For a reaction, $A ⇌ P,$ the plots of [A] and [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 $Δ H^{0}$ and $Δ S^{0}$ are independent of temperature and ratio of lnK at $T_{1}$ to lnK 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.)

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$Δ G^{θ} < 0, Δ S^{θ} > 0$

$Δ H^{θ} < 0, Δ S^{θ} < 0$

$Δ G^{θ} < 0, Δ H^{θ} > 0$

$Δ G^{θ} < 0, Δ S^{θ} < 0$

answer is A, C.

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Detailed Solution

$\begin{array}{ll} A ⇌ P given T_{2} > T_{1} \\ \frac{\ln K_{1}}{\ln K_{2}} > \frac{T_{2}}{T_{1}} \Rightarrow T_{1} \ln k_{1} > T_{2} \ln k_{2} \Rightarrow - Δ G_{1}^{0} > - Δ G_{2}^{0} \\ \Rightarrow (- Δ H^{\circ} + T_{1} Δ S^{\circ}) > (- Δ H^{\circ} + T_{2} Δ S^{\circ}) \Rightarrow T_{1} Δ S^{\circ} > T_{2} Δ S^{\circ} \Rightarrow Δ S^{\circ} < 0 \end{array}$