The equilibrium constant for $H_2 (g) + CO_2 (g) \leftrightarrow H_2O(g) + CO(g)$ is 1.80 at 1000^oC. If 1 mole of H₂ and 1 mole of CO₂ are placed in one-liter flask, the final equilibrium concentration of CO at 1000*C will be

A. 0.573 M

B. 0.385 M

C. 5.73 M

D. 0.295 M

Answers (1)

$\\\text{The given reaction is }:- \qquad \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ \text { Initial moles: } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \1 \ \ \ \ \ \ \ \ \quad 1 \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0 \\ \text { At eqm: } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \1-\mathrm{x} \ \ \ \ \quad 1-\mathrm{x} \quad \ \ \ \ \ \mathrm{x} \ \ \ \ \ \ \ \ \quad \mathrm{x}$

$\begin{aligned} \mathrm{Now}, \mathrm{K}_{\mathrm{C}} &=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right][\mathrm{CO}]}{\left[\mathrm{H}_{2}\right]\left[\mathrm{CO}_{2}\right]} \\ \mathrm{K}_{\mathrm{C}} &=\frac{\mathrm{x} \cdot \mathrm{x}}{(1-\mathrm{x})(1-\mathrm{x})}=1.8 \\ \Rightarrow \quad & \mathrm{x}^{2}=1.8(1-\mathrm{x})(1-\mathrm{x}) \end{aligned}$

$\\\Rightarrow \quad x=1.34(1-x) & \Rightarrow 2.34 x=1.34 \\ & \Rightarrow x=0.573 \\ \text{So, concentration of }\mathrm{CO}$ at $\mathrm{eq} \mathrm{m}=0.573 \mathrm{M}$$

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The equilibrium constant for is 1.80 at 1000o C. If 1 mole of H2 and 1 mole of CO2 are placed in one-liter flask, the final equilibrium concentration of CO at 1000*C will be A. 0.573 M B. 0.385 M C. 5.73 M D. 0.295 M

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The equilibrium constant for $H_2 (g) + CO_2 (g) \leftrightarrow H_2O(g) + CO(g)$ is 1.80 at 1000^oC. If 1 mole of H₂ and 1 mole of CO₂ are placed in one-liter flask, the final equilibrium concentration of CO at 1000*C will be

A. 0.573 M

B. 0.385 M

C. 5.73 M

D. 0.295 M