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

Correct Answer: 141.34

The value of $\Delta S^{0}$ (in $J\, K ^{-1} mol ^{-1}$ ) for the given reaction, at $1000\, K$ is 141.34.

Answer 2

The equation that relates the standard Gibbs free energy change (\(\Delta G\degree\)), the enthalpy change (\(\Delta H\degree\)), and the entropy change (\(\Delta S\degree\)) is:
\(\Delta G\degree=\Delta H\degree-T\Delta S\degree.....(i)\)
where T is the temperature in kelvin and R is the gas constant.
The equation that relates the equilibrium constant (K) and the standard Gibbs free energy change is:

\(\Delta G\degree=-RT\ In(K).....(ii)\)

We can combine these two equations to eliminate ΔG° and solve for K.
Given that \(K=\frac{p^2}{p\degree}\) and P° = 1 bar

\(ln(\frac{P}{1}) = \frac{-\Delta H\degree}{RT} + \frac{\Delta S\degree}{R}.....(iii)\)

The slope of the graph in the image is equal to \(\frac{-\Delta H\degree}{R}\).

Slope = \(10^4(\frac{Y_2-Y_1}{X_2-X_1})=10^4(\frac{-7+3}{12-10})=-2\times10^4\)

We can use the gas constant value (R = 8.314 J/mol*K) to find ΔH°.
\(-2\times10^4=\frac{\Delta H\degree}{R}=\gt2\times10^4\times8.314\)
\(\Delta H\degree=166280Jmol^{-1}=166.28kJ\ mol{-1}\)

We can also use the graph to find the temperature (T) at which the pressure is 10^7. The temperature is 10^3 K.

We can substitute the values of ΔH° and T into equation (iii) to solve for ΔS°.

\(ln(\frac{P}{1}) = \frac{-\Delta H\degree}{RT} + \frac{\Delta S\degree}{R}\)

\(-3 = -\frac{2\times10^4\times R}{R\times 10^3} + \frac{\Delta S\degree}{R}\)

\(\frac{\Delta S\degree}{R}=20-3=17\)
\(\frac{\Delta S\degree}{R}=17\times R=17\times8.314=141.34JK^{-1}\ mol^{-1}\)

So, the correct answer is \(141.34JK^{-1}\ mol^{-1}\)