Question

An equilibrium mixture for the reaction $H_{2}S\left(g\right)\rightleftharpoons H_2\left(g\right)+S_2\left(g\right)$ had 1 mol of hydrogen sulphide, 0.2 mol of $H_2$ and 0.8 mol of $S_2$ in a 2 L vessel. The value of $K_c$ in M is

• A. 0.004
• B. 0.016
• C. 0.08
• D. 0.16

Answer 1

The correct option is D 0.16

For a reaction at equilibrium the relation between the concentration of product raised to the power of their coefficient to the concentration of reactant raised to the power of their coefficient is known as equilibrium constant.
$aA+bB\rightleftharpoons cC+dD$
Equilibrium Constant $\left(K_c\right)=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$
$2H_{2}S\left(g\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$
No. of moles of $H_{2}S=1$ mol
volume of flask $=2L$
No. of mole of $H_2=0.2$ mol
No. of mole of $S_2=0.8$ mol
Concentration of $H_{2}S,\left[H_{2}S\right]=\frac{\text{Mole}}{\text{Volume}}=\frac{\text{1 mol}}{2L}=0.5M$
Concentration of $H_2,\left[H_2\right]=\frac{\text{Mole}}{\text{Volume}}=\frac{\text{0.2 mol}}{2L}=0.1M$
Concentration of $S_2,\left[S_2\right]=\frac{\text{Mole}}{\text{Volume}}=\frac{\text{0.8 mol}}{2L}=0.4M$
$K_c=\frac{\left[H_2{]}^2\right[S_2]}{[H_{2}S{]}^2}=\frac{[0.1M{]}^2\times [0.4M]}{[0.5M{]}^2}$
$=\frac{4\times {10}^{-3}}{0.25}=0.016M$