Consider the following reaction approaching equilibrium at $27^{\circ} \mathrm{C}$ and 1 atm pressure $\mathrm{A}+\mathrm{B} \underset{\mathrm{K}_{\mathrm{r}}=10^{2}}{\stackrel{\mathrm{~K}_{\mathrm{f}}=10^{3}}{\rightleftharpoons}} \mathrm{C}+\mathrm{D}$ The standard Gibb's energy change ( $\Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ ) at $27^{\circ} \mathrm{C}$ is ( - ) ______ $\mathrm{kJ} \mathrm{mol}^{-1}$ (Nearest integer). (Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and $\ln 10=2.3$ )

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Chemical Thermodynamics

Consider the following reaction approaching equilibrium at

2 7^{\circ} C

and 1 atm pressure

A + B K_{r} = 1 0^{2} ⇌ K_{f} = 1 0^{3} C + D

The standard Gibb's energy change (

Δ_{r} G^{\circ}

) at

2 7^{\circ} C

is ( - ) ______

kJ mol^{- 1}

(Nearest integer).
(Given :

R = 8.3 J K^{- 1} mol^{- 1}

and

ln 10 = 2.3

)

Answer: 6

The equilibrium constant is the ratio of the forward and reverse rate constants,

K = \frac{k _{f}}{k _{r}} = \frac{1 0 ^{3}}{1 0 ^{2}} = 10

. The standard free-energy change follows from

Δ_{r} G^{\circ} = - R T ln K = - (8.3) (300) ln 10 = - (8.3) (300) (2.3) J mo l^{- 1} = - 5.73 \times 1 0^{3} J mo l^{- 1}

. To the nearest integer this is a decrease of about

6 kJ mo l^{- 1}

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer