Equilibrium Question 727
Question: (1) $ N_2(g)+3H_2(g)\rightarrow 2NH_3(g),K_1 $
(2) $ N_2(g)+O_2(g)\rightarrow 2NO(g),K_2 $
(3) $ H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g),K_3 $
The equation for the equilibrium constant of the reaction $ 2NH_3(g)+\frac{5}{2}O_2(g)\rightarrow 2NO(g)+3H_2O(g), $
$ (K_4) $ in terms of $ K_1,K_2 $ and $ ~K_3 $ is:
Options:
A) $ \frac{K_1.K_2}{K_3} $
B) $ \frac{K_1.K_3^{2}}{K_2} $
C) $ K_1K_2K_3 $
D) $ \frac{K_2.K_3^{3}}{K_1} $
Show Answer
Answer:
Correct Answer: D
Solution:
To calculate the value of $ K_4 $ in the given equation we should apply: $ eqn.( 2 )+eqn.( 3 )\times 3-eqn.( 1 ) $ hence $ K_4=\frac{K_2K_3^{3}}{K_1} $
BETA
