Question 15.9: At 448 °C, the equilibrium constant Kc for the reaction H2(g...
At 448 °C, the equilibrium constant K_{c} for the reaction
H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 × 10^{-2} mol of HI, 1.0 × 10^{-2} mol of H_{2}, and 3.0 × 10^{-2} mol of I_{2} in a 2.00 L container.
Analyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium.
Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium expression to calculate the reaction quotient, Q_{c}. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed.
Solve
The initial concentrations are
[HI] = 2.0 × 10^{-2} mol/2.00 L = 1.0 × 10^{-2} M
[H_{2}] = 1.0 × 10^{-2} mol/2.00 L = 5.0 × 10^{-3} M
[I_{2}] = 3.0 × 10^{-2} mol/2.00 L = 1.5 × 10^{-2} M
The reaction quotient is therefore
Q_{c} =\frac{[HI]²}{[H_{2}][I_{2}]}=\frac{(1.0 × 10^{-2})²}{(5.0 × 10^{-3})(1.5 × 10^{-2})}= 1.3Because Q_{c} < K_{c}, the concentration of HI must increase and the concentrations of H_{2} and I_{2} must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium.