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If K_(1)andK_(2)are the respective equilibrium constants for the two reactions :XeF_(6)(g)+H_(2)O(g)hArr XeOF_(4)(g)+2HF(g)XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g)The equilibrium constant for the reaction :XeO_(4)(g)+2HF(g)hArr XeO_(3)F_(2)(g)+H_(2)O(g)will be : |
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Answer» `K_(1)//K_(2)^(2)` |
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