Question
For the reaction, $\text{CH}_4(\text{g})+2\text{H}_2\text{S(g)}\rightleftharpoons\text{CS}_2\text{(g)}+4\text{H}_2(\text{g})$ at 1173K, the magnitude of the equilibrium constant, $K_c$​​​​​​​ is 3.6. For each of the following compositions, decide whether reaction mixture is at equilibrium. If it is not, decide in which direction the reaction should go?
  1. $[\text{CH}_4]=1.07\text{M},[\text{H}_2\text{S}]=1.20\text{M},[\text{CS}_2]$
$=0.90\text{M},[\text{H}_2]=1.78\text{M}$
  1. $[\text{CH}_4]=1.45\text{M},[\text{H}_2\text{S}]=1.29\text{M},[\text{CS}_2]$
$=1.25\text{M},[\text{H}_2]=1.75\text{M}$

Answer

  1. $\text{CH}_4(\text{g})+2\text{H}_2\text{S(g)}\rightleftharpoons\text{CS}_2\text{(g)}+4\text{H}_2(\text{g})$
$\text{Q}_{\text{c}}=\frac{[\text{CS}_4][\text{H}_2]^4}{[\text{CH}_4][\text{H}_2\text{S}]^2}=\frac{[0.90][1.78]^4}{[1.07][1.20]^2}$
$=\frac{10.03\times0.9}{1.44\times1.07}=\frac{9.027}{1.54}$
$=5.86\text{ K}_{\text{c}}=3.6$
Since $Q_c > K_c$​​​​​​​, the equilibrium will shift in backward direction.
  1. $\text{Q}_{\text{c}}=\frac{[1.25][1.75]^4}{[1.45][1.29]^2}$
$=\frac{9.38\times1.25}{1.45\times1.66}=\frac{11.725}{2.407}=4.87$
Since $Q_c > K_c$​​​​​​​, the equilibrium will shift in backward direction.

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