Factors Affecting The Rate of a Chemical Reaction

The various factors which influences the rate of reaction are

(i) concentration of reactants

(ii) temperature of reactants

(iii) nature of reacting substances

(iv) presence of catalyst

(v) exposure to radiations

1. Concentration of the reactants

Greater is the concentration of reactants, more will be the chances of collisions between the reacting particles, consequently, larger is the rate of the reaction. For gaseous reactants as the concentration is related to the pressure, therefore, greater is the pressure more will be number of molecules per unit volume and consequently, greater will be the rate of encounters between the molecules.

2. Effect of temperature on reaction rate

The rates of almost all reactions increase with the increase in temperature. In most of the cases the rate of the reaction becomes almost double for every 10° rise of temperature. This is also expressed in the terms of Temperature co-efficient. Which is the ratio of rate constant of the reaction at two temperatures differing by 10°. The two temperature generally selected are 298K and 308 K. Thus

$Temperaturecoefficient=\frac{Ratecons\tan tat308K}{Ratecons\tan tat298K}$

Explanation of the effect of temperature

According to collision theory of reaction rates.

Rate of a reaction = f x z

Where f = no. of effective collision

z = frequency of collision

Thus, the increase in rate is due to either of the above factors is f or z or due to both of these.

It may be observed that the increase in the total number of collision per unit volume per unit time (collision frequency) is not so much responsible for the higher reaction rate as is the increase in the fraction of effective collisions. Let us, for eg, calculate the increase in collision frequency when temperature increases from 298 to 308 K. As we know that collision frequency is directly proportional to the square root of absolute temperature, therefore, the rate of collision frequencies at these temperatures follows as

$\frac{Z_{308}}{Z_{298}}=\sqrt{\frac{308}{298}}=1.016$

From the above ratio it is clear that there is a insignificant increase in the collision frequency. Hence it can not explain the observed increase in the ratio of the reaction with increase in temperature.

Let us now consider the effect of increase in temperature on the number of effective collisions.

Now, as we know that the rise in temperature increases the kinetic energy of the molecules. $\left(\because K.E\propto T\right)$therefore the energy distribution curve gets flattened and shifts towards higher energy region. A close revels examination of the curves in the graph clearly reveals that the fraction of molecules possessing higher kinetic energy i.e. energy greater than threshold energy, as indicated by shaded portion becomes almost double and therefore, the rate of reaction almost doubles for 10° rise in temperature. Thus, increase in the rate of reaction with increase in temperature is mainly due to increase in no. of collisions which are energetically effective.

Note: The reaction rate dependence on temperature can also explained by Vant Hoff's equation.

$Log\frac{K_2}{K_1}=\frac{\Delta H}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

Nature of reacting substance

The nature of reacting substances affect, the rates significantly. For e.g. the oxidation of ferrous (Fe⁺²) by KMnO₄ in acidic medium is practically instaneous. On the other hand, oxidation of oxalate ions $\left(C_2{O}_4^{-2}\right)$ by KMnO₄ in acidic medium is comparatively much slower

\begin{gathered} MnO_4^{ - 1} + 8{H^ + } + 5F{e^{ + 2}}\xrightarrow[{}]{{}}M{n^{ + 2}} + 4F{e^{ + 3}} + 4{H_2}O \hfill \\ \hspace{9cm} \left( {fast} \right) \hfill \\ \end{gathered}

\begin{gathered} 2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{-2}+16{{H}^{+}}\xrightarrow[{}]{{}}2M{{n}^{+2}}+10C{{O}_{2}}+8{{H}_{2}}O \hfill \\ \hspace{8cm} \left( {Slow} \right) \hfill \\ \end{gathered}

Catalyst

A catalyst is a substance, which increases the rate of a reaction without itself being consumed at the end of the reaction, and the phenomenon is called catalysis. There are some catalysts which decrease the rate of reaction and such catalysts are called negative catalyst. Obviously, the catalyst accelerating the rate will be positive catalyst. However, the term positive is seldom used and catalyst itself implies positive catalyst.

Catalyst are generally foreign substances but sometimes one of the product formed may act as a catalyst and such catalyst is called "auto catalyst" and the phenomenon is called auto catalysis.

