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The equilibrium constant for the following two reactions are `K_(1)` and `K_(2)` respectively. `XeF_(6)(g)+H_(2)O(g) hArrXeOF_(4)(g)+2HF(g)` `XeO_(4)(g)+XeF_(6)(g)hArrXeOF_(4)(g)+XeO_(3)F_(2)(g)` The equilibrium constant for the given reactiono is `:`A. `K_(1)K_(2)^(2)`B. `K_(1)-K_(2)`C. `K_(2)//K_(1)`D. `K_(2)//K_(2)` |
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Answer» Correct Answer - 3 Substracting `1^(st)` reaction from second, we will get desired reaction `3^(rd)`. So `K=K_(2)//K_(1)` |
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