question_answer

If \[{{K}_{1}}\] and \[{{K}_{2}}\] are respective equilibrium constants for the two reactions \[Xe{{F}_{6}}(g)+{{H}_{2}}O(g)\rightleftharpoons XeO{{F}_{4}}(g)+2HF(g)\] \[Xe{{O}_{4}}(g)+Xe{{F}_{6}}(g)\rightleftharpoons XeO{{F}_{4}}(g)+Xe{{O}_{3}}{{F}_{2}}(g)\] the equilibrium constant for the reaction \[Xe{{O}_{4}}(g)+2HF(g)\rightleftharpoons Xe{{O}_{3}}{{F}_{2}}(g)+{{H}_{2}}O(g)\] will be

A) \[\frac{{{K}_{1}}}{K_{2}^{2}}\]                                

B) \[{{K}_{1}}.{{K}_{2}}\]

C) \[\frac{{{K}_{1}}}{{{K}_{2}}}\]                                  

D) \[\frac{{{K}_{2}}}{{{K}_{1}}}\]