Conce. Of thio, N= 0.0967.
Example2: Rate constant for saponification of ethyl acetate with sodium hydroxide at 283 K equal to (2,83) , concentration of ethyl acetate 0.05 N .calculate the time required for saponification of 50% of ethyl acetate 1 m3 for the following cases:
1-mixing with 1 m3 of NaOH (0.05 N),
2-mixing with 1 m3 of NaOH (0.1 N),
Answer:
Upon mixing solutions together, concentrations would change and the new concs. Can be calculated from dilution law for each case as follows:
(N1xV1=N2 XV2)
Case a: in this case, the two concentrations are equal:
N1 x V1=N2 x V2 (0.05 x 1= N2 x 2), N2= a= 0.025
X= a x 50%= 0.025 x 50%= 0.0125
K02= x/at(a-x)
T=x/ak(a-x), t= 0.0125/2.38(0.025-0.0125)= 16.8 minute,
Case b: in this case we have different concentrations between substances a and b:

Time = 2.303/k2(a-b) Log b(a-x)/a(b-x), (a= 0.05, b=0.025)
Time= 2.303/2.38(0.025- 0.05)Log 0.05(0.025-0.05)/0.025(0.05-0.0125)= 6.81 minute





Third order reaction
There are three cases for the third order reaction:

Case 1: a?b?c:
dx/dt=k3(a-x)(b-x)(c-x),
?dx/(a-x)(b-x)(c-x)= k3 ?dt
This relation can be integrated using special integration processes to give:

Case 2: a=b=c:

Using special integration methods we get:


Case 3: a=b?c or a?b=c or a=c?b:
In this case, two substances are equal while the third is different:


Examples about 3rd ordered reactions:
1-the reaction between FeCl3 and SnCl2 was investigated by taking stoichiometric quantities of the reacting materials (0.0625 M) in a thermostatic conical flask with drawing different amounts at interval periods with passing in FeCl3 to remove any excess in SnCl2. The obtained FeCl2 was estimated by titration with K2Cr2O7 as shown in table below:
Time (min.)
1
3
7
40

Conc. 0.01434 0.02664 0.03612 0.05058

Prove that this reaction is 3rd ordered in its kinetics?
For this case, a= 0.0625 and x = 0.01434, 0.02664, 0.03612, and 0.05058 and by applying third order reaction for these values we obtain:



As we obtained same values for k3 for each case, this proves that it s third ordered reaction.



Determination of reaction order
The order of a reaction cannot be deduced from the chemical equation of the reaction. It must be determined by experiment.
1-Trial method:
This method is based on applying 1st order reaction and evaluating rate constant for each substitution. If values of k1 are similar or the same in values then this means that reaction is 1st order in its kinetics. If not applying 2nd order reaction and so on …..
2-graphical methods
By plotting the obtained values according to zero order, 1st order, 2nd order, and 3rd order as shown below:
















3-Method of initial rates
The order of a reaction for each reactant can be estimated from the variation in initial rate with the concentration of that reactant, using the natural logarithm of the typical rate equation For example, the initial rate can be measured in a series of experiments at different initial concentrations of reactant A with all other concentrations [B], [C], ... kept constant, so that The slope of a graph of as a function of then corresponds to the order x with respect to reactant A.
However, this method is not always reliable because
1. measurement of the initial rate requires accurate determination of small changes in concentration in short times (compared to the reaction half-life) and is sensitive to errors, and
2. the rate equation will not be completely determined if the rate also depends on substances not present at the beginning of the reaction, such as intermediates or products.
Integral method
The tentative rate equation determined by the method of initial rates is therefore normally verified by comparing the concentrations measured over a longer time (several half-lives) with the integrated form of the rate equation.
For example, the integrated rate law for a first-order reaction is where [A] is the concentration at time t and [A]0 is the initial concentration at zero time. The first-order rate law is confirmed if is in fact a linear function of time. In this case the rate constant is equal to the slope with sign reversed.
Method of flooding
The partial order with respect to a given reactant can be evaluated by the method of flooding (or of isolation) of Ostwald. In this method, the concentration of one reactant is measured with all other reactants in large excess so that their concentration remains essentially constant. For a reaction a·A + b·B ? c·C with rate law: , the partial order ? with respect to A is determined using a large excess of B. In this case with , and ? may be determined by the integral method. The order ? with respect to B under the same conditions (with B in excess) is determined by a series of similar experiments with a range of initial concentration [B]0 so that the variation of k can be measured.
5-Isolation method
This method is applied for complex reactions, which involve more than two materials and it is based on separation of these substances each from other:
A+B+C P
According to this method, A can be separated from B and C and then finding order of reaction with respect to A only. This can be achieved by following one of above methods. Then same way is applied for B and C separately. Then the order of the whole reaction will be summation of these orders as shown below:

Order of reaction= n1+n2+n3 ,
Whereas, n1 is the order of substance A, n2 is the order of B and n3 is the order of C.

