Question

The equilibrium constant for the following reaction is K = 8.0 × 10-2 M2 at a given temperature.
CH4(g) + 2H2S(g) ⇌ CS2(g) + 4 H2(g)
a) The following equilibrium amounts of substances were observed in gas chamber of volume V at this temperature: 0.10 mol CS2, 0.20 mol H2, 0.15 mol CH4 and 0.05 mol H2S. Determine the value of V.
b) In which direction will this reaction shift if the volume of is decreased. Explain your reasoning.
c) The reaction is exothermic in the forward direction. How will the concentration of CS2 change if the temperature is decreased? Explain your reasoning.

Answer 1

1.

$$K = \frac{[\mathrm{CS}_2][\mathrm{H}_2]^4}{[\mathrm{CH}_4][\mathrm{H}_2\mathrm{S}]^2} = 8 \times 10^{-2}$$ $$c = n/V$$ $$K = \frac{\frac{0.1}{V} \times \left(\frac{0.2}{V}\right)^4}{\frac{0.15}{V} \times \left(\frac{0.05}{V}\right)^2} = \frac{0.427}{V^2} = 8 \times 10^{-2}$$ $$V = \sqrt{\frac{0.427}{8 \times 10^{-2}}} = 2.31\ \mathrm{L}$$

2.

If the volume is decreased, the reaction will be shifted in forward direction (according to Le Chatelier’s principle, because the volume of reaction mixture is increased with the time).

3.

The concentration of $\mathrm{CS}_2$ will be increased (according to Le Chatelier’s principle, because the temperature of reaction mixture is increased with the time).