Q12 / 90 · KVPY 2010 (SB/SX) · Chemistry
From equations 1 and 2 ,
$$
\begin{array}{l}
\mathrm{CO}_{2} \rightleftharpoons \mathrm{CO}+1 / 2 \mathrm{O}_{2}\left[\mathrm{~K}_{\mathrm{cl}}=9.1 \times 10^{-13} \text { at } 1000^{\circ} \mathrm{C}\right](\text { eq. } 1) \\
\mathrm{H}_{2} \mathrm{O} \rightleftharpoons{\mathrm{H}_{2}+1 / 2} \mathrm{O}_{2}\left[\mathrm{~K}_{\mathrm{cl}}=7.1 \times 10^{-12} \text { at } 1000^{\circ} \mathrm{C}\right]
\end{array}
$$
the equilibrium constant for the reaction $\mathrm{CO}_{2}+\mathrm{H}_{2} \rightleftharpoons \mathrm{CO}+\mathrm{H}_{2} \mathrm{O}$ at the same temperature, is-
- A$0.78$
- B$2.0$
- C$16.2$
- Correct answerD$1.28$
Solution
(i) $\mathrm{CO}_{2} \rightleftharpoons \mathrm{CO}+\frac{1}{2} \mathrm{O}_{2} \quad \mathrm{~K}_{1}=9.1 \times 10^{-13}$
(ii) $\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \mathrm{~K}_{2}=7.1 \times 10^{-12}$
Object $\quad \mathrm{CO}_{2}+\mathrm{H}_{2} \rightleftharpoons \mathrm{CO}+\mathrm{H}_{2} \mathrm{O}$
equation (i) - (ii) $\therefore \mathrm{K}^{1}=\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=1.28$ $\mathrm{C}_{\mathrm{t}}=\mathrm{C}_{0} \mathrm{e}^{-\mathrm{kt}}$ $\therefore \mathrm{R}=\mathrm{R}_{0} \mathrm{e}^{-\mathrm{kt}}$
Hence, the correct answer is option: 'D'.