For the reaction, $\mathrm{NO}_2+\mathrm{CO} \rightleftharpoons \mathrm{NO}+\mathrm{CO}_2$ one mole of $\mathrm{NO}_2$ and 2 moles of $\mathrm{CO}$ were kept in a vessel. Calculate the equilibrium constant $K_p$, if at equilibrium $25 \%$ of initial amount CO is consumed.

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Aicharya · Accepted Answer

In this question, we have to assume that volume of the vessel is $1 \mathrm{~L}$. $\begin{array}{lcccc} & \mathrm{NO}_2(g)+\mathrm{CO}(g) & \rightleftharpoons & \mathrm{NO}(g)+\mathrm{CO}_2(g) \ t=0 & 1 & 2 & 0 & 0 \ \text {mole } t=t_{\text {eq }} & 1-0.5 & 2-0.5 & 0.5 & 0.5 \ & =0.5 & =1.5 & & \ \text {Molar conc. }\left(\mathrm{mol} \mathrm{L}^{-1}\right) & \frac{0.5}{1} & \frac{0.5}{1} & \frac{0.5}{1} & \frac{0.5}{1} \ & =0.5 & =1.5 & 0.5 & 0.5\end{array}$ Given, $\alpha=2 \times \frac{25}{100}=0.5$ $\Rightarrow K_C=\frac{[\mathrm{NO}]\left[\mathrm{CO}_2\right]}{\left[\mathrm{NO}_2\right][\mathrm{CO}]}=\frac{0.5 \times 0.35}{0.5 \times 1.5}=\frac{1}{3}=K_p$ Because, $\Delta n_g=(\mathrm{l}+\mathrm{l})-(\mathrm{l}+\mathrm{l})=0$ $\begin{array}{ll} \Rightarrow & K_p=K_C(R T)^{\Delta n_s}=K_C(R T)^0=K_C \ \Rightarrow & K_p=\frac{1}{3} \end{array}$

For the reaction, \(\mathrm{NO}_2+\mathrm{CO} \rightleftharpoons \mathrm{NO}+\mathrm{CO}_2\) one mole of…

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