Question 17.6: Calculating Kc from Concentration Data Problem In order to s......
Calculating K_c from Concentration Data
Problem In order to study hydrogen halide decomposition, a researcher fills an evacuated 2.00-L flask with 0.200 mol of HI gas and allows the reaction to proceed at 453°C:
2HI(g) \xrightleftharpoons[]{} H_2(g) + I_2(g)At equilibrium, [HI] = 0.078 M. Calculate K_c.
Plan To calculate K_c, we need the equilibrium concentrations. We can find the initial [HI] from the amount (0.200 mol) and the flask volume (2.00 L), and we are given [HI] at equilibrium (0.078 M). From the balanced equation, when 2x mol of HI reacts, x mol of H_2 and x mol of I_2 form. We set up a reaction table, use the known [HI] at equilibrium to solve for x (the change in [H_2] or [I_2]), and substitute the concentrations into Q_c.
Solution Calculating initial [HI]:
[HI] = \frac{0.200 \text{ mol}}{2.00 L} = 0.100 MSetting up the reaction table (Table 1), with x = [H_2] or [I_2] that forms and 2x = [HI] that reacts:
Solving for x, using the known [HI] at equilibrium:
[HI] = 0.100 M − 2x = 0.078 M \text{ so } x = 0.011 M
Therefore, the equilibrium concentrations are
[H2] = [I_2] = 0.011 M\text{ and we were given }[HI] = 0.078 M
Substituting into the reaction quotient:
Q_c = \frac{[H_2][I_2]}{[HI]^2}
Thus, K_c = \frac{(0.011)(0.011)}{0.078^2 } = 0.020
Check Rounding gives ∼0.01²/0.08² = 0.02. Because the initial [HI] of 0.100 M fell slightly at equilibrium to 0.078 M, relatively little product formed; so we expect K_c < 1.
Table 1
| Concentration (M) | \mathbf{2HI(g) \xrightleftharpoons[]{} H_2(g) + I_2(g)} |
| Initial | 0.100 0 0 |
| Change | -2x +x +x |
| Equilibrium | 0.100 – 2x x x |