Chapter 13, Problem 13.135QP

The rate constants for the first-order decomposition of an organic compound in solution are measured at several temperatures:

$\begin{array}{cccccc}k(s^{-1})& 0.00492& 0.0216& 0.0950& 0.326& 1.15\\ T(K)& 278& 288& 298& 308& 318\end{array}$

Determine graphically the activation energy and frequency factor for the reaction.

Interpretation Introduction

Interpretation:

The activation energy and frequency factor for the given reaction has to be determined graphically.

Concept introduction:

Arrhenius equation:

Arrhenius equation is a formula that represents the temperature dependence of reaction rates

The Arrhenius equation has to be represented as follows

$k=A{e}^{-E_a/RT}$

$E_a$ is the activation energy and it’s unit is kJ/mol

R represents the universal gas constant and it has the value of 8.314 J/K.mol

T represents the absolute temperature

A is the frequency factor

Activation energy can be determined through graphical method,

In the graph of $\ln k$ and,$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$T$}\right.$, slope will give the activation energy.

$\begin{array}{c}\text{slope}\text{=}\frac{{\text{-E}}_{\text{a}}}{\text{R}}\\ {\text{-E}}_{\text{a}}\text{=}\text{slope}\text{×}\text{R}\end{array}$

Answer to Problem 13.135QP

The activation energy and frequency factor for the given reaction are $101kJ/mol$ and $A=3\times 10^{16}{s}^{-1}$ respectively.

Explanation of Solution

Given,

Rate constants for the first order decomposition of an organic compound in solution are measured at several temperatures, and they are listed below,

$\begin{array}{cccccc}k\left(s^{-1}\right)& 0.00492& 0.0216& 0.0950& 0.326& 1.15\\ T\left(K\right)& 278& 288& 298& 308& 318\end{array}$

To draw the graph, we need $\ln kand\frac{1}{T}$ values.

$\begin{array}{cccccc}\ln k& -5.31& -3.84& -2.35& -1.12& 0.140\\ \frac{1}{T}\left(K^{-1}\right)& 3.60\times {10}^{-3}& 3.47\times {10}^{-3}& 3.36\times {10}^{-3}& 3.25\times {10}^{-3}& 3.14\times {10}^{-3}\end{array}$

 Form the given data we can draw the graph by plotting $\ln k$ value on Y-axis, $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$T$}\right.$ value on X-axis. The obtained graph is,

Fig (1)

According to the straight-line equation in mathematics,

$\begin{array}{l}\text{y}=\text{mx+c}\\ \text{m}\to \text{slope}\\ \text{c}\to \text{intercept}\end{array}$

Herein,

$y=-1.21\times {10}^{4}x+e^{3.80}$

The slope of the line is $-1.24\times {10}^4\text{K}$ and is $-\raisebox{1ex}{$E_a$}\!\left/ \!\raisebox{-1ex}{$R$}\right.$

Now, the activation energy for the given reaction can be calculated as follows,

$\begin{array}{c}\text{slope}\text{=}\frac{{\text{-E}}_{\text{a}}}{\text{R}}\\ {\text{-E}}_{\text{a}}\text{=}\text{slope}\text{×}\text{R}\end{array}$

$-E_a=(-1.21\times {10}^{4}K)\times (\text{8}\text{.314 J/K}\text{.mol})$

$E_a=1.01\times 10^{5}J/mol=101kJ/mol$

The y-intercept equals the natural logarithm of the frequency factor (A).

Therefore,

$\begin{array}{c}\text{Intercept}=\text{A}\left(\text{Frequency}\text{factor}\right)\text{=}{\text{e}}^{\text{3}\text{.80}}\\ \text{ln}\text{A}=\text{38}\text{.0}\\ A=3\times 10^{16}{s}^{-1}\end{array}$

Conclusion

The activation energy and frequency factor for the given reaction were graphically determined.