Question

At 1400 K, $K_c=2.5\times {10}^3$ for the reaction $C{H}_4\left(g\right)+2H_{2}S\rightleftharpoons C{S}_2\left(g\right)+4H_2\left(g\right)$. A 10.0 L reaction vessel at 1400 K contains 2.0 mol of $C{H}_4$, 3.0 mol of $C{S}_2$, 3.0 mol of $H_2$ and 4.0 mol of $H_{2}S$.The reaction will proceed from right to left to reach equilibrium.

• A. True
• B. False
• C. Ambiguous
• D. None of the above

Answer 1

The correct option is A True

At 1400 K, $K_c=2.5\times {10}^3$ for the reaction $C{H}_4\left(g\right)+2H_{2}S\rightleftharpoons C{S}_2\left(g\right)+4H_2\left(g\right)$. A 10.0 L reaction vessel at 1400 K contains 2.0 mol of $C{H}_4$, 3.0 mol of $C{S}_2$, 3.0 mol of $H_2$ and 4.0 mol of $H_{2}S$.The reaction will proceed from right to left to reach equilibrium.

The reaction quotient $Q=\frac{\left[C{S}_2\right][H_2{]}^2}{\left[C{H}_4\right][H_{2}S{]}^2}$

$Q=\frac{3.0\times (3.0{)}^2}{2.0\times (4.0{)}^2\times (10{)}^2}=0.0084$

But the equilibrium constant $K_c=2.5\times {10}^3$

Hence, the reaction quotient Q is smaller than the equilibrium constant K

$Q<K$

Hence, the concentration of products is smaller than the equilibrium concentration and the concentration of reactants is higher than the equilibrium concentration.

The system will shift right to reach equilibrium till Q becomes equal to K.