An equilibrium mixture of the reaction 2 H 2 S ( s ) leftharpoons 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 litre -1 is

Question

An equilibrium mixture of the reaction 2 H 2 S ( s ) leftharpoons 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 litre -1 is (A) 0.016 (B) 0.008 (C) 0.004 (D) 0.160

Tardigrade Admin · Accepted Answer

2 H 2 S (g) leftharpoons 2 H 2(g)+ S 2(g) Equilibrium constant is, K=([ H 2]2[ S 2]/[ H 2 S ]2)=([0.1]2[0.4]/[0.5]2) =0.016 mol L -1

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

0.016

0.008

0.004

0.160

$2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)$

Equilibrium constant is,

$K = \frac{[ H _{2} ] ^{2} [ S _{2} ]}{[ H _{2} S ] ^{2}} = \frac{[ 0.1 ] ^{2} [ 0.4 ]}{[ 0.5 ] ^{2}}$

$= 0.016 m o l L^{- 1}$

Q. An equilibrium mixture of the reaction 2H2​S(s)​⇌2H2(g)​+S2(g)​, had 0.5moleH2​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