Consider the reaction $\mathrm{X} \rightleftharpoons \mathrm{Y}$ at 300 K. If $\Delta \mathrm{H}^{\theta}$ and K are $28.40 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta \mathrm{S}^{\theta}$ for the reaction in $\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ is ______. (Nearest Integer). (Given: $\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, $\ln 10=2.3$, $\log 3=0.48$, $\log 2=0.30$)

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Chemical Thermodynamics

Consider the reaction

X ⇌ Y

at 300 K. If

Δ H^{θ}

and K are

28.40 kJ mol^{- 1}

and

1.8 \times 1 0^{- 7}

at the same temperature, then the magnitude of

Δ S^{θ}

for the reaction in

J K^{- 1} mol^{- 1}

is ______. (Nearest Integer).
(Given:

R = 8.3 JK^{- 1} mol^{- 1}

ln 10 = 2.3

lo g 3 = 0.48

lo g 2 = 0.30

)

Answer: 34

Combining

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}

with

Δ G^{\circ} = - R T ln K

gives

Δ S^{\circ} = \frac{Δ H ^{\circ}}{T} + R ln K .

With

ln K = 2.3 lo g K

and

lo g (1.8 \times 1 0^{- 7}) = lo g 1.8 - 7

, where

lo g 1.8 = lo g 2 + 2 lo g 3 - 1 = 0.30 + 0.96 - 1 = 0.26

ln K = 2.3 (0.26 - 7) = - 15.50.

Therefore

Δ S^{\circ} = \frac{28400}{300} + 8.3 (- 15.50) = 94.67 - 128.65 = - 33.98 J K^{- 1} mo l^{- 1} .

The magnitude of

Δ S^{\circ}

34 J K^{- 1} mo l^{- 1}

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Integer answer