# 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_______ $\times 10^{-2}$. (Round off to the Nearest integer)

$\left[\right.$ 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:

Using formula

$\Delta_{r} G^{0}=-R T \ln K_{p}$

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

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

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

$=5.01 \times 10^{-9} \mathrm{~Pa}^{-1}$

$=\frac{\mathrm{K}_{\mathrm{C}}^{8}}{8.3 \times 400}$

$\mathrm{K}_{\mathrm{C}}=1.66 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{mole}$

$=1.66 \times 10^{-2} \mathrm{~L} / \mathrm{mol}$

Ans $=2$