For a reaction, $\mathrm{A} \rightleftharpoons \mathrm{P}$, the plots of $[\mathrm{A}]$ and $[\mathrm{P}]$ with time at temperatures $T_{1}$ and $T_{2}$ are given below. If $T_{2}>T_{1}$, the correct statement(s) is (are) (Assume $\Delta H^{\theta}$ and $\Delta S^{\theta}$ are independent of temperature and ratio of $\ln K$ at $T_{1}$ to $\ln K$ at $T_{2}$ is greater than $\frac{T_{2}}{T_{1}}$. Here $H$, $S$, $G$ and $K$ are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

(A) $\Delta H^{\theta}<0, \Delta S^{\theta}<0$; (C) $\Delta G^{\theta}<0, \Delta S^{\theta}<0$

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Chemical Thermodynamics

For a reaction,

A ⇌ P

, the plots of

[A]

and

[P]

with time at temperatures

T_{1}

and

T_{2}

are given below.

If

T_{2} > T_{1}

, the correct statement(s) is (are)
(Assume

Δ H^{θ}

and

Δ S^{θ}

are independent of temperature and ratio of

ln K

T_{1}

ln K

T_{2}

is greater than

\frac{T _{2}}{T _{1}}

. Here

H

S

G

and

K

are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

Answer: (A), (C)

From the plots,

[P]_{eq} > 5

and

[A]_{eq} < 5

, so

K_{eq} = [P] / [A] > 1

, giving

Δ G^{\circ} = - R T ln K_{eq} < 0

. Since

\frac{l n K _{T_{1}}}{l n K _{T_{2}}} > \frac{T _{2}}{T _{1}} > 1

, we have

K_{T_{2}} < K_{T_{1}}

, indicating the reaction is exothermic (

Δ H^{\circ} < 0

). From

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

and the fact that

Δ G^{\circ} < 0

while

Δ H^{\circ} < 0

, for the inequality to hold with increasing temperature reducing

K

, we need

Δ S^{\circ} < 0

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · One or more correct options