Equilibrium Question 622
Question: A mixture of 0.3 mole of $ H_2 $ and 0.3 mole of $ I_2 $ is allowed to react in a 10 litre evacuated flask at $ 500^{o}C $ . The reaction is $ H_2+I_2 $ ⇌ $ 2HI $ , the $ K $ is found to be 64. The amount of unreacted $ I_2 $ at equilibrium is [KCET 1990]
Options:
A) 0.15 mole
B) 0.06 mole
C) 0.03 mole
D) 0.2 mole
Show Answer
Answer:
Correct Answer: B
Solution:
$ K_{c}=\frac{{{[HI]}^{2}}}{[H_2][I_2]} $ ; $ 64=\frac{x^{2}}{0.03\times 0.03} $
$ x^{2}=64\times 9\times {10^{-4}} $
$ x=8\times 3\times {10^{-2}}=0.24 $ x is the amount of HI at equilibrium amount of $ I_2 $ at equilibrium will be $ 0.30-0.24=0.06 $
