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If the equilibrium constant of the reaction `2HI hArrH_(2) + I_(2)` is `0.25` , then the equilibrium constant of the reaction `H_(2) + I_(2)hArr2HI` would beA. `1.0`B. `2.0`C. `3.0`D. `4.0` |
Answer» Correct Answer - D `K_(1)` for reaction `2HI hArrH_(2) + I_(2)` is `0.25,K_(2)` for reaction `H_(2) + I_(2)hArr2HI` will be `K_(2) = (1)/(K_(1)) = (1)/(0.25) = 4` Because second reaction is reverse of first. |
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