The following reaction is performed at 298 K . $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$ The standard free energy of formation of $\mathrm{NO}(\mathrm{g})$ is $86.6 \mathrm{~kJ} / \mathrm{mol}$ at 298 K . What is the standard free energy of formation of $\mathrm{NO}_{2}(\mathrm{~g})$ at 298 K ? ( $\mathrm{K}_{\mathrm{P}}=1.6 \times 10^{12}$ )

(C) $0.5\left[2 \times 86,600-\mathrm{R}(298) \ln \left(1.6 \times 10^{12}\right)\right]$

(A) $86600+\mathrm{R}(298) \ln \left(1.6 \times 10^{12}\right)$

(B) $86600-\frac{\ln \left(1.6 \times 10^{12}\right)}{\mathrm{R}(298)}$

(D) $R(298) \ln \left(1.6 \times 10^{12}\right)-86600$

JEE Main · 2015 · 04 Apr 2015 · Offline · Chemical Thermodynamics

The following reaction is performed at 298 K .

2 NO (g) + O_{2} (g) ⇌ 2 NO_{2} (g)

The standard free energy of formation of

NO (g)

86.6 kJ / mol

at 298 K . What is the standard free energy of formation of

NO_{2} (g)

at 298 K ? (

K_{P} = 1.6 \times 1 0^{12}

)

Answer: (C)

Write the reaction free energy through the formation energies:

Δ G^{\circ} = 2Δ G_{f}^{\circ} (N O_{2}) - 2Δ G_{f}^{\circ} (NO) - Δ G_{f}^{\circ} (O_{2})

, with

Δ G_{f}^{\circ} (O_{2}) = 0

. At equilibrium

Δ G^{\circ} = - R T ln K_{P}

, so

- R T ln K_{P} = 2Δ G_{f}^{\circ} (N O_{2}) - 2 (86600)

. Solving for the single unknown,

Δ G_{f}^{\circ} (N O_{2}) = \frac{1}{2} [2 \times 86600 - R (298) ln (1.6 \times 1 0^{12})]

J/mol.

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option