Complex reactions التفاعلات المعقده:
Complex reactions can be defined as reactions that occur in more than one-step and involve the following types:
1-reversible reactions (Opposed reactions) ,
2-consecutive reactions,
3-parallel reactions,
4-chain reactions,
5-side reactions,
6-double reactions,
7-catalytic reactions.

1-Opposed reactions(reversible reactions)التفاعلات العكسية:
A pair of forward and reverse reactions may occur simultaneously with comparable speeds. For example, A and B react into X and Y and vice versa (s, t, u, and v are the stoichiometric coefficients):
The reaction rate expression for the above reactions (assuming each one is elementary) can be expressed as:

where: k1 is the rate coefficient for the reaction that consumes A and B; k2 is the rate coefficient for the backwards reaction, which consumes X and Y and produces A and B. The constants k1 and k2 are related to the equilibrium coefficient for the reaction (K) by the following relationship (set r=0 in balance):



Concentration of A (A0 = 0.25 mole/l) and B versus time reaching equilibrium kf = 2 min?1 and kr = 1 min?1
Simple example
In a simple equilibrium between two species: Where the reactions starts with an initial concentration of A, , with an initial concentration of 0 for B at time t=0.
Then the constant K at equilibrium is expressed as: Where and are the concentrations of A and B at equilibrium, respectively. The concentration of A at time t, is related to the concentration of B at time t, , by the equilibrium reaction equation: Note that the term is not present because, in this simple example, the initial concentration of B is 0. his applies even when time t is at infinity; i.e., equilibrium has been reached:
These equations allow us to uncouple the system of differential equations, and allow us to solve for the concentration of A alone.
The reaction equation, given previously as:
derivative is negative because this is the rate of the reaction going from A to B, and therefore the concentration of A is decreasing. To simplify annotation, let x be, the concentration of A at time t. Let be the concentration of A at equilibrium. Then: The reaction rate becomes: which results in:
A plot of the negative natural logarithm of the concentration of A in time minus the concentration at equilibrium versus time t gives a straight line with slope kf + kb. By measurement of Ae and Be the values of K and the two reaction rate constants will be known. This type of reaction containing two directions one of them is the forward reaction (kf) and the other is the backward reaction (kr). In a desired case these two reactions become in equilibrium which in it (kf=kr).

In above equation each of a,b,c, and d represent elementary order of reaction for each of substances A,B,C and D and the total order of whole reaction will be equal to the summation:
(n=a+b+c+d).for this type of reaction we have the following cases:
Case1:
when forward and backward reaction are both of them from 1st order then:


In order to find integrated form for rate constant of this type of reaction:



Half time for 1st order reversible reaction will be:

Time1/2= 0.692/kf+kr

















Case 2:
In this case forward reaction is 1st order and the reverse reaction is 2nd order:








Case 3:
In this case forward reaction is 2nd order and the reverse reaction is 1st order:




By integration, this equation we obtain:




The value of kr for this case can be calculated from the relation between k2 aand kr:

k2(a-xe)2=kr(xe) ,

kr= k2(a-xe)2/xe






Case 4:
In this case both of forward and reverse (backward) reaction are second order kinetics.





Examples:
Write rate equation for the following reversible reaction:

Solution:
As the reaction is second order in both sides:

So that reaction rate is written as follows:



2-Consecutive reactions: التفاعلات المتعاقبة (المتتابعة)
Constitutive reactions, are a type of reactions that can occur in more than one step due to involving of formation of some intermediates before yielding of the final product as presented below:


From above equations, a is the initial concentration of A, x is the initial concentration of B and y is the concentration of the final product C.
Rate equation for reaction of A can be written as follows:

Rate equation for B is:

By integration of equation 1:

In this context, (a-x) represents the remaining concentration of A at time=t, and x can be calculated from a. by substituting the value of x in equation number (2) above, and using special integration method the value of (b) can be calculated as follows:


(a=CA+CB+CC)
Concentration of C (Cc=y) can be calculated as follows:

In case of k2 is too greater than k1 (k2??k1) then k1 can be ignored as follows:



3-chain reactions: (التفاعلات المتسلسلة)
This type of reactions involves a series of reactions to yield the final product, this type of reactions occur as a result of light, heat or any other energy source. A chain reaction is a sequence of reactions where a reactive product or by-product causes additional reactions to take place. In a chain reaction, positive feedback leads to a self-amplifying chain of events.Chain reactions are one way that systems which are not in thermodynamic equilibrium can release energy or increase entropy in order to reach a state of higher entropy. For example, a system may not be able to reach a lower energy state by releasing energy into the environment, because it is hindered or prevented in some way from taking the path that will result in the energy release. If a reaction results in a small energy release making way for more energy releases in an expanding chain, then the system will typically collapse explosively until much or all of the stored energy has been released.
A macroscopic metaphor for chain reactions is thus a snowball causing a larger snowball until finally an avalanche results ("snowball effect"). This is a result of stored gravitational potential energy seeking a path of release over friction. Chemically, the equivalent to a snow avalanche is a spark causing a forest fire. In nuclear physics, a single stray neutron can result in a prompt critical event, which may finally be energetic enough for a nuclear reactor meltdown or (in a bomb) a nuclear explosion..Numerous chain reactions can be represented by a mathematical model based on Markov chains.
In this type of reactions reactive species (free radicals) are responsible on continuous of chain reactions and reaction is continued as these free radicals are available. Important example of this type is the formation of HCl(g) by reaction of Cl2(g) with (H2(g) in presence of light as follows:

