An equilibrium mixture of the reaction 2H2S(g) leftharpoons 2H2(g) + S2(g) had 0.5 mole H2S, 0.10 mole H2 and 0.4 mole S2 in one litre vessel. The value of equilibrium constant (K) in mole litre-1 is

Question

An equilibrium mixture of the reaction 2H2S(g) leftharpoons 2H2(g) + S2(g) had 0.5 mole H2S, 0.10 mole H2 and 0.4 mole S2 in one litre vessel. The value of equilibrium constant (K) in mole litre-1 is (A) 0.004 (B) 0.008 (C) 0.016 (D) 0.160

Tardigrade Admin · Accepted Answer

K =([H2]2[S2]/[H2S]2) =([0.10]2[0.4]/[0.5]2)=0.016

Q. An equilibrium mixture of the reaction
$2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)$ had 0.5 mole $H_{2} S, 0.10$ mole $H_{2}$ and 0.4 mole $S_{2}$ in one litre vessel. The value of equilibrium constant (K) in mole $l i t r e^{- 1}$ is

$0.004$

$0.008$

$0.016$

$0.160$

$K = \frac{[ H _{2} ] ^{2} [ S _{2} ]}{[ H _{2} S ] ^{2}} = \frac{[ 0.10 ] ^{2} [ 0.4 ]}{[ 0.5 ] ^{2}} = 0.016$

Q. An equilibrium mixture of the reaction 2H2​S(g)⇌2H2​(g)+S2​(g) had 0.5 mole H2​S,0.10 mole H2​ and 0.4 mole S2​ in one litre vessel. The value of equilibrium constant (K) in mole litre−1 is

0.004

0.008

0.016

0.160