Question

For a reaction $A\rightleftharpoons P$, the plots of [A] and [P] with time temperatrues $T_1$ 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 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.)

• A. $\Delta H^{\theta }<0,\Delta S^{\theta }<0$
• B. $\Delta G^{\theta }<0,\Delta H^{\theta }>0$
• C. $\Delta G^{\theta }<0,\Delta S^{\theta }<0$
• D. $\Delta G^{\theta }<0,\Delta S^{\theta }>0$

Answer 1

Answer: (A), (C)

The correct options are
A $\Delta H^{\theta }<0,\Delta S^{\theta }<0$
C $\Delta G^{\theta }<0,\Delta S^{\theta }<0$

$A\rightleftharpoons P$
$(P{)}_{eq}>5,(A{)}_{eq}<5$
$K_{eq}=\frac{\left[P\right]}{\left[A\right]}>1$
$\Delta G^0=-RTln{K}_{eq},\Delta G^0<0$
$\frac{ln{K}_{T_1}}{ln{K}_{T_2}}>\frac{T_2}{T_1}>1$
$\Rightarrow \frac{K_{T_1}}{K_{T_2}}>1$
$\Rightarrow K_{T_2}<K_{T_1}$ (exothermic)
$\Delta H^0<0$, since (P) at $T_2$ < at $T_1$
$\Delta G^0=\Delta H^0-T\Delta S^0$
$T\Delta S^0=\Delta H^0-\Delta G^0$
$\Delta S^0=\frac{\Delta H^0-\Delta G^0}{T};\left(\Delta H^0\right)>\left(\Delta G^0\right)$
$\Delta S^0<0$
Also, $-T_{1}ln{K}_{T_1}<-T_{2}ln{K}_{T_2}$
$\Delta G_{T_1}^0<\Delta G_{T_2}^0$
$\Delta H_{T_1}^0-T\Delta S_{T_1}^0<\Delta H_{T_2}^0-T\Delta S_{T_2}^0$
It is possible only if $\Delta S^0<0.$