(ઉપયોગ $R =8.31\, J\, K ^{-1}\, mol ^{-1} ; \log 2=0.3010$. In $10=$ $2.3, \log 3=0.477$ )
$t _{ i } \,\quad\quad 1$
$\text { teq } \quad 1-0.4 \quad\quad \frac{0.4}{2} \quad\quad\quad \frac{0.4}{2}$
$K _{ p }= \frac{(0.2)^{\frac{1}{2}}(0.2)^{\frac{1}{2}}}{1-0.4}=\frac{0.2}{0.6}=\frac{1}{3}$
$\Delta G =\Delta G ^{\circ}+ RT \ln K =0$
$\Delta G ^{\circ} =- RT \ln K \Rightarrow-8.31 \times 300 \times 2.3 \times \log \left(\frac{1}{3}\right)$
$=2735\, J / mol$
$II: C + D $ $\rightleftharpoons$ $ 3A ; K_{eq}= K_2, $
$III: 6B + D $ $\rightleftharpoons$ $2C; K_{eq} = K_3$ જેથી,
$\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29} \text {. }$
(આપેલ : R = $\left.8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$