Equilibrium Question 724
Question: Consider following reaction in equilibrium concentration 0.01 M of every species
(I) $ PCl_5(g) \leftrightharpoons PCl_3(g)+Cl_2(g) $
(II) $ 2HI(g) \leftrightharpoons H_2(g)+I_2(g) $
(III) $ N_2(g)+3H_2(g) \leftrightharpoons 2NH_3(g) $
Extent of the reactions taking place is
Options:
A) I > II > III
B) I < II < III
C) II < III < I
D) III < I < II
Show Answer
Answer:
Correct Answer: B
Solution:
Extent of reaction can be calculated by the value of $ K_{C} $ . The higher value of $ K_{C} $ , the larger the extent of reactions. $ PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g) $
$ \frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.01\times 0.01}{0.01}=0.01 $ (II) $ 2HI\rightarrow H_2+I_2,K_{C}=\frac{[H_2][I_2]}{{{[HI]}^{2}}} $
$ =\frac{(0.01)(0.01)}{{{(0.01)}^{2}}}=1 $ (III) $ N_2(g)+3H_2(g)\rightarrow 2NH_3(g) $
$ K_{C}=\frac{{{[NH_3]}^{2}}}{[N_2]{{[H_2]}^{3}}}=\frac{{{(0.01)}^{2}}}{{{(0.01)}^{4}}}=\frac{1}{0.01\times 0.01}=10000 $ Extent of reaction I<II<III.