Generally, this type of reactions occur in three main steps, they are1- initiation step,2- propagation step and 3-termination step as follows:
1-Initiation step: خطوة البدء
This step involves, formation of active particles or chain carriers, often free radicals, in either a thermal or a photochemical step (equations 1and 2),
2- Propagation step: خطوة النمو او الانتشار
(This step involves, may comprise several elementary steps in a cycle, where the active particle through reaction forms another active particle which continues the reaction chain by entering the next elementary step). In effect the active particle serves as a catalyst for the overall reaction of the propagation cycle. Particular cases are:chain branching (a propagation step which forms more new active particles than enter the step);
chain transfer (a propagation step in which the active particle is a growing polymer chain which reacts to form an inactive polymer whose growth is terminated and an active small particle (such as a radical), which may then react to form a new polymer chain (equations 2 and 3) .
3- Termination step: خطوة الانتهاء
This step involves, elementary step in which the active particle loses its activity; e. g. by recombination of two free radicals).The chain length is defined as the average number of times the propagation cycle is repeated, and equals the overall reaction rate divided by the initiation rate.[1]Some chain reactions have complex rate equations with fractional order or mixed order kinetics.
Another example:
The reaction H2 + Br2 ? 2 HBr proceeds by the following mechanism:
1-Initiation step:
Br2 ? 2 Br• (thermal) or Br2 + h? ? 2 Br• (photochemical)
Each Br atom is a free radical, indicated by the symbol « • » representing an unpaired electron.
2-Propagation (here a cycle of two steps
Br• + H2 ? HBr + H•
H• + Br2 ? HBr + Br•
The sum of these two steps corresponds to the overall reaction H2 + Br2 ? 2 HBr, with catalysis by Br• which participates in the first step and is regenerated in the second step
3- Retardation (inhibition)
H• + HBr ? H2 + Br•
This step is specific to this example, and corresponds to the first propagation step in reverse.
4-Termination step:
2 Br• ? Br2
Recombination of two radicals, corresponding in this example to initiation in reverse. As can be explained using the steady-state approximation, the thermal reaction has an initial rate of fractional order (3/2), and a complete rate equation with a two-term denominator (mixed-order kinetics).
Kinetics of chain reactions
For this type of very quick reactions, its kinetics can be investigated by using steady state approximation which means the average of change of concentrations of free radicals with time is equal to zero.
For example the following reaction:

Mechanism of this reaction involves:

From above equations, steps from 2-5 can be written as follows:

Rate equation for formation of HBr can be written as follows:

Applying steady state approximation for reactive species (H., Br.) for above equations:

By solving these equations with respect to Br and H radicals we obtain:


By substituting of (H.) and (Br.) from above equations in equation (1), we obtain:


Theories interesting of chemical kinetics
There are two main theories that interepting of chemical kinetics:
1-Collision theory,
2-Activated complex theory,
1- Collision theory,
By the end of this section, you will be able to:
Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
Define the concepts of activation energy and transition state
Use the Arrhenius equation in calculations relating rate constants to temperature
We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.
Collision theory is based on the following postulates:
1-The rate of a reaction is proportional to the rate of reactant collisions:

2-The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
3-The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species .
We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy of collisions, when we consider the reaction of carbon monoxide with oxygen:
Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient quantity, the reaction is spontaneous at high temperature and pressure.
The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:
Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure 1. In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms . This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.

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Reaction rate tends to increase with concentration phenomenon explained by collision theory
Collision theory is a theory proposed independently by Max Trautz in 1916[1] and William Lewis in 1918, that qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions.
The collision theory states that when suitable particles of the reactant hit each other, only a certain fraction of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions. The successful collisions must have enough energy, also known as activation energy, at the moment of impact to break the preexisting bonds and form all new bonds. This results in the products of the reaction. Increasing the concentration of the reactant particles or raising the temperature - which brings about more collisions and hence more successful collisions - therefore increases the rate of a reaction.
When a catalyst is involved in the collision between the reactant molecules, less energy is required for the chemical change to take place, and hence more collisions have sufficient energy for reaction to occur. The reaction rate therefore increases.
Collision theory is closely related to chemical kinetics.
Rate constant
The rate constant for a bimolecular gas-phase reaction, as predicted by collision theory is:

where: k(T), is a rate constant that is dependent on T ,Z is the collision frequency,,p is the steric factor, Ea is the activation energy of the reaction,T is the temperature, R is the gas constant.
The collision frequency is:
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The rate constant for a bimolecular gas phase reaction as predicted by collision theory is:

