An equilibrium mixture of the reaction, $2H_{2}S(g) \rightleftharpoons 2H_{2}(g) + S_{2}(g)$ has 0.5 mole $H_{2}S$, 0.10 mole $H_{2}$ and 0.4 mole $S_{2}$ in 1 L vessel. The value of equilibrium constant (K) in $mol\ L^{- 1}$, is

Question

Dr. Abubacker · Accepted Answer

(c)
$K_{c} = \frac{\left\lbrack H_{2} \right\rbrack^{2}\lbrack S_{2}\rbrack}{\left\lbrack H_{2}S \right\rbrack^{2}} = \frac{\lbrack 0.10\rbrack^{2}\lbrack 0.4\rbrack}{\lbrack 0.5\rbrack^{2}} = 0.016$

Q. An equilibrium mixture of the reaction, $2H_{2}S(g) \rightleftharpoons 2H_{2}(g) + S_{2}(g)$ has 0.5 mole $H_{2}S$, 0.10 mole $H_{2}$ and 0.4 mole $S_{2}$ in 1 L vessel. The value of equilibrium constant (K) in $mol\ L^{- 1}$, is

Explanation

Reviewed by Dr. Abubacker

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