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, $R=8.314 \,J \,K ^{-1} mol ^{-1}$ )
The value of $\Delta S^{0}$ (in $J\, K ^{-1} mol ^{-1}$ ) for the given reaction, at $1000\, K$ is ______.

Answer 1

$ 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)

$(1) \Rightarrow \ln \left(\frac{p_{z}}{p^{0}}\right)=-\left(\frac{\Delta H^{0}}{10^{4} R}\right) \times \frac{10^{4}}{T}+\frac{\Delta S^{0}}{T} \ldots(2)$
Putting the value of $\Delta H ^{0}$ in equation (2), we get
$-3=-\left(\frac{2 R \times 10^{4}}{10^{4} R }\right) \times \frac{10^{4}}{7}+\frac{\Delta S ^{0}}{ R } $
$-3=-2 R \times \frac{10^{4}}{ T }+\frac{\Delta S ^{0}}{ R }$
$-3=-2 \times \frac{10^{4}}{1000}+\frac{\Delta S ^{0}}{ R } $
$-3=-20+\frac{\Delta S ^{0}}{ R }$
$\therefore \frac{\Delta S ^{0}}{ R }=17$
$\therefore \Delta S ^{0}=17 \times 8.314$
$=141.34 \,J\,K ^{-1} mol ^{-1}$