Questions: A mixture containing 0.500 mol of carbon monoxide and 0.400 mol of bromine were placed in a rigid 1.00 L container and the system was allowed to come to equilibrium. The equilibrium concentration of COBr2 was found to be 0.233 M. What is the value of Kc for this reaction? CO(g) + Br2(g) ↔ COBr2(g)

Solution Steps

The initial moles of CO and Br\(_2\) are given as 0.500 mol and 0.400 mol, respectively, in a 1.00 L container. Therefore, the initial concentrations are: \[ [\text{CO}]_0 = \frac{0.500 \, \text{mol}}{1.00 \, \text{L}} = 0.500 \, \text{M} \] \[ [\text{Br}_2]_0 = \frac{0.400 \, \text{mol}}{1.00 \, \text{L}} = 0.400 \, \text{M} \]

At equilibrium, the concentration of COBr\(_2\) is given as 0.233 M. Since the reaction produces COBr\(_2\), the change in concentration for COBr\(_2\) is +0.233 M. The stoichiometry of the reaction indicates that the change in concentration for CO and Br\(_2\) is -0.233 M each.

Using the initial concentrations and the changes, calculate the equilibrium concentrations: \[ [\text{CO}] = 0.500 - 0.233 = 0.267 \, \text{M} \] \[ [\text{Br}_2] = 0.400 - 0.233 = 0.167 \, \text{M} \] \[ [\text{COBr}_2] = 0.233 \, \text{M} \]

The equilibrium constant \(K_c\) is given by the expression: \[ K_c = \frac{[\text{COBr}_2]}{[\text{CO}][\text{Br}_2]} \] Substitute the equilibrium concentrations: \[ K_c = \frac{0.233}{0.267 \times 0.167} \] \[ K_c = \frac{0.233}{0.044589} \approx 5.23 \]

Final Answer