JEE Advanced · Chemistry · 6. Thermodynamics (C)

For a reaction, $A ⇌ P$ , the plots of $[A] a n d [P]$ with time at temperatures $T_{1} a n d T_{2}$ are given below.

If $T_{2} > T_{1}$ , the correct statement(s) is (are)
(Assume $∆ H^{⊝} a n d ∆ S^{⊝}$ are independent of temperature and ratio of $\ln K a t T_{1} t o \ln K a t T_{2}$ is greater than $T_{2} / T_{1}$ . Here H, S , G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

A $∆ H^{⊝} < 0, ∆ S^{⊝} < 0$
B $∆ G^{⊝} < 0, ∆ H^{⊝} < 0$
C $∆ G^{⊝} < 0, ∆ S^{⊝} < 0$
D $∆ G^{⊝} > 0, ∆ S^{⊝} > 0$

Verified Solution

Answer & Solution

Correct Answer

(C) $∆ G^{⊝} < 0, ∆ S^{⊝} < 0$

Step-by-step Solution

Detailed explanation

$A \rightleftharpoons P$
Given $T_2>T_1$
$\frac{\ln K_1}{\ln K_2}>\frac{T_2}{T_1}$
$\Rightarrow T_1 \ln k_1>T_2 \ln k_2 $
$ \Rightarrow -\Delta G_1^0>-\Delta G_2^0 $
$ \Rightarrow \left(-\Delta H^0+T_1 \Delta S^0\right)>\left(-\Delta H^0+T_2 \Delta S^o\right) $
$ \Rightarrow T_1 \Delta S^0>T_2 \Delta S^0 $
$ \Rightarrow \Delta S^o<0$
At $T_2$ temperature concentration of product is decreases that shows reaction is exothermic.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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