Question

Consider the following reversible reaction, $A(g) + B(g) \rightleftharpoons AB(g)$. The activation energy of the backward reaction exceeds that of the forward reaction by $2RT(Jmo{l^{ - 1}})$. If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta {G^o}(Jmo{l^{ - 1}})$ for the reaction at 300 K is ______.

Given $ln 2 = 0.7,RT = 2500Jmo{l^{ - 1}}$ at 300 K and $G$ is the Gibbs energy

Answer 1

To solve this question you must recall the relation between rate constant of a reaction with its free energy change. Gibbs free energy is a thermodynamic potential that is used to calculate the maximum work that can be performed by a system at a constant pressure and temperature.

Formula used:

$\Delta {G^\theta } = - RT\ln {K_{eq}}$

And,

${K_f} = {A_f}{e^{\dfrac{{ - {{\left( {{E_a}} \right)}_f}}}{{RT}}}}$

${K_b} = {A_b}{e^{\dfrac{{ - {{\left( {{E_a}} \right)}_b}}}{{RT}}}}$

${K_f} = {A_f}{e^{\dfrac{{ - {{\left( {{E_a}} \right)}_f}}}{{RT}}}}$

${K_b} = {A_b}{e^{\dfrac{{ - {{\left( {{E_a}} \right)}_b}}}{{RT}}}}$

Where, $\Delta {G^o}$ is the Gibbs free energy, ${K_f}$ is the rate constant of forward reaction, ${K_b}$ is the rate constant for the backward reaction, ${K_{eq}}$ is the rate constant of reaction, ${E_a}$ is the activation energy, $T$ is the temperature and $R$ is the gas constant.

Complete step by step answer:

For the given reaction, $A(g) + B(g) \rightleftharpoons AB(g)$, we know that

${\left( {{E_a}} \right)_f} - {\left( {{E_a}} \right)_b} = 2RT$

And, $\dfrac{{{A_f}}}{{{A_b}}} = 4$

We know that, the rate constant of the reaction can be given by, ${K_{eq}} = \dfrac{{{K_f}}}{{{K_b}}}$.

Substituting the values of ${K_f}$and ${K_b}$, we get,

${K_{eq}} = \dfrac{{{A_f}}}{{{A_b}}} \times {e^{\dfrac{{ - {{\left( {{E_a}} \right)}_f}}}{{RT}} + }}^{\dfrac{{{{\left( {{E_a}} \right)}_b}}}{{RT}}}$

$\Rightarrow {K_{eq}} = 4 \times {e^{\dfrac{{{{\left( {{E_a}} \right)}_b} - {{\left( {{E_a}} \right)}_f}}}{{RT}}}}$

$\Rightarrow {K_{eq}} = 4{e^2}$ , since ${\left( {{E_a}} \right)_f} - {\left( {{E_a}} \right)_b} = 2RT$

Now using this value to find the Gibbs free energy we get,

$\Delta {G^\theta } = - RT\ln \left( {4{e^2}} \right)$

$\Rightarrow \Delta {G^\theta } = - RT\left( {\ln 4 + 2\ln e} \right)$

$\Rightarrow \Delta {G^\theta } = - RT(1.40 + 2) = - 2500 \times 3.40$

$\therefore \Delta {G^\theta } = - 8500J$

Note:

Activation energy is the energy that we need to provide to compounds in order for a chemical reaction to take place. The activation energy $\left( {{E_a}} \right)$ is commonly measured in joules per mole (J/mol). Activation energy can be considered as the magnitude of the energy barrier separating the initial and final thermodynamic states, namely the reactants and products. For a chemical reaction to occur at a good rate, the temperature of the system should be high enough so that there are an appreciable number of molecules with energy greater than or equal to the activation energy. $k = A{e^{\dfrac{{ - \left( {{E_a}} \right)}}{{RT}}}}$ is known as the Arrhenius equation, which gives a relation between the activation energy of the reaction and the rate at which the reaction proceeds. The activation energy of a reaction can be reduced by addition of suitable catalysts that make the reaction more feasible and faster.