Suppose that the preceding reaction is carried out at a different temperature and that the initial concentrations of the reactants are \( [\mathrm{CO}]=0.500 \mathrm{M} \) and \( \left[\mathrm{H}_{2}\right]=1.00 \mathrm{M} \). Assuming there is no product at the beginning of the reaction, at equilibrium find the equilibrium \( [\mathrm{CO}]=0.15 \mathrm{M} \) constant at this new temperature. (Hint: Use the stoichiometric relationships from the balanced equation to find the equilibrium concentrations of \( \mathrm{H}_{2} \) and \( \mathrm{CH}_{3} \mathrm{OH} \).) Express your answer numerically.

To find the equilibrium concentrations of \( \mathrm{H}_2 \) and \( \mathrm{CH}_3\mathrm{OH} \) given the equilibrium concentration of \( \mathrm{CO} \), we start with the balanced chemical equation for the reaction: \[ \mathrm{CO} + 2\mathrm{H}_2 \rightleftharpoons \mathrm{CH}_3\mathrm{OH} \] ### Initial Concentrations: - \( [\mathrm{CO}]_0 = 0.500 \, \mathrm{M} \) - \( [\mathrm{H}_2]_0 = 1.00 \, \mathrm{M} \) - \( [\mathrm{CH}_3\mathrm{OH}]_0 = 0 \, \mathrm{M} \) ### Change in Concentrations: Let \( x \) be the change in concentration of \( \mathrm{CO} \) that reacts to reach equilibrium. According to the stoichiometry of the reaction: - For every 1 mole of \( \mathrm{CO} \) that reacts, 2 moles of \( \mathrm{H}_2 \) react and 1 mole of \( \mathrm{CH}_3\mathrm{OH} \) is produced. ### Equilibrium Concentrations: At equilibrium, we have: - \( [\mathrm{CO}] = [\mathrm{CO}]_0 - x = 0.500 - x \) - \( [\mathrm{H}_2] = [\mathrm{H}_2]_0 - 2x = 1.00 - 2x \) - \( [\mathrm{CH}_3\mathrm{OH}] = 0 + x = x \) ### Given Equilibrium Concentration: We are given that at equilibrium, \( [\mathrm{CO}] = 0.15 \, \mathrm{M} \). Therefore, we can set up the equation: \[ 0.500 - x = 0.15 \] Solving for \( x \): \[ x = 0.500 - 0.15 = 0.350 \] ### Now, we can find the equilibrium concentrations of \( \mathrm{H}_2 \) and \( \mathrm{CH}_3\mathrm{OH} \): 1. **Equilibrium concentration of \( \mathrm{H}_2 \)**: \[ [\mathrm{H}_2] = 1.00 - 2x = 1.00 - 2(0.350) = 1.00 - 0.700 = 0.300 \, \mathrm{M} \] 2. **Equilibrium concentration of \( \mathrm{CH}_3\mathrm{OH} \)**: \[ [\mathrm{CH}_3\mathrm{OH}] = x = 0.350 \, \mathrm{M} \] ### Summary of Equilibrium Concentrations: - \( [\mathrm{CO}] = 0.15 \, \mathrm{M} \) - \( [\mathrm{H}_2] = 0.300 \, \mathrm{M} \) - \( [\mathrm{CH}_3\mathrm{OH}] = 0.350 \, \mathrm{M} \) ### Equilibrium Constant \( K_c \): The equilibrium constant \( K_c \) for the reaction is given by: \[ K_c = \frac{[\mathrm{CH}_3\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_2]^2} \] Substituting the equilibrium concentrations: \[ K_c = \frac{0.350}{(0.15)(0.300)^2} \] Calculating \( K_c \): \[ K_c = \frac{0.350}{(0.15)(0.09)} = \frac{0.350}{0.0135} \approx 25.93 \] Thus, the equilibrium constant \( K_c \) at this new temperature is approximately **25.93**.