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)` isA. `0.004`B. `0.008`C. `0.016`D. `0.160`

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)` isA. `0.004`B. `0.008`C. `0.016`D. `0.160`
Answer» Correct Answer - C `K = [H_(2]^(2)[S_(2)])/([H_(2)S]^(2)) = ([0.10]^(2)[0.4])/([0.5]^(2)) = 0.016`

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