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InterviewSolution
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An equilibrium mixture for the reaction `2H_(2)S(g) hArr 2H_(2)(g) + S_(2)(g)` had 1 mole of `H_(2)S, 0.2` mole of `H_(2)` and 0.8 mole of `S_(2)` in a 2 litre flask. The value of `K_(c)` in mol `L^(-1)` isA. 0.004B. 0.08C. 0.016D. 0.16 |
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Answer» Correct Answer - C `K_(c) =[[H_(2)]^(2)[S_(2)]]/[[H_(2)S]^(2)]` `[H_(2)] =(2xx0.2)/(2) =0.2 " mole"//L` `[S_(2)]=(0.8)/(2) =0.4 "mole"//L ,[H_(2)S]=1 "mole"//L` `:. K_(c) =[[0.2]^(2)[0.4]]/(1) =0.016 "mole"//L` |
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