Equilibrium Question 586
Question: If equilibrium constant for reaction $ 2AB $ ⇌ $ A_2+B_2 $ , is 49, then the equilibrium constant for reaction AB ⇌ $ \frac{1}{2}A_2+\frac{1}{2}B_2 $ , will be [EAMCET 1998; MP PMT 2003]
Options:
A) 7
B) 20
C) 49
D) 21
Show Answer
Answer:
Correct Answer: A
Solution:
$ 2AB $ ⇌ $ A_2+B_2 $
$ K_{c}=\frac{[A_2][B_2]}{{{[AB]}^{2}}} $ For reaction $ AB $ ⇌ $ \frac{1}{2}A_2+\frac{1}{2}B_2 $
$ K_c^{’}=\frac{{{[A_2]}^{{1}/{2};}}{{[B_2]}^{{1}/{2};}}}{[AB]} $ ; $ K_c^{’}=\sqrt{K_{c}}=\sqrt{49}=7 $ .
