InterviewSolution
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InterviewSolution
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If `K_(1)` and `K_(2)` are respective equilibrium constants for two reactions `:` `XeF_(6)(g) +H_(2)O hArr XeOF_(4)(g) +2HF_(g)` `XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g)` Then equilibrium constant for the reaction `XeO_(4)(g)+2HF(g) hArr XeO_(3)F_(2)(g)+H_(2)O(g)` will beA. `K_(1)K_(2)`B. `K_(1)//(K_(2))^(2)`C. `K_(1)(K_(2))^(-1)`D. `K_(2)//K_(1)` |
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Answer» Correct Answer - D `XeF_(6)(g)+H_(2)O(g) hArr XeOF_(4)(g)+2HF(g)` `K_(1)=([XeOF_(4)][HF]^(2))/([XeF_(6)][H_(2)O])` `XeO_(4)(g)+XeF_(6)(g) hArr XeOF_(4)(g)` `K_(2)=([XeOF_(4)][XeO_(3)])/([XeO_(4)][XeF_(6)])` `XeO_(4)(g)+2HF(g) hArr XeO_(3)(g)+H_(2)O(g)` Let its eqn. constant `=K` `:. K=([XeO_(3)F_(2)][H_(2)O])/([XeO_(4)][HF]^(2))=(K_(2))/(K_(1))` |
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