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Chemistry Question on Aldehydes, Ketones and Carboxylic Acids

If K1K_1 and K2K_2 are the respective equilibrium constants for the two reactions, XeF6(g)+H2O(g)<=>XeOF4(g)+2HF(g){ XeF_6(g)+H_2O(g) <=> XeOF_4(g)+2HF(g)} XeO4(g)+XeF6(g)<=>XeOF4(g)+XeO3F2(g){ XeO_4(g)+XeF_6(g) <=> XeOF_4(g)+ XeO_3F_2(g)} The equilibrium constant of the reaction, XeO4(g)+2HF(g)<=>XeO3F2(g)+H2O(g){XeO_4(g)+2HF(g) <=> XeO_3F_2(g)+H_2O(g)}

A

K1/(K2)2K_1/(K_2)^2

B

K1.K2K_1.K_2

C

K1/K2K_1/K_2

D

K2/K1K_2/K_1

Answer

K2/K1K_2/K_1

Explanation

Solution

XeF6(g)+H2O(g)<=>XeOF4(g)+2HF(g){XeF_6(g)+H_2O(g) <=> XeOF_4(g)+2HF(g)}
K1=[XeOF4][HF]2[XeF6][H2O]{K_1=\frac {[XeOF_4][HF]^2}{[XeF_6][H_2O]}} \hspace10mm ...(i)
XeO4(g)+XeF6(g)<=>XeOF4(g)+XeO3F2(g){XeO_4(g)+XeF_6(g) <=> XeOF_4(g)+ XeO_3F_2(g)}
K2=[XeOF4][XeO3F2][XeO4][XeF6]K_2=\frac {[XeOF_4][XeO_3F_2]}{[XeO_4][XeF_6]} \hspace10mm ...(ii)
For the reaction,
XeO4(g)+2HF(g)<=>XeO3F2(g)+H2O(g){ XeO_4(g)+2HF(g) <=> XeO_3F_2(g)+H_2O(g)}
K=[XeO3F2][H2O][XeO4][HF]2K=\frac {[XeO_3F_2][H_2O]}{[XeO_4][HF]^2} \hspace10mm ...(iii)
∴\therefore From Eqs. (i), (ii) and (iii)
K=K2K1K=\frac {K_2}{K_1}