Arrhenius Equation - Practice Questions & MCQ

We know that on increasing the temperature, the rate of the reaction or rate constant increases.
The rate equation is given as follows:

Rate $=\mathrm{k}[\text { conc }]^{\mathrm{n}}$

Here k is the rate constant

Now, we will see the relation between k and T or also known as 'approximate dependency of k on T'.

Generally on 10^oC rise in temperature, rate constant nearly doubles.

Temperature Coefficient: It is the ratio of two rate constants. Thus, mathematically it is given as:

$T_{\text {coeff }}=\frac{\mathrm{k}_{(\mathrm{t}+10)^{\circ} \mathrm{C}}}{\mathrm{k}_{\mathrm{t}^{\circ} \mathrm{C}}}$

Thus, the temperature coefficient is showing the dependency of the rate constant(k) on temperature(T).

NOTE: The standard value of the temperature coefficient is given at t = 25^oC and (t+10) = 35^oC.

The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation. It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation.

        $\mathrm{k}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}}$

where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and E_a is activation energy measured in joules/mole(J mol^–1).

We have the rate constant K₁ at temperature T₁ and rate constant K₂ at temperature T₂. 

We know that the Arrhenius equation is given as follows:

$
\log _{10} \mathrm{~K}_1=\log _{10} \mathrm{~A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT} \mathrm{~T}_1}
$

$
\log _{10} \mathrm{~K}_2=\log _{10} \mathrm{~A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}_2}
$

On subtracting equation (i) from (ii), we get:

$\log _{10} \mathrm{~K}_2-\log _{10} \mathrm{~K}_1=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}_1}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}_2}$

Thus, $\log \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{Ea}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]$

Although the Arrhenius equation explains the exact relationship between the rate of reaction and the temperature but there are still some exceptions in this theory. Actually, on increasing the temperature rate may decrease sometimes and may not follow Arrhenius equation. Following examples will illustrate these exceptions.

• Bacterial decomposition
  
  From the graph, it is clear that first on increasing the temperature the rate also increases but at T_i (also known as inversion temperature) rate starts to decrease on further increasing the temperature.
•  

On the basis of mechanism, we have two types of reactions:

• Simple or elementary reaction
• Complex or multi-step reaction

Simple or Elementary reaction

• The reactions, which occur in single step, are called simple or elementary reactions. For example:
  
  $\mathrm{NO}(\mathrm{g})+\mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$
• An elementary reaction is an individual molecular event that involves breaking or making of chemical bonds. The overall reaction describes the stoichiometry of the overall process but provides no information how the reaction occurs.

Complex Reaction

• A complex reaction takes place in a sequence of a number of elementary steps. 
• Molecularity of complex reaction is not defined. Molecularity of each step can be defined but not for overall.
• Overall rate of reaction is given by slowest step of the complex reaction.
  For example, combination of NO₂ and CO occurs in a sequence of elementary steps.
  $\begin{array}{ll}\mathrm{NO}_2(\mathrm{~g})+\mathrm{NO}_2(\mathrm{~g}) \rightarrow \mathrm{NO}_3(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) & \text { (elementary step I) } \\ \mathrm{NO}_3(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{NO}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) & \text { (elementary step II) }\end{array}$

Important Facts:

• The numbeer of reactant molecules taking part in an elementary step or in an elementary reaction is expressed as molecularity of that step of molecularity of that reaction respectively.
• For elementary reactions usually order of reaction and molecularity are same. Thus, it can be said that if order of a reaction for a change is fractional it cannot be an elementary reaction.

This example will illustrate how to determine the rate law when the intermediate is involved in the rate-determining step.

$2 \mathrm{O}_3 \rightarrow 3 \mathrm{O}_2$

Mechanism
$\mathrm{O}_3 \rightleftharpoons \mathrm{O}_2+\mathrm{O}$
$K_1$ and $K_2$ are the forward and backward reaction constants respectively
$\mathrm{O}+\mathrm{O}_3 \rightarrow 2 \mathrm{O}_2$
$\mathrm{K}_3$ is the rate constant

In this case, step 1 is fast and step 2 is slow.

