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An equlibrium mixture of the reaction `2H_(2)S_((g))hArr2H_(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 vlaue of equlibrium constant (K) in mole `"litre"^(-1)` isA. `0.004`B. `0.008`C. `0.016`D. `0.160` |
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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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