Chapter 11, Problem 11.59PAE

Interpretation Introduction

Interpretation:

$E_a$ should be determined in kJ/mol for the reaction of production of nitrogen oxides in internal combustion engines, when the rate constants of that reaction, measured at three different temperatures are given.

Concept introduction:

Rate of a reaction can be explained using either growth of products or reduction of reactants. Both give the same rate, but our concern is different.

Every reaction has activation energy. To overcome this energy sometimes we should provide energy from outside. But for some reactions, we do not have to supply energy. Ambient temperature is enough that reactions. Those reactions are called spontaneous reactions.

Two methods can be used to find out the activation energy of a reaction when the rate constants at two or more temperatures are known.

1. Two point method by solving simultaneous equations (If rate constants are known only for two temperatures
2. Graphical method where activation energy can be obtained using the gradient of the graph (When rate constants for more than two temperatures are known)

Answer to Problem 11.59PAE

Solution:

$E_a={\text{ 316 kJmol}}^{-1}$

Given:

• Chemical reaction

$O(g)+N_2(g)\to NO(g)+N(g)$

$k/{\text{ Lmol}}^{-1}{s}^{-1}$	Temperature/K
$4.4\times {10}^2$	2000
$2.5\times {10}^5$	3000
$5.9\times {10}^6$	4000

Explanation of Solution

$O(g)+N_2(g)\to NO(g)+N(g)$

• The rate of the equation for the reaction can be written as follows.

$R=-k[{{\rm N}}_2][{\rm O}]$

• The only equation relating activation energy and rate constant is Arrhenius equation which is given below. The frequency factor doesn’t depend on the temperature.

It can be written as

$\ln k=\ln A-\frac{E_a}{RT}$

Therefore at two different temperatures at $T_1{\text{ and T}}_2$ ;

$\begin{array}{l}ln{k}_1=lnA–\raisebox{1ex}{$\Delta E_a$}\!\left/ \!\raisebox{-1ex}{$R{T}_1$}\right.\to 1\hfill \\ ln{k}_2=lnA–\raisebox{1ex}{$\Delta E_a$}\!\left/ \!\raisebox{-1ex}{$R{T}_2\to 2$}\right.\hfill \end{array}$

When equation 1 is subtracted from equation 2,

$ln\left(k_2/k_1\right)=\frac{\Delta E_a}{R}\left(\frac{1}{T_1}–\frac{1}{T_2}\right)$

Formula used:

$ln\left(k_2/k_1\right)=\frac{\Delta E_a}{R}\left(\frac{1}{T_1}–\frac{1}{T_2}\right)$

Calculation:

$\begin{array}{l}ln\left(k_2/k_1\right)=\frac{\Delta E_a}{R}\left(\frac{1}{T_1}–\frac{1}{T_2}\right)\hfill \\ \hfill \\ \text{Substitution of values}\hfill \\ ln\left(2.5\times {10}^5/4.4\times {10}^2\right)=\frac{\Delta E_a}{8.314}\left(\frac{1}{2000}-\frac{1}{3000}\right)\hfill \\ \begin{array}{l}ln\left(5.68\times {10}^2\right)=\frac{\Delta E_a}{8.314}\left(1.67\times {10}^{-4}\right)\\ 6.34=\frac{\Delta E_a}{8.314}\left(1.67\times {10}^{-4}\right)\\ \Delta E_a=3.16\times {10}^{5}Jmo{l}^{-1}\\ \Delta E_a=\text{ 316}\text{.38 }kJmo{l}^{-1}\end{array}\hfill \\ \hfill \end{array}$

Conclusion

$E_a={\text{ 316 kJmol}}^{-1}$