An equilibrium mixture of the reaction 2H_(2)S(g)("number of moles of" O_(2))/("volume" ("in litre")) = (96)/(32)xx(1)/(2) = 1.5 mol//L2H_(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

An equilibrium mixture of the reaction 2H_(2)S(g)("number of moles of"

1.	An equilibrium mixture of the reaction 2H_(2)S(g)("number of moles of" O_(2))/("volume" ("in litre")) = (96)/(32)xx(1)/(2) = 1.5 mol//L2H_(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
Answer» `0.004` `0.008` `0.016` `0.160` SOLUTION :`K = [H_(2]^(2)[S_(2)])/([H_(2)S]^(2)) = ([0.10]^(2)[0.4])/([0.5]^(2)) = 0.016`