Question

The equilibrium constant for the reaction between $C{H}_4(g)$ and $H_{2}S(g)$ to form $C{S}_2(g)$ and $H_2(g)$, at $1173K$ is $\mathrm{3.6.}$ For the following composition of the reaction mixture,
$\left[C{H}_4\right]=1.0M,\left[H_{2}S\right]=1.20M$
$\left[C{S}_2\right]=2.0M,\left[H_2\right]=1.8M$
decide which of the following option is correct?
$\left[C{H}_4(g)+2H_{2}S(g)\rightleftharpoons C{S}_2(g)+4H_2(g)\right]$

• A. Reaction is in equilibrium
• B. Reaction will shift to form more $C{S}_2$
• C. Reaction will shift to form more $H_{2}S$
• D. No reaction takes place

Answer 1

Answer: (C)

$k_c=3.6$ (given)
$C{H}_{4(g)}+2H_{2}S\rightleftharpoons C{S}_{2(g)}+4H_{2(g)}$
$Q_c=\frac{{\left[H_2\right]}^4\left[C{S}_2\right]}{{\left[H_{2}S\right]}^2\left[C{H}_4\right]}$
$Q_c=\frac{(1.8{)}^4}{(1.2{)}^2}\cdot 2=14.5g$
$Q_c>k_c$
Reaction will shift to form more $H_{2}S$