Question

Solve the above problem !!!​

Answer 1

Explanation:

$\tt De{F}_{6} + {H}_{2}O \rightleftharpoons XeO{F}_{4} + 2HF$ \\ \\ \tt {K}_{1} = \frac{ \left [ XeO{F}_{4} \right ]{ \left [ HF \right ]}^{2}}{\left [ Xe{F}_{6} \right ] \left [ {H}_{2}O \right ] } \\ \\ \tt Xe{O}_{4} + Xe{F}_{6} \rightleftharpoons XeO{F}_{4} + De{O}_{2}{F}_{2} \\ \\ \tt {K}_{2} = \frac{ \left [ XeO{F}_{4} \right ] \left [ Xe{O}_{3}{F}_{2} \right ]}{\left [ Xe{O}_{4} \right ] \left [ Xe{F}_{6} \right ] } \\ \\ \tt Xe{O}_{4} + 2HF \rightleftharpoons De{O}_{3}{F}_{2} + {H}_{2}O \\ \\ \tt K = \frac{ \left [ Xe{O}_{3{{F}_{2} \right ] \left [ {H}_{2}O \right ]}{\left [ Xe{O}_{4} \right ] {\left [ HF \right ]}^{2} } \\ \\ \tt If \: we \: divide \: {K}_{2} \:by \:{K}_{1} \\ \\ \tt \frac{{K}_{2}}{{K}_{1}} = \frac{ \left [ XeO{F}_{4} \right ] \left [ Xe{O}_{3}{F}_{2} \right ]}{\left [ Xe{O}_{4} \right ] \left [ Xe{F}_{6} \right ] } × \frac{ \left [ Xe{F}_{6} \right ]\left [ {H}_{2}O \right ]}{\left [ XeO{F}_{4} \right ] {\left [ HF \right ]}^{2} } \\ \\ \tt \frac{{K}_{2}}{{K}_{1}} = \frac{ \left [ Xe{O}_{3{{F}_{2} \right ] \left [ {H}_{2}O \right ]}{\left [ Xe{O}_{4} \right ] {\left [ HF \right ]}^{2} } \\ \\ \tt \boxed{\bold{\underline{\red{\tt \frac{{K}_{2}}{{K}_{1}} = K }}}} [/tex]