Question

At $1400K$, for the reaction $C{H}_4\left(g\right)+2H_{2}S\left(g\right)\rightleftharpoons C{S}_2\left(g\right)+4H_2\left(g\right)$, $K_c=2.5\times {10}^{-3}$. A $10L$ reaction vessel contains 2 moles of $C{H}_4,$ 3 moles of $C{S}_2$ , 3 moles of $H_2$ and 4 moles of $H_{2}S$. Choose the correct statement from the following:

• A. This reaction is at equilibrium with given concentrations.
• B. The reaction will proceed in forward direction
• C. The reaction will proceed in backward direction
• D. The information is insufficient to decide the direction of progress of reaction

Answer 1

The correct option is C The reaction will proceed in backward direction

For the reaction,
$Q_C=\frac{{\left(\frac{n_{H_2}}{V}\right)}^4\left(\frac{n_{C{S}_2}}{V}\right)}{\left(\frac{n_{C{H}_4}}{V}\right){\left(\frac{n_{H_{2}S}}{V}\right)}^2}$

putting the values,
$Q_C=\frac{{\left(\frac{3}{10}\right)}^4\left(\frac{3}{10}\right)}{\left(\frac{2}{10}\right){\left(\frac{4}{10}\right)}^2}=\frac{243}{32}\times {10}^{-2}=7.59\times {10}^{-2}>K_c$

So, reaction will proceed in backward direction.