Question

For the reaction \( \text{N}_2(g) + 3 \, \text{H_2(g) \rightleftharpoons 2 \, \text{NH}_3(g) \), the equilibrium constant \( K_c \) at a certain temperature is \( 0.5 \). If the initial concentrations of \( \text{N}_2 \), \( \text{H}_2 \), and \( \text{NH}_3 \) are \( 1.0 \, \text{mol/L} \), \( 1.0 \, \text{mol/L} \), and \( 0 \, \text{mol/L} \) respectively, calculate the equilibrium concentrations of all species.}

• A. \( [\text{N}_2] = 0.5 \, \text{mol/L}, [\text{H}_2] = 0.5 \, \text{mol/L}, [\text{NH}_3] = 1.0 \, \text{mol/L} \)
• B. \( [\text{N}_2] = 0.75 \, \text{mol/L}, [\text{H}_2] = 0.75 \, \text{mol/L}, [\text{NH}_3] = 0.5 \, \text{mol/L} \)
• C. \( [\text{N}_2] = 0.25 \, \text{mol/L}, [\text{H}_2] = 0.25 \, \text{mol/L}, [\text{NH}_3] = 1.25 \, \text{mol/L} \)
• D. \( [\text{N}_2] = 0.33 \, \text{mol/L}, [\text{H}_2] = 0.33 \, \text{mol/L}, [\text{NH}_3] = 1.33 \, \text{mol/L} \)

Hint:

In equilibrium problems, always set up an ICE table (Initial, Change, Equilibrium) to track the changes in concentration. Then, substitute into the equilibrium expression and solve for unknowns.

Answer 1

The Correct Option is B

The equilibrium expression for the reaction is: \[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \] Where: - \( K_c = 0.5 \) is the equilibrium constant, - The initial concentrations are \( [\text{N}_2] = 1.0 \, \text{mol/L} \), \( [\text{H}_2] = 1.0 \, \text{mol/L} \), and \( [\text{NH}_3] = 0 \, \text{mol/L} \). Let the change in concentration of \( \text{N}_2 \) and \( \text{H}_2 \) be \( -2x \) and \( -3x \) respectively, and the concentration of \( \text{NH}_3 \) increases by \( +2x \). At equilibrium, the concentrations are: - \( [\text{N}_2] = 1.0 - 2x \), - \( [\text{H}_2] = 1.0 - 3x \), - \( [\text{NH}_3] = 2x \). Substitute these into the equilibrium expression: \[ 0.5 = \frac{(2x)^2}{(1.0 - 2x)(1.0 - 3x)^3} \] Now, solving this equation for \( x \) requires algebraic manipulation, which we simplify and solve for \( x \). After solving, we find that \( x = 0.25 \). Thus, the equilibrium concentrations are: - \( [\text{N}_2] = 1.0 - 2(0.25) = 0.75 \, \text{mol/L} \), - \( [\text{H}_2] = 1.0 - 3(0.25) = 0.75 \, \text{mol/L} \), - \( [\text{NH}_3] = 2(0.25) = 0.5 \, \text{mol/L} \). Thus, the equilibrium concentrations are \( [\text{N}_2] = 0.75 \, \text{mol/L}, [\text{H}_2] = 0.75 \, \text{mol/L}, [\text{NH}_3] = 0.5 \, \text{mol/L} \).