Chapter 6: Problem 7
\(\mathrm{XeF}_{6}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{XeOF}_{4}+2 \mathrm{HF}\) equilibrium constant \(=K_{1}\). \(\mathrm{XeO}_{4}+\mathrm{XeF}_{6} \rightleftharpoons \mathrm{XeOF}_{4}+\mathrm{XeO}_{3} \mathrm{~F}_{2}\) equilibrium constant \(=K_{2} .\) Then equilibrium constant for the following reaction will be: \(\mathrm{XeO}_{4}+2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) \(\frac{K_{1}}{K_{2}}\) (b) \(K_{1}+K_{2}\) (c) \(\frac{K_{2}}{K_{1}}\) (d) \(K_{2}-K_{1}\)
Short Answer
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Key Concepts
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