For the reaction, $\mathbf{X}(s) ightleftharpoons \mathbf{Y}(s)+\mathbf{Z}(g)$, the plot of $\ln \frac{p_{\mathbf{Z}}}{p^{ heta}}$ versus $\frac{10^{4}}{T}$ is given below (in solid line), where $p_{\mathbf{Z}}$ is the pressure (in bar) of the gas $\mathbf{Z}$ at temperature $T$ and…

Question

For the reaction, $\mathbf{X}(s) ightleftharpoons \mathbf{Y}(s)+\mathbf{Z}(g)$, the plot of $\ln \frac{p_{\mathbf{Z}}}{p^{	heta}}$ versus $\frac{10^{4}}{T}$ is given below (in solid line), where $p_{\mathbf{Z}}$ is the pressure (in bar) of the gas $\mathbf{Z}$ at temperature $T$ and $p^{	heta}=1$ bar. (Given, $\frac{\mathrm{d}(\ln K)}{\mathrm{d}\left(\frac{1}{T}ight)}=-\frac{\Delta H^{	heta}}{R}$, where the equilibrium constant, $K=\frac{p_{z}}{p^{	heta}}$ and the gas constant, $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ) The value of standard enthalpy, &#8710; H o (in kJ &#160; mol - 1 ) for the given reaction is ___.

Aicharya · Accepted Answer

Slop g the Given graph d &#160; ln p z p &#981; d &#160; 10 4 T = &#8722; 7 + 3 12 &#8722; 10 = &#8722; 2 d &#160; ln p z p &#981; d 1 T &#8722; 2 &#215; 10 4 = &#8722; &#916;H o R &#916;H o = 2 &#215; 10 4 &#215; R = 2 &#215; 8 .314 &#215; 10 &#160; kJ / mol = 166 . 28

For the reaction, $\mathbf{X}(s) \rightleftharpoons \mathbf{Y}(s)+\mathbf{Z}(g)$, the plot of $\ln…

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