$400\, K$ $\Delta G ^{\circ}=+25.2\, kJ mol ^{-1}$. એ $2 A ( g ) \rightleftharpoons A _{2}( g )$
આ પ્રકિયા માટે સંતુલન અચળાંક $K _{ C }$$...... \times 10^{-2}$
$\left.\log _{10} 2=0.30,1\, atm =1\, bar \right]$
$[$ antilog $(-0.3)=0.501]$
$\Delta_{ r } G ^{0}=- RT \ln K _{ p }$
$25200=-2.3 \times 8.3 \times 400 \log \left( K _{ p }\right)$
$K _{ p }=10^{-3.3}=10^{-3} \times 0.501$
$=5.01 \times 10^{-4} \,Bar ^{-1}$
$=5.01 \times 10^{-9}\, Pa ^{-1}$
$=\frac{ K _{ C }}{8.3 \times 400}$
$K _{ C }=1.66 \times 10^{-5}\, m ^{3} / mole$
$=1.66 \times 10^{-2}\, L / mol$
Ans $=2$