Question

For the reaction;
$A\left(s\right)\to B\left(g\right) + C\left(g\right)$
$\Delta_{r}G^{\circ} =\left(1000 -50 T ln T\right)$
$\Delta_{r}G^{\circ}_{p}$ of the above reaction in kJ/mol is $10 X$. Find the value of $X$

Answer 1

$\Delta_{r}G^{\circ} = 1000 - 50 T\, ln\, T$
$\frac{d\left(\Delta_{r} G^{\circ}\right)}{dT} = - \Delta_{r}S^{\circ} = -50\, ln \,T - 50$
$\therefore \Delta_{r}H^{\circ} = \Delta_{r}G^{\circ} + T\Delta_{r}S^{\circ}$
or $\Delta_{r}H^{\circ} = \Delta_{r}G^{\circ} + T\left(+50 ln T + 50\right)$
or $\Delta_{r}H^{\circ} = 1000 - 50 T\, ln \,T +50 T\, ln \,T + 50 T$
or $\Delta_{r}H^{\circ} = 1000 + 50 T$
or $\Delta_{r}C^{\circ}_{p} =\frac{d}{dT}\left(\Delta_{r}H^{\circ}\right)$
$ = 50 kJ/mol = 10X$
$\Rightarrow X=5 $