.
where: r, is the rate of reaction, Z is the collision frequency, is the steric factor, Ea is the activation energy of the reaction, T is the temperature., R is gas constant.
The collision frequency for A and B is:

whereas:NA is the Avogadro constant, ?AB is the reaction cross section, kB is Boltzmann s constant, ?AB is the reduced mass of the reactants.
Quantitative insights
Consider the reaction:
A + B ? C
In collision theory it is considered that two particles A and B will collide if their nuclei get closer than a certain distance. The area around a molecule A in which it can collide with an approaching B molecule is called the cross section (?AB) of the reaction and is, in principle, the area corresponding to a circle whose radius ( ) is the sum of the radii of both reacting molecules, which are supposed to be spherical. A moving molecule will therefore sweep a volume (? r2ABCA) per second as it moves, where CA is the average velocity of the particle.
From kinetic theory it is known that a molecule of A has an average velocity (different from root mean square velocity) of,
where is Boltzmann constant and is the mass of the molecule.
The solution of the two body problem states that two different moving bodies can be treated as one body which has the reduced mass of both and moves with the velocity of the center of mass, so, in this system must be used instead of .
Therefore, the total collision frequency of all A molecules, with all B molecules, is:

From Maxwell- Boltzmann distribution, it can be deduced that the fraction of collisions with more energy than the activation energy .Therefore the rate of a bimolecular reaction for ideal gases will be:

Where: r, is the rate of reaction, Z is the collision frequency. is the steric factor, Ea is the activation energy of the reaction,T is the absolute temperature, R is gas constant.
Effect of temperature, pressure and density on collision frequency
Temperature is evident from the collisional frequency equation, when temperature increases, the collisional frequency increases. Density from a conceptual point, if the density is increased, the number of molecules per volume is also increased. If everything else remains constant, a single reactant comes in contact with more atoms in a denser system. Thus if density is increased, the collisional frequency must also increase.
Size of Reactants
Increasing the size of the reactants increases the collisional frequency. This is directly due to increasing the radius of the reactants as this increases the collisional cross section. This in turn increases the collisional cylinder. Because radius term is squared, if the radius of one of the reactants is doubled, the collisional frequency is quadrupled. If the radii of both reactants are doubled, the collisional frequency is increased by a factor of 16.
Validity of the theory and steric factor
Once a theory is formulated, its validity must be tested, that is, compare its predictions with the results of the experiments.
When the expression form of the rate constant is compared with the rate equation for an elementary bimolecular reaction:
rate=k(T)[A][B] , it is noticed that

So that rate equation can be written as follows:

This expression is similar to the Arrhenius equation and gives the first theoretical explanation for the Arrhenius equation on a molecular basis. The weak temperature dependence of the pre-exponential factor is so small compared to the exponential factor that it cannot be measured experimentally, that is, "it is not feasible to establish, on the basis of temperature studies of the rate constant, whether the predicted T½ dependence of the pre-exponential factor is observed experimentally.

Steric factor
If the values of the predicted rate constants are compared with the values of known rate constants, it is noticed that collision theory fails to estimate the constants correctly, and the more complex the molecules are, the more it fails.
The reason for this is that particles have been supposed to be spherical and able to react in all directions, which is not true, as the orientation of the collisions is not always proper for the reaction. For example, in the hydrogenation reaction of ethylene the H2 molecule must approach the bonding zone between the atoms, and only a few of all the possible collisions fulfill this requirement.
To alleviate this problem, a new concept must be introduced: the steric factor ?. It is defined as the ratio between the experimental value and the predicted one (or the ratio between the frequency factor (A) and the collision frequency(Z):

The value of p is often less than unity (i.e.1? p).
Usually, the more complex the reactant molecules, the lower the steric factor. Nevertheless, some reactions exhibit steric factors greater than unity: the harpoon reactions, which involve atoms that exchange electrons, producing ions. The deviation from unity can have different causes: the molecules are not spherical, so different geometries are possible; not all the kinetic energy is delivered into the right spot; the presence of a solvent (when applied to solutions), etc.Collision theory can be applied to reactions in solution; in that case, the solvent cage has an effect on the reactant molecules, and several collisions can take place in a single encounter, which leads to predicted preexponential factors being too large. ? values greater than unity can be attributed to favorable entropic contributions.
Example of application of collision theory for a 2nd order reaction:

For this reaction, rate equation is written as follows:

Herein, n represents number of the produced molecules due to reacting of A and B. NA* and NB* are the number of reacting molecules of A and B.





2-Activated complex theory
Reaction intermediates are chemical species, often unstable and short-lived (however sometimes can be isolated), which are not reactants or products of the overall chemical reaction, but are temporary products and/or reactants in the mechanism s reaction steps. Reaction intermediates are often free radicals or ions.
The kinetics (relative rates of the reaction steps and the rate equation for the overall reaction) are explained in terms of the energy needed for the conversion of the reactants to the proposed transition states (molecular states that corresponds to maxima on the reaction coordinates, and to saddle points on the potential energy surface for the reaction).