$2KCl{O}_3+\left[Mn{O}_2\right]\to 2KCl+3O_2\uparrow +\left[Mn{O}_2\right]$

MnO₂ can be received in the same composition and mass at the end of the reaction. In the permanganate titration of oxalic acid in the presence of bench H₂SO₄ (acid medium), it is found that the titration in the beginning there is slow discharge of the colour of permanganate solution but after sometime the discharge of the colour become faster. This is due to the formation of MnSO₄ during the reaction which acts as a catalyst for the same reaction. Thus, MnSO₄ is an "auto catalyst" for this reaction. This is an example of auto catalyst.

$2KMn{O}_4+3H_{2}S{O}_4+5H_2{C}_2{O}_4\to K_{2}S{O}_4+8H_{2}O+10C{O}_2+2MnS{O}_4$

General characteristics of catalyst

$\bullet$A catalyst does not initiate the reaction. It simply fastens it.

$\bullet$Only a small amount of catalyst can catalyse the reaction.

$\bullet$A catalyst does not alter the position of equilibrium i.e. magnitude of equilibrium constant and hence $\Delta G^0$. It simply lowers the time needed to attain equilibrium. This means if a reversible reaction in absence of catalyst completes to go to the extent of 75% till attainment of equilibrium, and this state of equilibrium is attained in 20 minutes then in presence of a catalyst also the reaction will go to 75% of completion before the attainment of equilibrium but the time needed for this will be less than 20 minutes.

$\bullet$A catalyst drives the reaction through a different route for which energy barrier is of shortest height and hence E_a is of lower magnitude. That is, the function of the catalyst is to lower down the activation.

E_a = Energy of activation in absence of catalyst.

E'a = Energy of activation in presence of catalyst.

Ea – E'_a = lowering of activation energy by catalyst.

If k and k_cat be the rate constant of a reaction at a given temperature T, and E_a and E'_a are the activation energies of the reaction in absence and presence of catalyst, respectively, the

$\frac{k_{cat}}{k}=\frac{A{e}^{-E^{\prime }a/RT}}{A{e}^{-E_a/RT}}$

$\frac{k_{cat}}{k}=e^{\left(E_a-{E^{\prime }}_a\right)/RT}$

Since E_a > E_a' so k_cat > k. the ratio $\frac{k_{cat}}{k}$ gives the number of times the rate of reaction will increase by the use of catalyst at a given temperature and this depends upon E_a -$E_a^{{}^{\prime }}$. Greater the value of E_a –$E_a^{{}^{\prime }}$, more number of times k_cat is greater than k.

The rate of reaction in the presence of catalyst at any temperature T₁ may be made equal to the rate of reaction in absence of catalyst but for this sake we will have to raise the temperature. Let this temperature be T₂ then

$e^{-{E^{\prime }}_a/R{T}_1}=e^{-E_a/R{T}_2}$

or$\frac{{E^{\prime }}_a}{T_1}=\frac{E_a}{T_2}$

Illustration 1. Let k₁:k₂ = 1 :15. Calculate the ratio, $\frac{\left[\text{C}\right]}{\left[\text{A}\right]}$at the end of one hour assuming that k₁ = x hr^–1

Solution: $\frac{-\text{d}\left[\text{A}\right]}{\text{dt}}$= $\left(k_1+k_{\text{2}}\right)\left[A\right]$

$\therefore$$\frac{-\text{d}\left[\text{A}\right]}{\left[\text{A}\right]}$ = (k₁+k₂) dtIntegrating with in the required limits, we get$\text{ln}\frac{{\left[\text{A}\right]}_{\text{0}}}{{\left[\text{A}\right]}_{\text{t}}}$ = (k₁ + k₂ )t$\therefore$ ln $\frac{{\left[\text{A}\right]}_{\text{t}}+\left[\text{B}\right]+\left[\text{C}\right]}{{\left[\text{A}\right]}_{\text{t}}}$ = (k₁+k₂) tSince$\frac{\left[\text{B}\right]}{\left[\text{C}\right]}$ = $\frac{k_1}{k_2}$ = $\frac{1}{15}$$\therefore$ ln $\frac{{\left[\text{A}\right]}_{\text{t}}+\frac{\left[C\right]}{15}+\left[C\right]}{{\left[\text{A}\right]}_{\text{t}}}$ = 16x$\frac{\left[C\right]}{{\left[A\right]}_t}=\frac{15}{16}\left(e^{16x}-1\right)$

