At \(1000 \mathrm{~K}\), the partial pressures of \(\mathrm{CO}_2(g)\) and \(\mathrm{CO}(g)\) for the…
At \(1000 \mathrm{~K}\), the partial pressures of \(\mathrm{CO}_2(g)\) and \(\mathrm{CO}(g)\) for the reaction \(\mathrm{CO}_2(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\) in a closed vessel at equilibrium are 0.15 and 0.60 bar respectively. The \(K_c\) for this reaction at the same temperature is approximately
\(2.0 \times 10^{-4}\)
\(2.89 \times 10^{-2}\)
\(2.89 \times 10^{-3}\)
\(5.78 \times 10^{-3}\)
Solution
For the reaction,
\(\begin{aligned}
\mathrm{CO}_2(g)+\mathrm{C}(s) & \rightleftharpoons 2 \mathrm{CO}(g) \\
K_{\mathrm{sp}}= & \frac{\left[\mathrm{CO}^2(g)\right.}{\left[\mathrm{CO}_2\right](g)}=\frac{p\left[\mathrm{CO}^2(g)\right.}{p\left[\mathrm{CO}_2\right](g)} \\
& {\left[\therefore p V=n R T \text { or } \frac{n}{V}=\frac{p}{R T}, \frac{n}{V} \propto p(R, T)\right] }
\end{aligned}\)
Given,
\(\begin{aligned}
p[\mathrm{CO}] & =0.6 \text { bar } \\
p\left[\mathrm{CO}_2\right] & =0.15 \text { bar } \\
K_p & =\frac{[0.6]^2}{[0.15]}=2.4
\end{aligned}\)
Also, \(K_p=K_c \cdot(R T)^{\Delta n}\)
where, \(\Delta n=\) moles of gaseous products moles of gaseous reactants \(=2-1=1\)
\(\therefore \quad K_p=K_c \times R \times T\)
where,
\(\begin{aligned}
K_c & =\frac{K_p}{R T} \\
K_p & =2.4 \\
R & =0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}
\end{aligned}\)
or \(K_c=2.89 \times 10^{-2}\)
Hence, option (b) is the correct answer.