Equilibrium Question 604
Question: On a given condition, the equilibrium concentration of $ HI,,H_2 $ and $ I_2 $ are 0.80, 0.10 and 0.10 mole/litre. The equilibrium constant for the reaction $ H_2+I_2 $ ⇌ $ 2HI $ will be [MP PET 1986]
Options:
A) 64
B) 12
C) 8
D) 0.8
Show Answer
Answer:
Correct Answer: A
Solution:
$ H_2+I_2 $ ⇌ 2HI; [HI] = 0.80, $ [H_2]=0.10 $ , $ [I_2]=0.10 $
$ K_{c}=\frac{{{[HI]}^{2}}}{[H_2][I_2]}=\frac{0.80\times 0.80}{0.10\times 0.10}=64 $
