For the reaction, X(s)⇌Y(s)+Z(g), the plot of lnpzp0 versus 104T is given below (in solid line), where pz is the pressure (in bar ) of the gas Z at temperature T and p0=1 bar.(Given, d(ln K)d(1T)=−ΔH0R, where the equilibrium constant, K=pzp0 and the gas constant. R=8.314 JK−1 mol−1The value of standard enthalpy, ΔH0,(in kJ mol−1) for the given reaction is

For the reaction, X(s)⇌Y(s)+Z(g), the plot of lnpzp0 versus 104T is

1.	For the reaction, X(s)⇌Y(s)+Z(g), the plot of lnpzp0 versus 104T is given below (in solid line), where pz is the pressure (in bar ) of the gas Z at temperature T and p0=1 bar.(Given, d(ln K)d(1T)=−ΔH0R, where the equilibrium constant, K=pzp0 and the gas constant. R=8.314 JK−1 mol−1The value of standard enthalpy, ΔH0,(in kJ mol−1) for the given reaction is
Answer» For the reaction, $X (s) ⇌ Y (s) + Z (g)$ , the plot of $l n \frac{p_{z}}{p^{0}} versus \frac{10^{4}}{T}$ is given below (in solid line), where $p_{z}$ is the pressure (in $b a r$ ) of the gas $Z$ at temperature $T$ and $p^{0} = 1 b a r$ . (Given, $\frac{d (l n K)}{d (\frac{1}{T})} = - \frac{Δ H^{0}}{R}$ , where the equilibrium constant, $K = \frac{p_{z}}{p^{0}}$ and the gas constant. $R = 8.314 J K^{- 1} m o l^{- 1}$ The value of standard enthalpy, $Δ H^{0}, (i n k J m o l^{- 1})$ for the given reaction is