Kinetics
The speed at which reactions takes place is studied by reaction kinetics. The rate depends on various parameters, such as:

1-Reactant concentrations, which usually make the reaction happen at a faster rate if raised through increased collisions per unit time. Some reactions, however, have rates that are independent of reactant concentrations. These are called zero order reactions.
2-Surface area available for contact between the reactants, in particular solid ones in heterogeneous systems. Larger surface areas lead to higher reaction rates.
3-Pressure – increasing the pressure decreases the volume between molecules and therefore increases the frequency of collisions between the molecules.
4-Activation energy, which is defined as the amount of energy required to make the reaction start and carry on spontaneously. Higher activation energy implies that the reactants need more energy to start than a reaction with a lower activation energy.
5-Temperature, which hastens reactions if raised, since higher temperature increases the energy of the molecules, creating more collisions per unit time,
6-The presence or absence of a catalyst. Catalysts are substances which change the pathway (mechanism) of a reaction which in turn increases the speed of a reaction by lowering the activation energy needed for the reaction to take place. A catalyst is not destroyed or changed during a reaction, so it can be used again.
7-For some reactions, the presence of electromagnetic radiation, most notably ultraviolet light, is needed to promote the breaking of bonds to start the reaction. This is particularly true for reactions involving radicals.
Several theories allow calculating the reaction rates at the molecular level. This field is referred to as reaction dynamics. The rate v of a first-order reaction, which could be disintegration of a substance A, is given by:

Its integration yields:

Here k is first-order rate constant having dimension 1/time, [A](t) is concentration at a time t and [A]0 is the initial concentration. The rate of a first-order reaction depends only on the concentration and the properties of the involved substance, and the reaction itself can be described with the characteristic half-life. More than one time constant is needed when describing reactions of higher order. The temperature dependence of the rate constant usually follows the Arrhenius equation:

where Ea is the activation energy and kB is the Boltzmann constant.
Transition state theory (TST):
Explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.








The basic ideas behind transition state theory are as follows:
1-Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. 2-The details of how these complexes are formed are not important. The saddle point itself is called the transition state.
3-The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.
4-The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion
In general according to this theory:

Herein, (X*) is the activated complex which is less stable than both of reactants and products and it will be in a quasi-equilibrium between these materials.
If we suppose that, there is an equilibrium between AC(X*) and the reactants (A) and B. then equilibrium can be written:


Herein, CXx, CA and CB are the concentrations of activated complex, A and B.

Equilibrium constant, can be related to partion functions:

k*= (?X*)/ (?A.?B) . e –(Ea/RT) ---3
(?X*) is the partion function for X*, (?A. for A and ?B for B.

CX*= (?X*)/ (?A.?B) . e –(Ea/RT) . CA.CB ---4

Assumption 1:
AC theory suppose that, one vibrational degree of freedom is responsible for the dissociation of AC(X*) to yield a final product. Vibrational function that is responsible on dissociation of AC is denoted as (qv) and it is given by:

qv= (kT)/(h?), ----5
herein, k, is a Boltzmann constant, T is the absolute temperature, h is Blank constant and ? is the frequency.
Vibration function (qv) that leads to dissociate of AC(X*) can be written as :
qv= (?X* for AC)/ (?total) ------- ---6
?X*= qv . ?total -----------6
?X*= ?total . (kT/h?) -----------7
By substitution equation (7) in (4), it gives,

CX*= CA.CB. (kT/h?). [(?X*)/ (?A.?B)] . e –(Ea/RT) -----8
Equilibrium between AC and the product:

Rate of reaction depends on concentration of AC and degree of vibration (?):
Rate: ?. CX* ---9
Rate= ?. CA.CB. (kT/h?). [(?X*)/ (?A.?B)] . e –(Ea/RT) --10
Rate= CA.CB. (kT/h). [(?X*)/ (?A.?B)] . e –(Ea/RT) --------11

Rate= k. CA.CB -------------------12
By comparing equation (11) and (12),

K= (kT/h). [(?X*)/ (?A.?B)] . e –(Ea/RT) ------------13

Thermodynamics of chemical kinetics:
Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. Chemical thermodynamics involves not only laboratory measurements of various thermodynamic properties, but also the application of mathematical methods to the study of chemical questions and the spontaneity of processes.
The structure of chemical thermodynamics is based on the first two laws of thermodynamics. Starting from the first and second laws of thermodynamics, four equations called the "fundamental equations of Gibbs" can be derived. From these four, a multitude of equations, relating the thermodynamic properties of the thermodynamic system can be derived using relatively simple mathematics. This outlines the mathematical framework of chemical thermodynamics

Thermodynamic reaction control or kinetic reaction control in a chemical reaction can decide the composition in a reaction product mixture when competing pathways lead to different products and the reaction conditions influence the selectivity or stereoselectivity. The distinction is relevant when product A forms faster than product B because the activation energy for product A is lower than that for product B, yet product B is more stable. In such a case A is the kinetic product and is favoured under kinetic control and B is the thermodynamic product and is favoured under thermodynamic control.[1][2][3]
The conditions of the reaction, such as temperature, pressure, or solvent, affect which reaction pathway may be favored: either the kinetically controlled or the thermodynamically controlled one. Note this is only true if the activation energy of the two pathways differ, with one pathway having a lower Ea (energy of activation) than the other.
Prevalence of thermodynamic or kinetic control determines the final composition of the product when these competing reaction pathways lead to different products. The reaction conditions as mentioned above influence the selectivity of the reaction - i.e., which pathway is taken.











