Question

An equilibrium mixture of the reaction $2H_{2}S_{(g)}\rightleftharpoons 2H_{2(g)} + S_{2(g)}$ had 0.5 mol ${{H}_{2}}S$, 0.10 mol ${{H}_{2}}$, and 0.4 mol ${{S}_{2}}$in a one-litre vessel. The value of the equilibrium constant (K) in $mol/l$ is:
(A) 0.004
(B) 0.008
(C) 0.016
(D) 0.160

Answer 1

Hint: An equilibrium constant (K) is the relationship between the concentration of reactants and products present at equilibrium in a reversible chemical process at a given temperature. It is the ratio of the concentration of products and reactants.

Formula Used: For a reaction: $2A\rightleftharpoons 2B + C$
The equilibrium constant is $K=\frac{{{[B]}^{2}}[C]}{{{[A]}^{2}}}$

Complete Step by Step Answer:
The given reaction is:
$2H_{2}S_{(g)}\rightleftharpoons 2H_{2(g)} + S_{2(g)}$
The concentration of the products and reactants in a one-litre vessel is: $\left[ {{H}_{2}} \right]=0.10M$,$\left[ {{S}_{2}} \right]=0.4M$ and $\left[ {{H}_{2}}S \right]=0.50M$
The equilibrium constant, K, is given by $K=\frac{{{[{{H}_{2}}]}^{2}}[{{S}_{2}}]}{{{[{{H}_{2}}S]}^{2}}}$
                                                                  $K=\frac{{{(0.10)}^{2}}\times (0.40)}{{{(0.50)}^{2}}}$
                                                                  $K=\frac{0.01\times 0.40}{0.25}$
                                                                  $K=0.016mol/l$
Thus, the value of the equilibrium constant (K) = $0.016mol/l$
Correct Option: (C) 0.016.

Additional Information: Chemical equilibrium is a state in which the rates of both the forward and backward reactions are equal and the concentration of both the reactants and products is constant. At equilibrium, there is no net change in the number of moles, although conversion from reactants to products or products to reactants is still occurring.

Note: The units of equilibrium constant depend upon the number of moles of reactant and product. Its units are given by${{\left( mol/l \right)}^{\Delta n}}$ . As in this case, $\Delta n={{n}_{P}}-{{n}_{R}}=3-2=1$ . So, units of K is $mol/l$. To calculate the equilibrium constant, you must first understand the entire reaction and its stoichiometric coefficients. If K>1, the equilibrium favours products, and if K<1, then the equilibrium favours reactants. But, if K = 1, then both reactants and products are present in the mixture in significant amounts.