The rate law is given as follows:

rate $=\mathrm{K}_3[\mathrm{O}]^1\left[\mathrm{O}_3\right]^1$

We know from equilibrium theory that:

$
\begin{aligned}
& \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{K}_1}{\mathrm{~K}_2}=\frac{\left[\mathrm{O}_2\right][\mathrm{O}]}{\left[\mathrm{O}_3\right]} \\
& {[\mathrm{O}]=\frac{\mathrm{K}_1\left[\mathrm{O}_3\right]}{\mathrm{K}_2\left[\mathrm{O}_2\right]}}
\end{aligned}
$
$
\text { Thus, rate }=\frac{\mathrm{K}_3 \cdot \mathrm{~K}_1}{\mathrm{~K}_2} \frac{\left[\mathrm{O}_3\right]\left[\mathrm{O}_3\right]}{\left[\mathrm{O}_2\right]}=\frac{\mathrm{K}_3 \cdot \mathrm{~K}_1}{\mathrm{~K}_2} \frac{\left[\mathrm{O}_3\right]^2}{\left[\mathrm{O}_2\right]}
$

Thus, Order $=2-1=1$
rate constant $=\frac{\mathrm{K}_3 \cdot \mathrm{~K}_1}{\mathrm{~K}_2}$

In this situation, B and C both are forming. These types of reactions are known as parallel reactions. Both these reactions are first order reactions with rate constants K₁ and K₂ respectively and half-lives as t_(1/2)1 and t_(1/2)2.

For these parallel reactions, we need to find:

• Effective order
• Effective rate constant
• Effective t_1/2
• Effective Activation energy
• [A], [B], [C] with time (t) variation
• % of [B] and % of [C]

We know that the rate equations are given as follows:

$\begin{aligned} & \mathrm{r}_1=\frac{-\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}_1[\mathrm{~A}] \\ & \mathrm{r}_2=\frac{-\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}_2[\mathrm{~A}]\end{aligned}$

Thus, overall rate of reaction is :

$
\frac{-\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}_1[\mathrm{~A}]+\mathrm{K}_2[\mathrm{~A}]=\left(\mathrm{K}_1+\mathrm{K}_2\right)[\mathrm{A}]
$

Thus, rate $=\left(\mathrm{K}_1+\mathrm{K}_2\right)[\mathrm{A}]^1$

Effective Rate Constant $\left(\mathrm{K}_{\mathrm{eff}}\right)=\left(\mathrm{K}_1+\mathrm{K}_2\right)$

Effective order of reaction $=1$

$\begin{aligned} \text { Now, effective half }- \text { life }\left(\mathrm{t}_{1 / 2}\right) & =\frac{0.693}{\mathrm{~K}_{\mathrm{eff}}}=\frac{0.693}{\mathrm{~K}_1+\mathrm{K}_2} \\ \Rightarrow & \frac{0.693}{\frac{0.693}{\left(\mathrm{t}_{1 / 2}\right)_1}+\frac{0.693}{\left(\mathrm{t}_{1 / 2}\right)_2}}\end{aligned}$

Thus, effective half life is given as :

$
\frac{1}{\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{eff}}}=\frac{1}{\left(\mathrm{t}_{1 / 2}\right)_1}+\frac{1}{\left(\mathrm{t}_{1 / 2}\right)_2}
$

NOTE: Effective activation energy, [A], [B], [C] with time (t) variation and % of [B] and % of [C] will be discussed in later concepts.

We know that Arrhenius equation is given as:

$\mathrm{K}=\mathrm{A} \cdot \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}}$

$\mathrm{K}_{\mathrm{eff}}=\mathrm{K}_1+\mathrm{K}_2$

Thus, $\mathrm{A}_{\mathrm{eff}} \cdot \mathrm{e}^{-\mathrm{Ea}_{\mathrm{eff}} / \mathrm{RT}}=\mathrm{A}_1 \cdot \mathrm{e}^{-\mathrm{Ea}_1 / \mathrm{RT}}+\mathrm{A}_2 \cdot \mathrm{e}^{-\mathrm{Ea}_2 / \mathrm{RT}}$

Differentiate this equation with respect to temperature ' T '

Thus, we have:

$\mathrm{A}_{\mathrm{eff}} \cdot \mathrm{e}^{-\mathrm{Ea}_{\mathrm{eff}} / \mathrm{RT}}\left(\frac{+\mathrm{Ea}_{\mathrm{eff}}}{\mathrm{RT}^2}\right)=\mathrm{A}_1 \cdot \mathrm{e}^{-\mathrm{Ea}_1 / \mathrm{RT}}\left(\frac{+\mathrm{Ea}_1}{\mathrm{RT}^2}\right)+\mathrm{A}_2 \cdot \mathrm{e}^{-\mathrm{Ea}_2 / \mathrm{RT}}\left(\frac{+\mathrm{Ea}_2}{\mathrm{RT}^2}\right)$