Energy profile diagram for kinetic versus thermodynamic product reaction

Kinetics of enzymatic reactions:
The reaction catalysed by an enzyme uses exactly the same reactants and produces exactly the same products as the uncatalysed reaction. Like other catalysts, enzymes do not alter the position of equilibrium between substrates and products.[1] However, unlike uncatalysed chemical reactions, enzyme-catalysed reactions display saturation kinetics. For a given enzyme concentration and for relatively low substrate concentrations, the reaction rate increases linearly with substrate concentration; the enzyme molecules are largely free to catalyse the reaction, and increasing substrate concentration means an increasing rate at which the enzyme and substrate molecules encounter one another. However, at relatively high substrate concentrations, the reaction rate asymptotically approaches the theoretical maximum; the enzyme active sites are almost all occupied by substrates resulting in saturation, and the reaction rate is determined by the intrinsic turnover rate of the enzyme.[2] The substrate concentration midway between these two limiting cases is denoted by KM. Thus, Km is the substrate concentration value in which the substrate concentration is reaching halfway of the maximum reaction velocity.[2]
The two most important kinetic properties of an enzyme are how easily the enzyme becomes saturated with a particular substrate, and the maximum rate it can achieve. Knowing these properties suggests what an enzyme might do in the cell and can show how the enzyme will respond to changes in these conditions.


As larger amounts of substrate are added to a reaction, the available enzyme binding sites become filled to the limit of V max {\displaystyle V_{\max }} . Beyond this limit the enzyme is saturated with substrate and the reaction rate ceases to increase.
Single-substrate reactions
Enzymes with single-substrate mechanisms include isomerases such as triosephosphateisomerase or bisphosphoglycerate mutase, intramolecular lyases such as adenylate cyclase and the hammerhead ribozyme, an RNA lyase.[9] However, some enzymes that only have a single substrate do not fall into this category of mechanisms. Catalase is an example of this, as the enzyme reacts with a first molecule of hydrogen peroxide substrate, becomes oxidised and is then reduced by a second molecule of substrate. Although a single substrate is involved, the existence of a modified enzyme intermediate means that the mechanism of catalase is actually a ping–pong mechanism, a type of mechanism that is discussed in the Multi-substrate reactions section below.

Michaelis–Menten kinetics
The Michaelis–Menten kinetic model of a single-substrate reaction is shown on the right. There is an initial bimolecular reaction between the enzyme E and substrate S to form the enzyme–substrate complex ES. The rate of enzymatic reaction increases with the increase of the substrate concentration up to a certain level called Vmax; at Vmax, increase in substrate concentration does not cause any increase in reaction rate as there no more enzyme (E) available for reacting with substrate (S). Here, the rate of reaction becomes dependent on the ES complex and the reaction becomes a unimolecular reaction with an order of zero. Though the enzymatic mechanism for the unimolecular reaction ES ? k c a t E + P {\displaystyle {\ce {ES ->[k_{cat}] E + P}}} can be quite complex, there is typically one rate-determining enzymatic step that allows this reaction to be modelled as a single catalytic step with an apparent unimolecular rate constant kcat. If the reaction path proceeds over one or several intermediates, kcat will be a function of several elementary rate constants, whereas in the simplest case of a single elementary reaction (e.g. no intermediates) it will be identical to the elementary unimolecular rate constant k2. The apparent unimolecular rate constant kcat is also called turnover number and denotes the maximum number of enzymatic reactions catalysed per second.
The Michaelis–Menten equation[10] describes how the (initial) reaction rate v0 depends on the position of the substrate-binding equilibrium and the rate constant k2.


Figure xx: A chemical reaction mechanism with or without enzyme catalysis. The enzyme (E) binds substrate (S) to produce product (P).


Figure xx: Saturation curve for an enzyme reaction showing the relation between the substrate concentration and reaction rate.

