Question 17.10: Predicting Reaction Direction and Calculating Equilibrium Co......
Predicting Reaction Direction and Calculating Equilibrium Concentrations
Problem The research and development unit of a chemical company is studying the reaction of CH_4 and H_2S, two components of natural gas:
CH_4(g) + 2H_2S(g) \xrightleftharpoons[] CS_2(g) + 4H_2(g)In one experiment, 1.00 mol of CH_4, 1.00 mol of CS_2, 2.00 mol of H_2S, and 2.00 mol of H_2 are mixed in a 250.-mL vessel at 960°C. At this temperature, K_c = 0.036.
(a) In which direction will the reaction proceed to reach equilibrium?
(b) If [CH_4] = 5.56 M at equilibrium, what are the equilibrium concentrations of the other substances?
Plan (a) To find the direction, we convert the given initial amounts and volume (0.250 L) to concentrations, calculate Q_c, and compare it with K_c. (b) Based on this information, we determine the sign of each concentration change for the reaction table and then use the known [CH_4] at equilibrium (5.56 M) to determine x and the other equilibrium concentrations.
Solution (a) Calculating the initial concentrations:
[CH_4] = \frac{1.00\text{ mol}}{0.250 L} = 4.00 M
Similarly, [H_2S] = 8.00 M, [CS_2] = 4.00 M, and [H_2] = 8.00 M.
Calculating the value of Q_c:
Q_c = \frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2} = \frac{(4.00)(8.00)^4}{(4.00)(8.00)^2} = 64.0
Comparing Q_c and K_c: Q_c > K_c (64.0 > 0.036), so the reaction proceeds to the left. Therefore, concentrations of reactants increase (+x) and those of products decrease (–x).
(b) Setting up a reaction table (Table 1), with x = [CS_2] that reacts, which equals [CH_4] that forms:
Solving for x: At equilibrium,
[CH_4] = 5.56 M = 4.00 M + x\text{ so }x = 1.56 M
Thus, [H_2S] = 8.00 M + 2x = 8.00 M + 2(1.56 M) = 11.12 M
[CS_2] = 4.00 M − x = 4.00 M − 1.56 M = 2.44 M
[H_2] = 8.00 M − 4x = 8.00 M − 4 (1.56 M) = 1.76 M
Check The comparison of Q_c and K_c showed the reaction proceeding to the left. The given data from part (b) confirm this because [CH_4] increases from 4.00 M to 5.56 M during the reaction. As always, you should check that the concentrations give the known K_c:
\frac{(2.44)(1.76)^4}{(5.56)(11.12)^2} = 0.0341,\text{ which is close to }0.036
Table 1
| Concentration (M) | \mathbf{CH_4(g) + 2H_2S(g) \xrightleftharpoons[]{} CS_2(g) + 4H_2(g)} |
| Initial | 4.00 8.00 4.00 8.00 |
| Change | +x +2x –x –4x |
| Equilibrium | 4.00 + x 8.00 + 2x 4.00 – x 8.00 – 4x |