Question

The following reaction is performed at 298 K.

$2NO(g)+{{O}_{2}}(g)\rightleftharpoons 2N{{O}_{2}}(g)$

The standard free energy of formation of NO (g) is 86.6 KJ/mol at 298 K. What is the standard free energy of formation of $N{{O}_{2}}$ (g) at 298 K? (${{K}_{p}}=1.6\times {{10}^{12}}$ ).

A. $R(298)\log (1.6\times {{10}^{12}})J/mol$

B. $86600+R(298)\log (1.6\times {{10}^{12}})J/mol$

C. $86600-\dfrac{1.6\times {{10}^{12}}}{R(298)}J/mol$

D. $0.5\times [2\times 86600-R(298)\log 1.6\times {{10}^{12}}]J/mol$

Answer 1

We should know the relationship between equilibrium constant and Gibbs free energy of the reaction to find the standard energy of the formation of the product.

The relationship between equilibrium constant and Gibbs free energy of a reaction is as follows.

$\Delta G_{rxn}^{0}=-RT\log {{K}_{p}}\to (1)$

Where $\Delta G_{rxn}^{0}$ = free energy of the reaction

R = Gas constant

T = Temperature of the gas

${{K}_{p}}$ = equilibrium constant of the reaction.

Complete step by step answer:

- In the question the given chemical reaction is as follows.

$2NO(g)+{{O}_{2}}(g)\rightleftharpoons 2N{{O}_{2}}(g)$

- In the above reaction two moles of nitrogen monoxide reacts with one mole of oxygen and forms 2 moles of nitrogen dioxide.

- The standard Gibbs free energy change for the above reaction is as follows.

$\Delta G_{rxn}^{0}=2\Delta G_{f}^{0}(N{{O}_{2}})-2\Delta G_{f}^{0}(NO)\to (2)$

- Substitute equation 2 in equation 1 and we will get the following expression.

$2\Delta G_{f}^{0}(N{{O}_{2}})-2\Delta G_{f}^{0}(NO)=-RT\log {{K}_{p}}$

$2\Delta G_{f}^{0}(N{{O}_{2}})=2\Delta G_{f}^{0}(NO)-RT\log {{K}_{p}}$

$\Delta G_{f}^{0}(N{{O}_{2}})=0.5\times [2\Delta G_{f}^{0}(NO)-RT\log {{K}_{p}}]\to (3)$

- In the question it is given that the standard free energy of formation of NO (g) is 86.6 KJ/mol = 86600 J/mol at 298 K.

- ${{K}_{p}}=1.6\times {{10}^{12}}$ , means equilibrium constant of the reaction.

- Substitute all the known parameters in the equation (3) to get the answer.

$\Delta G_{f}^{0}(N{{O}_{2}})=0.5\times [2\Delta G_{f}^{0}(NO)-RT\log {{K}_{p}}]$

$\Delta G_{f}^{0}(N{{O}_{2}})=0.5\times [2\times 86600-R(298)\log 1.6\times {{10}^{12}}]J/mol$

So, the correct answer is “Option D”.

Note: In the solution the terms
${{K}_{p}}$ is the equilibrium constant of the reaction
$\Delta G_{f}^{0}(N{{O}_{2}})$ is the Gibbs free energy of the product (nitrogen dioxide)
$\Delta G_{f}^{0}(NO)$ is the Gibbs free energy of the reactant (nitrogen monoxide).