Linear plots of the Michaelis–Menten equation
The plot of v versus [S] above is not linear; although initially linear at low [S], it bends over to saturate at high [S]. Before the modern era of nonlinear curve-fitting on computers, this nonlinearity could make it difficult to estimate KM and Vmax accurately. Therefore, several researchers developed linearisations of the Michaelis–Menten equation, such as the Lineweaver–Burk plot, the Eadie–Hofstee diagram and the Hanes–Woolf plot. All of these linear representations can be useful for visualising data, but none should be used to determine kinetic parameters, as computer software is readily available that allows for more accurate determination by nonlinear regression methods.[19] The Lineweaver–Burk plot or double reciprocal plot is a common way of illustrating kinetic data. This is produced by taking the reciprocal of both sides of the Michaelis–Menten equation. As shown on the right, this is a linear form of the Michaelis–Menten equation and produces a straight line with the equation y = mx + c with a y-intercept equivalent to 1/Vmax and an x-intercept of the graph representing ?1/KM.

Lineweaver–Burk or double-reciprocal plot of kinetic data, showing the significance of the axis intercepts and gradient
. Kinetics of Enzyme-Catalyzed Reactions
Reaction Velocity. The time course of an enzymatic reaction permits one to deduce the substrate affinity, the catalytic mechanism in the active center, and the efficiency of the enzyme (maximum rate, turnover number). The rate of an enzyme-catalyzed single reactant reaction depends on the concentration of substrate and product, respectively. The velocity of the reaction V is:
Rate(v)=d(s)/dt



where first term is the rate of dissapearance of substrate S and second term is the rate of appearance of product P (both S and P are in concentration).

Behavior of Initial Rates. The initial rate (Vo) is determined by extrapolating the slope of the time course of the substrate or product concentration to time zero (Fig. 3.5). The dependence of Vo on the substrate concentration, S (at constant enzyme concentration), is shown in Fig. 3.6. It reflects the typical substrate saturation. At first, Vo increases proportionally to the amount of substrate. Upon further enhancement of substrate concentration Vo levels off. The curve asymptotically approaches a maximum value, Vmax. When this plateau is reached, a change of S does not lead to a measurable change of Vo: the enzyme is saturated by substrate and has thus reached the limit of its efficiency.

Micahaelis-Menten Kinetics. These kinetics result from the fast and reversible formation of an enzyme-substrate complex, ES, which dissociates in a second, slower reaction under liberation of the product, P (Fig. 3.7):

E + S ES E + P -----------3

Because the second reaction is rate-limiting, at very high substrate concentration almost all enzyme is present as enzyme-substrate com-plex. Under these conditions a steady state is reached in which the enzyme is steadily saturated by substrate and the initial rate is at a maximum (Vmax). This relation between substrate concentration and reaction rate may be described by the Michaelis-Menten equation:





where KM is the Michaelis constant of the enzyme for the given sub-strate. KM may also be described by:


Meaning of KM The relevance of KM becomes evident at S = KM. Then Vo = Vmax /2, i.e., KM is the substrate concentration at which the reaction rate is half maximum (Fig. 3.6). The KM value characterizes the affinity between the substrate and the enzyme. At known KM and Vmax, Vo can be calculated for each value of substrate concentration. A low KM value reflects high affinity. At substrate concentrations S << KM, the reaction rate is directly proportional to the substrate concentration (first order reaction); at high substrate concentration (S >> KM) the reaction is zero order and is no longer dependent on the substrate concentration but only on the enzyme activity.

Fig. 3.5. Determination of initial rates at different substrate concentrations.




Fig. 3.6. A plot of Vo vs. substrate concentration S.
Total enzyme sites = occupied sites + free sites:
Enzyme Kinetics



Lineweaver-Burk Plot. To calculate KM and Vmax (and inhibitor constants) it is advantageous to transform the Michaelis-Menten relation so as to obtain linear relationships between S and Vo that can be evaluated graphically. An example is the Lineweaver-Burk equation, containing the reciprocal values of Vo and S:


Total enzyme sites = occupied sites + free sites:

Effect of temperature and pressure on the kinetics of reaction
1-Effect of temperature
To get to the other end of the road, an object must roll with enough speed to completely roll over the hill of a certain height. The faster the object moves, the more kinetic energy (Ek=1/2mV2) ,
V=(8RT/?m)1/2
If the object moves too slowly, it does not have enough kinetic energy necessary to overcome the barrier; as a result, it eventually rolls back down. In the same way, there is a minimum amount of energy needed in order for molecules to break existing bonds during a chemical reaction. If the kinetic energy of the molecules upon collision is greater than this minimum energy, then bond breaking and forming occur, forming a new product (provided that the molecules collide with the proper orientation.


The activation energy (Ea), labeled ?G‡?G‡ in Figure is the energy difference between the reactants and the activated complex, also known as transition state. In a chemical reaction, the transition state is defined as the highest-energy state of the system. If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction occurs and products form. In other words, the higher the activation energy, the harder it is for a reaction to occur and vice versa.
Overcoming the energy barrier from thermal energy involves addressing the fraction of the molecules that possess enough kinetic energy to react at a given temperature. According to kinetic molecular theory, a population of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the Maxwell-Boltzman distribution law.