Arrhenius Equation

The variation of equilibrium constant of a reaction with temperature is described by Van't Hoff equation of thermodynamics which is as follows:

$\frac{d\ln K_p}{dT}=\frac{\Delta H}{R{T}^2}$

If k₁ and k₂ be the rate constants of forward reaction and backward reaction, respectively then K_p = k₁/k₂. Further,

$\Delta$H = Ea₁ – Ea₂ .Putting these in the above equation we get,

$\frac{d\ln k_1}{dT}-\frac{d\ln k_2}{dT}=\frac{E_{a_1}}{R{T}^2}-\frac{E_{a_2}}{R{T}^2}$

Splitting into two parts

$\frac{d\ln k_1}{dT}=\frac{E_{a{}_1}}{R{T}^2}+Z$ (For FR)

$\frac{d\ln k_2}{dT}=\frac{E_{a_2}}{R{T}^2}+Z$ (For BR)

where Z is constant

Arrhenius sets Z equal to zero and without specifying FR and BR, he gave the following equation called Arrhenius equation.

$\frac{d\ln k}{dT}=\frac{Ea}{R{T}^2}$ …(i)

From this equation it is evident that rate of change of logarithm of rate constant with temperature depends upon the magnitude of energy of activation of the reaction. Higher the Ea smaller the rate of change of logarithm of rate constant with temperature. That is, rate of the reaction with low E_a increases slowly with temperature while rate of the reaction with high E_a increases rapidly with temperature. It is also evident that rate of increase of logarithm of rate constant will go on decreasing with increase of temperature.

Integrating Equation 4 assuming Ea to be constant we get,

lnk =$-\frac{Ea}{RT}+\ln A$ ..(ii)

or $\ln \frac{k}{A}=-\frac{Ea}{RT}$

or k =$A{e}^{-Ea/RT}$ …(iii)

Equation (iii) is integrated form of Arrhenius equation. The constant A called pre-exponential factor is the frequency factor since it is somewhat related with collision frequency. It is a constant for a given reaction. From Equation (iii) it is evident that as T $\to$ $\infty$, k $\to$ A. Thus, the constant A is the rate constant of reaction at infinity temperature. The rate constant goes on increasing with temperature.

So, when T approaches infinity, k will be maximum. That is to say, A is the maximum rate constant of a reaction.

It is also to be noted that the exponential term i.e. e^–Ea/RT measures the fraction of total number of molecules in the activated state or fraction of the total number of effective collisions. If n_Ea and n be the number of molecules of reactant in the activated state and the total number of molecules of the reactant present in the reaction vessel respectively, then

$\frac{n_{E_a}}{n}=e^{-E_a/RT}$

Equation (ii) may also be put as

logk = $\left(-\frac{Ea}{2.303R}\right)\frac{1}{T}$ + logA …(iv)

Since $\frac{E_a}{2.303R}$ and logA both are constants for a given reaction. So from equation (iv) it is evident that a plot of log k vs. $\frac{1}{T}$will be a straight line of the slope equal to $-\frac{E_a}{2.303R}$ and intercept equal to logA as shown below.

\dfrac{-{{E}_{a}}}{2.303R}=\\tan \theta = - \tan \left( {180 - \theta } \right) = $-$\dfrac{OA}{OB}$

$\therefore$ E_a = $\frac{OA}{OB}\times 2.303R$

Thus, from this plot E_a and A both can be determined accurately.

If k₁ be the rate constant of a reaction at two different temperature T₁ and T₂ respectively then from equation (iv), we may write

log k₁ = $-\frac{E_a}{2.303R}$ × $\frac{1}{T_1}+\log A$ and log k₂ = $-\frac{E_a}{2.303R}$ × $\frac{1}{T_2}+\log A$

Subtracting former from the latter we get

$\log \frac{k_2}{k_1}$= $\frac{E_a}{2.303R}$$\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ … (v)

With the help of this equation it is possible to calculate E_a of a reaction provided, rate constants of reaction at two different temperatures are known. Alternatively one can calculate rate constant of a reaction at a given temperature provided that rate constant of the reaction at some other temperature and also E_a of the reaction is known.