The value of $\log \mathrm{K}$ for the reaction $\mathrm{A} \leftrightharpoons \mathrm{B}$ at 298 K is ______ (Nearest integer) Given: $\Delta \mathrm{H}^{0}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\Delta \mathrm{S}^{\circ}=10 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ (Take $2.303 \times 8.314 \times 298=5705$)

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Chemical Thermodynamics

The value of

lo g K

for the reaction

A ⇋ B

at 298 K is ______ (Nearest integer)

Given:

Δ H^{0} = - 54.07 kJ mol^{- 1}

Δ S^{\circ} = 10 JK^{- 1} mol^{- 1}

(Take

2.303 \times 8.314 \times 298 = 5705

)

Answer: 10

The standard free energy change is

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = - 54070 - 298 \times 10 = - 57050 J mo l^{- 1}

. This is related to the equilibrium constant by

Δ G^{\circ} = - 2.303 R T lo g K

. Using the supplied value

2.303 \times 8.314 \times 298 = 5705

- 57050 = - 5705 lo g K

, so

lo g K = \frac{57050}{5705} = 10

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Integer answer