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.



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