Equilibrium Question 762
Question: If $ K_1 $ and $ K_2 $ are respective equilibrium constants for the two reactions $ XeF_6(g)+H_2O(g)\rightarrow XeOF_4(g)+2HF(g) $
$ XeO_4(g)+XeF_6(g)\rightarrow XeOF_4(g)+XeO_3F_2(g) $ the equilibrium constant for the reaction $ XeO_4(g)+2HF(g)\rightarrow XeO_3F_2(g)+H_2O(g) $ will be
Options:
A) $ \frac{K_1}{K_2^{2}} $
B) $ K_1.K_2 $
C) $ \frac{K_1}{K_2} $
D) $ \frac{K_2}{K_1} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ K=K_2\times \frac{1}{K_1}=\frac{K_2}{K_1}. $
