(Solved): didnt understand these values corresponding 5 and 6 can you calculate again by writing thanks acc ...

(c) At \( 750^{\circ} \mathrm{C}(1023 \mathrm{~K}) \) for the reaction: \( 2 \mathrm{Zn}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{ZnO}(\mathrm{s}), K_{3}=\frac{1}{P_{0_{2}}} \) \[ \begin{aligned} \Delta G_{3}^{o} & =-118256 \mathrm{cals}=R T \ln p_{\mathrm{O}_{2}} \\ \log p_{\mathrm{O}_{2}} & =-25.3675 \end{aligned} \] Similarly for the reaction (4): \( \mathrm{ZnS}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g})=\mathrm{ZnO}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{~g}) \), we can express: \[ \log p_{\mathrm{O}_{2}}=-12.6507+\frac{2}{3} \log p_{\mathrm{SO}_{2}} \] On substituting the value of \( \log p_{\mathrm{O}_{2}}=-25.3675 \), from Eq. 5 in Eq. 6, we get: \[ \begin{aligned} \log p_{\mathrm{SO}_{2}} & =-19.0773, \\ \therefore p_{\mathrm{SO}_{2}} & =8.37 \times 10^{-20} \text { atm Ans. } \end{aligned} \]