$400\, K$ $\Delta G ^{\circ}=+25.2\, kJ mol ^{-1}$. 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 \(=2\)
$\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29} \text {. }$
(આપેલ : R = $\left.8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
$NO(g) \rightarrow \frac{1}{2} N_2(g)+ \frac{1}{2} O_2(g)$ સમાન તાપમાને શું થશે? :