$\mathrm{K}_{\mathrm{eff}} \mathrm{E}_{\mathrm{eff}}=\mathrm{K}_1 \mathrm{Ea}_1+\mathrm{K}_2 \mathrm{Ea}_2$

$\mathrm{E}_{\mathrm{eff}}=\left(\mathrm{K}_1 \mathrm{Ea}_1+\mathrm{K}_2 \mathrm{Ea}_2\right) / \mathrm{K}_{\mathrm{eff}}$

We know that the rate equations are given as follows:

$\begin{aligned} & \mathrm{r}_1=\frac{-\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}_1[\mathrm{~A}] \\ & \mathrm{r}_2=\frac{-\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}_2[\mathrm{~A}]\end{aligned}$

The overall rate of the reaction is:

$\frac{-\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}_1[\mathrm{~A}]+\mathrm{K}_2[\mathrm{~A}]=\left(\mathrm{K}_1+\mathrm{K}_2\right)[\mathrm{A}]$

$\begin{aligned} & \Rightarrow \frac{\mathrm{dA}}{\mathrm{dt}}=\left(\mathrm{K}_1+\mathrm{K}_2\right)[\mathrm{A}]^1 \\ & \Rightarrow \frac{\mathrm{dA}}{\mathrm{A}}=-\left(\mathrm{K}_1+\mathrm{K}_2\right) \mathrm{dt}\end{aligned}$

On integrating both sides of the above equation, we get:

$\begin{aligned} & \int_{\mathrm{A}_{\mathrm{o}}}^{\mathrm{A}} \frac{\mathrm{dA}}{\mathrm{A}}=-\int_0^{\mathrm{t}}\left(\mathrm{K}_1+\mathrm{K}_2\right) \mathrm{dt} \\ & \Rightarrow[\log \mathrm{A}]_{\mathrm{A}_{\mathrm{o}}}^{\mathrm{A}}=-\left(\mathrm{K}_1+\mathrm{K}_2\right)[\mathrm{t}]_0^t \\ & \Rightarrow \log \mathrm{~A}-\log \mathrm{A}_{\mathrm{o}}=-\mathrm{k}_{\mathrm{eff}} \mathrm{t} \\ & \Rightarrow \log \frac{\mathrm{A}}{\mathrm{A}_{\mathrm{o}}}=-\mathrm{k}_{\mathrm{eff}} \mathrm{t} \\ & \Rightarrow \frac{\mathrm{A}}{\mathrm{A}_0}=\mathrm{e}^{-\mathrm{k}_{\mathrm{eft}} \mathrm{t}}\end{aligned}$

Thus, $[\mathrm{A}]=\left[\mathrm{A}_{\mathrm{o}}\right] \mathrm{e}^{-\mathrm{k}_{\mathrm{eff}} \mathrm{t}}$ $\qquad$

Now, the rate equation for B can be given as follows:

$
\frac{\mathrm{dB}}{\mathrm{dt}}=\mathrm{k}_1[\mathrm{~A}]=\mathrm{k}_1\left[\mathrm{~A}_{\mathrm{o}}\right] \mathrm{e}^{-\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{t}}
$ ...........(From equation (I))

$\begin{aligned} & \Rightarrow \int_0^{\mathrm{B}} \mathrm{dB}=\int_0^{\mathrm{t}} \mathrm{k}_1\left[\mathrm{~A}_0\right] \mathrm{e}^{-\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{t}} \mathrm{dt} \\ & \text { Thus, }[\mathrm{B}]=\frac{\mathrm{k}_1 \mathrm{~A}_{\mathrm{o}}}{\mathrm{k}_1+\mathrm{k}_2}\left[1-\mathrm{e}^{-\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{t}}\right]\end{aligned}$

Similarly, [C] is given as follows:

$[\mathrm{C}]=\frac{\mathrm{k}_2 \mathrm{~A}_{\mathrm{o}}}{\mathrm{k}_1+\mathrm{k}_2}\left[1-\mathrm{e}^{-\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{t}}\right]$