Equilibrium Question 761

Question: For the equilibrium system $ 2HX(g)\rightarrow H_2(g)+X_2(g) $ the equilibrium constant is $ 1.0\times {10^{-5}} $ . What is the concentration of HX if the equilibrium concentration of $ H_2 $ and $ X_2 $ are $ 1.2\times {10^{-3}}M $ , and $ 1.2\times {10^{-4}}M $ respectively.

Options:

A) $ 12\times {10^{-4}}M $

B) $ 12\times {10^{-3}}M $

C) $ 12\times {10^{-2}}M $

D) $ 12\times {10^{-1}}M $

Show Answer

Answer:

Correct Answer: C

Solution:

At eqm. $ \begin{aligned} & 2HX(g)\rightarrow H_2(g),+,X_2(g) \\ & ,1.2\times {10^{-3}}M1.2\times {10^{-4}}M \\ \end{aligned} $

$ K_{eq}=\frac{[H_2][X_2]}{{{[HX]}^{2}}} $

$ {10^{-5}}=\frac{1.2\times {10^{-3}}\times 1.2\times {10^{-4}}}{{{[HX]}^{2}}} $

$ [HX]=\sqrt{\frac{1.2\times 1.2\times {10^{-7}}}{{10^{-5}}}} $

$ =1.2\times {10^{-1}} $

$ =12\times {10^{-2}}M $



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