Question

At a certain temperature, 2 moles of CO and 4 moles of Cl$_2$ gases were reacted to form COCl$_2$ in a 10 L vessel. At equilibrium if one mole of CO is present then equilibrium constant for the reaction is

• A. 4
• B. 3.3
• C. 1
• D. 2.5
• E. 4.5

Hint:

Always remember to divide the number of moles by the volume of the vessel to get the concentration before calculating the equilibrium constant $K_c$.

Answer 1

The Correct Option is B

Concept: The equilibrium constant ($K_c$) is calculated using the molar concentrations of reactants and products at equilibrium. For the reaction: $\text{CO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{COCl}_2(g)$.

Step 1: {Determine equilibrium moles using an ICE table.} Reaction: $\text{CO} + \text{Cl}_2 \rightleftharpoons \text{COCl}_2$
• Initial moles: $\text{CO} = 2$, $\text{Cl}_2 = 4$, $\text{COCl}_2 = 0$
• Equilibrium moles of $\text{CO}$ is 1. Therefore, change ($x$) = $2 - 1 = 1\text{ mole}$.
• Equilibrium $\text{Cl}_2 = 4 - x = 4 - 1 = 3\text{ moles}$.
• Equilibrium $\text{COCl}_2 = 0 + x = 1\text{ mole}$.

Step 2: {Calculate molar concentrations (Volume = 10 L).} $$[\text{CO}] = \frac{1\text{ mol}}{10\text{ L}} = 0.1\text{ M}$$ $$[\text{Cl}_2] = \frac{3\text{ mol}}{10\text{ L}} = 0.3\text{ M}$$ $$[\text{COCl}_2] = \frac{1\text{ mol}}{10\text{ L}} = 0.1\text{ M}$$

Step 3: {Solve for $K_c$.} $$K_c = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]}$$ $$K_c = \frac{0.1}{0.1 \times 0.3}$$ $$K_c = \frac{1}{0.3} = 3.33 \dots$$