Chapter 19.6, Problem 19.8WE

Interpretation Introduction

Interpretation:

For the given reaction with given temperature and rate constant data the activation energy should be determined.

Concept introduction:

In order to establish the plausibility of a mechanism, one must compare the rate law of the rate determining step to the experimentally determined rate law.

Rate determining step: In a chemical reaction the rate determining step is the slowest step in which the rate of the reaction depends on the rate of that slowest step.

Rate law: It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

Activation energy: It is defined as the minimum energy required by the reacting species in order to undergo chemical reaction.

Intermediate species: It is the species formed during the middle of the chemical reaction between the reactant and the desired product.

Arrhenius equation:

• Arrhenius equation is a formula that represents the temperature dependence of reaction rates
• The Arrhenius equation has to be represented as follows

$\begin{array}{l}k=A{e}^{-E_a/RT}\\ \ln k=\ln A{e}^{-E_a/RT}\\ \ln k=\left(-\frac{{\text{E}}_{\text{a}}}{\text{R}}\right)\left(\frac{1}{\text{T}}\right)+\ln \text{A}\end{array}$

• $E_a$ represents the activation energy and it’s unit is $\text{kJ/mol}$
• R represents the universal gas constant and it has the value of 8.314 $\text{J/K}\text{.mol}$
• T represents the absolute temperature
• A represents the frequency factor or collision frequency
• e represents the base of natural logarithm
•  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.

Answer to Problem 19.8WE

The activation energy for the given reaction is $\text{46kJ/mol}$.

Explanation of Solution

In order to find the activation energy we need to find the slope value of line which obtained from the graph plotted between $\text{lnK}$ and $\text{1/T}$.

First the following table should be generated for values $\text{lnK}$ and $\text{1/T}$ values using the given set of values

The values for $\text{lnK}$ are obtained as follows by using the value of given K.

$\begin{array}{l}k\left({\text{M}}^{\text{-1}}{\text{.s}}^{\text{-1}}\right)\ln k\\ 0.0521-2.95\\ 0.101-2.29\\ 0.184-1.69\\ 0.332-1.10\end{array}$

The values for $\text{1/T}$ are obtained as follows by using the given values $\text{T}\left(\text{K}\right)$.

$\begin{array}{l}\text{T}\left(\text{K}\right)\text{1/T}\left({\text{K}}^{\text{-1}}\right)\\ 2883.47\times {10}^{-3}\\ 2983.36\times {10}^{-3}\\ 3083.25\times {10}^{-3}\\ 3183.14\times {10}^{-3}\end{array}$

The resulting table is as follows,

$\begin{array}{l}\ln k\text{1/T}\left({\text{K}}^{\text{-1}}\right)\\ -2.953.47\times {10}^{-3}\\ -2.293.36\times {10}^{-3}\\ -1.693.25\times {10}^{-3}\\ -1.103.14\times {10}^{-3}\end{array}$

The graph for above table is drawn as follows,

Now the slope for the line in the graph is calculated by using the two marked points in the graph.

$\text{slope = }\frac{\text{-1}\text{.4-}\left(\text{-2}\text{.5}\right)}{\text{3}{\text{.2×10}}^{\text{-3}}{\text{K}}^{\text{-1}}\text{-3}{\text{.4×10}}^{\text{-3}}{\text{K}}^{\text{-1}}}\text{= -5}{\text{.5×10}}^{\text{3}}\text{K}$

Now, the activation energy is calculated as follows by using the expression $\text{slope = }\frac{-E_a}{R}$

Therefore,

$\begin{array}{l}{E}_a\text{ = }-\text{slope}\times R\\ =-\left(-5.5\times {10}^3\text{K}\right)\left(8.314\text{J/K}\text{. mol}\right)\\ =4.6\times {10}^4\text{J/mol}\\ \text{=}4.6\text{kJ/mol}\end{array}$

Therefore, the activation energy is $4.6\text{kJ/mol}$

Conclusion

The activation energy for the given reaction was determined using the given rate constant data.