Figure 9.5.39.5.3: Kinetic energy distributions (similar to Maxwell-Boltzman distributions for velocity) for a gas at two temperatures and critical energies for overcoming an activation barrier.
The two distribution plots shown here are for a lower temperature T1 and a higher temperature T2. The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of Ea that are shown. It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) proceed more rapidly at higher temperatures.
Arrhenius Equation
By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise, but the reasons for this were not clear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann distribution law into one of the most important relationships in physical chemistry:



Take a moment to focus on the meaning of this equation, neglecting the A factor for the time being. First, note that this is another form of the exponential decay law discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent –Ea?/RT. And what is the significance of this quantity? Recalling that RT is the average kinetic energy, it becomes apparent that the exponent is just the ratio of the activation energy Ea to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign). This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. Because these terms occur in an exponent, their effects on the rate are quite substantial.
The two plots in Figure 9.5.49.5.4 show the effects of the activation energy (denoted here by E‡) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 108.


Figure 9.5.49.5.4: Arrhenius plots. The logarithmic scale in the right-hand plot leads to nice straight lines. Looking at the role of temperature, a similar effect is observed. (If the x-axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.
Determining the activation ener
The Arrhenius equation ( k=A e?Ea/RT), can be written in a non-exponential form that is often more convenient to use and to interpret graphically). Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields
Lnk = ln(Ae?Ea/RT)
Lnk= lnA+ln(e?Ea/RT) = LnA+ (-Ea/R)(1/T)
which is the equation of a straight line whose slope is (–Ea?/R). This affords a simple way of determining the activation energy from values of k observed at different temperatures, by plotting (Lnk) as a function of (1/T).







- )Ea/R) = –3.27 × 104 K
Ea=– (8.314 J mol–1 K–1) (–3.27 × 104 K) = 273 kJ. mol–1

Calculating Ea without a plot
To calculate the value of Ea without plotting, oWe need to apply this equation at least at two different temperatures Because the ln k vs.-1/T plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. The ln A term is eliminated by subtracting the expressions for the two ln-k terms via the following steps:p


Lnk1= LnA- (Ea /RT1) -----------------1
Lnk2= LnA-(Ea /RT2) --------------------2
By re-writing equation 2, and substituting for LnA,
Lnk1= Lnk2+ (Ea/RT2)-(Ea/RT1)
Lnk1-Lnk2= (Ea/RT2)- (Ea/RT1),
Lnk2/k1= -Ea/R(1/T2-1/T1)

The Role of Collisions
What would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the frequency- or collision factor, Z:

In some reactions, the relative orientation of the molecules at the point of collision is important, so a geometrical or steric factor (commonly denoted by ? (Greek lower case rho) can be defined. In general, we can express A as the product of these two factors:
A=Z.p
Values of ? are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which A is assumed to be the same as Z. Usually, the more complex the reactant molecules, the lower the steric factors. The deviation from unity can have different causes: the molecules are not spherical, so different geometries are possible; not all the kinetic energy is delivered into the right spot; the presence of a solvent (when applied to solutions) and other factors




Figure 9.5.5: The Effect of Molecular Orientation on the Reaction of NO and O3. Most collisions of NO and O3 molecules occur with an incorrect orientation for a reaction to occur. Only those collisions in which the N atom of NO collides with one of the terminal O atoms of O3 are likely to produce NO2 and O2, even if the molecules collide with E > Ea.

2-Effect of pressure on reaction kinetics
Increasing the pressure in a gaseous reaction will increase the number of collisions between reactants, increasing the rate of reaction. This is because the activity of a gas is directly proportional to the partial pressure of the gas. This is similar to the effect of increasing the concentration of a solution.
In addition to this straightforward mass-action effect, the rate coefficients themselves can change due to pressure. The rate coefficients and products of many high-temperature gas-phase reactions change if an inert gas is added to the mixture; variations on this effect are called fall-off and chemical activation. These phenomena are due to exothermic or endothermic reactions occurring faster than heat transfer, causing the reacting molecules to have non-thermal energy distributions (non-Boltzmann distribution). Increasing the pressure increases the heat transfer rate between the reacting molecules and the rest of the system, reducing this effect.

Condensed-phase rate coefficients can also be affected by (very high) pressure; this is a completely different effect than fall-off or chemical-activation. It is often studied using diamond anvils.A reaction s kinetics can also be studied with a pressure jump approach. This involves making fast changes in pressure and observing the relaxation time of the return to equilibrium.
The relationship between pressure and concentration
Increasing the pressure of a gas is exactly the same as increasing its concentration. If you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller volume. If you have the same mass in a smaller volume, then its concentration is higher.
You can also show this relationship mathematically if you have come across the ideal gas equation


Rearranging this gives:

Because "RT" is constant as long as the temperature is constant, this shows that the pressure is directly proportional to the concentration. If you double one, you will also double the other.
The effect of increasing the pressure on the rate of reaction
Collisions involving two particles
The same argument applies whether the reaction involves collision between two different particles or two of the same particle.
In order for any reaction to happen, those particles must first collide. This is true whether both particles are in the gas state, or whether one is a gas and the other a solid. If the pressure is higher, the chances of collision are greater






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