Question

For the reaction, $X\left(s\right)\rightleftharpoons Y\left(s\right)+Z\left(g\right)$, the plot of $ln\frac{p_z}{p^0}\text{versus}\frac{{10}^4}{T}$ is given below (in solid line), where $p_z$ is the pressure (in $bar$) of the gas $Z$ at temperature $T$ and $p^0=1bar$.
(Given, $\frac{d\left(lnK\right)}{d\left(\frac{1}{T}\right)}=-\frac{\Delta H^0}{R}$, 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, $\Delta H^0,\left(inkJmo{l}^{-1}\right)$ for the given reaction is

Answer 1

$X\left(s\right)\rightleftharpoons Y\left(s\right)+Z\left(g\right)K_p=\frac{p_z}{p^0},\text{also}\Delta G^0=-RTln{k}_p\therefore \Delta G^0=-RTln\left(\frac{p_z}{p^0}\right)\text{Now,}\Delta G^0=\Delta H^0-T\Delta S\Rightarrow -RTln\left(\frac{p_z}{p^0}\right)=\Delta H^0-T\Delta S^0\Rightarrow ln\left(\frac{p_z}{p^0}\right)=-\left(\frac{\Delta H^0}{R}\right)\frac{1}{T}+\frac{\Delta S^0}{R}...\left(1\right)$
$\left(1\right)\Rightarrow ln\left(\frac{p_z}{p^0}\right)=-\left(\frac{\Delta H^0}{{10}^{4}R}\right)\times \frac{{10}^4}{T}+\frac{\Delta S^0}{T}....\left(2\right)\text{Slope of the line}=-\frac{\Delta H^0}{{10}^{4}R}=\frac{[-7-(-3\left)\right]}{12-10}=-2\therefore \Delta H^0=2R\times {10}^4=2\times 8.314\times {10}^{-3}\times {10}^4=166.28kJmo{l}^{-1}{K}^{-1}$