For a reaction A $\underset{}{⇌}$ P, the plots of [A] and [P] with time at temperature T₁and T₂ are given below.
If T₂ $>$ T₁, the correct statement(s) is(are)
(Assume ${ΔH}^{0} and {ΔS}^{0}$ are independent of temperature and ratio of lnK at T₁ to lnK at T₂ 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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${ΔG}^{0} < 0, {ΔH}^{0} > 0$

${ΔG}^{0} < 0, {ΔS}^{0} < 0$

${ΔH}^{0} < 0, {ΔS}^{0} < 0$

${ΔG}^{0} < 0, {ΔS}^{0} > 0$

answer is A, C.

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Detailed Solution

$\frac{\ln K_{T_{1}}}{\ln K_{T_{2}}} > 1 \Rightarrow \ln K_{T_{2}} so, K_{T_{1}} > K_{T_{2}}$
$Also, \frac{\ln K_{T_{1}}}{\ln_{T_{2}}} > \frac{T_{2}}{T_{1}}$
or $T_{1} \ln K_{T_{1}} > T_{2} \ln K_{T_{2}} \Rightarrow - {RT}_{1} \ln K_{T_{2}} \ln K_{T_{2}}$
$or {ΔG}_{T_{1}}^{0} < {ΔG}_{T_{2}}^{0} (∴ ΔG = - RTln K)$
$or {ΔH}^{0} - T_{1} {ΔS}^{0} < {ΔH}^{0} - T_{2} {ΔS}^{0}$
$(Also {ΔG}^{0} = {ΔH}^{0} - {TΔS}^{0},$ Gibbs Helmholtz equation)