Question

For the reaction, $X(s)\rightleftharpoons Y(s)+Z(g)$, the plot of $\ln \frac{p_Z}{p^0}$ versus $\frac{10^4}{T}$ is given below (in solid line), where $p_z$ is the pressure (in bar) of the gas $Z$ at temperature $T$ and $p^0=1$ bar.

(Given, $\frac{d(\ln K)}{d\left(\frac{1}{T}\right)}=-\frac{\Delta H^0}{R}$, where the equilibrium constant, $K=\frac{p_z}{p^0}$ and the gas constant, $R=8.314J{K}^{-1}mo{l}^{-1}$ )
The value of standard enthalpy, $\Delta H^0$ (in $kJmo{l}^{-1}$ ) for the given reaction is ______.

Answer 1

Answer: 166.28

$X(s)\rightleftharpoons Y(s)+Z(g)$
$K_p=\frac{p_z}{p^0},$ also$\Delta G^0=-RT\ln k_p$
$=-RT\ln \left(\frac{p_z}{p^0}\right)$
Now, $\Delta G^0=\Delta H^0-T\Delta S^0$
$-RT$ kn $\left(\frac{p_z}{p^0}\right)=\Delta H^0-T\Delta S^0$
$\ln \left(\frac{p_z}{p^0}\right)=-\left(\frac{\Delta H^0}{R}\right)\frac{1}{T}+\frac{\Delta S^0}{R}$ ....(1)

$(1)\Rightarrow \ln \left(\frac{p_z}{p^0}\right)=-\left(\frac{\Delta H^0}{10^{4}R}\right)\times \frac{10^4}{T}+\frac{\Delta S^0}{T}\dots (2)$
Slope of the line $=-\frac{\Delta H^0}{10^{4}R}$
$=\frac{[-7-(-3)]}{12-10}=-2$
$\therefore \Delta H^0=2R\times 10^4$
$=2\times 8.314\times 10^{-3}\times 10^4$
$=\mathrm{1.66.28}kJmo{l}^{-1}{K}^{-1}$