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 $
