The gas phase reaction

Question:

The gas phase reaction $2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{A}_{2}(\mathrm{~g})$ at $400 \mathrm{~K}$ has $\Delta \mathrm{G}^{\circ}=+25.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The equilibrium constant $\mathrm{K}_{\mathrm{C}}$ for this reaction is off to the Nearest integer) [ Use : $R=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \ln 10=2.3$$\left.\log _{10} 2=0.30,1 \mathrm{~atm}=1 \mathrm{bar}\right]$ $[$ antilog $(-0.3)=0.501]$

Solution:

(166)

Using formula

$\Delta_{\mathrm{r}} \mathrm{G}^{0}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{p}}$

$25200=-2.3 \times 8.3 \times 400 \log \left(\mathrm{K}_{\mathrm{p}}\right)$

$\mathrm{K}_{\mathrm{p}}=10^{-3.3}=10^{-3} \times 0.501$

$=5.01 \times 10^{-4} \mathrm{Bar}^{-1}$

$=\frac{\mathrm{K}_{\mathrm{C}}}{8.3 \times 400}$

$\mathrm{K}_{\mathrm{C}}=166 \times 10^{-2} \mathrm{~m}^{3} / \mathrm{mole}$

Ans $=166$

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