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 standard enthalpy, $\Delta H^{0}$ (in $kJ\, mol ^{-1}$ ) for the given reaction 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)

$(1) \Rightarrow \ln \left(\frac{p_{z}}{p^{0}}\right)=-\left(\frac{\Delta H^{0}}{10^{4} R}\right) \times \frac{10^{4}}{T}+\frac{\Delta S^{0}}{T} \ldots(2)$
Slope of the line $=-\frac{\Delta H ^{0}}{10^{4} R }$
$=\frac{[-7-(-3)]}{12-10}=-2$
$\therefore \Delta H ^{0}=2 R \times 10^{4}$
$=2 \times 8.314 \times 10^{-3} \times 10^{4}$
$=1.66 .28\, kJ \,mol ^{-1} K ^{-1}$