Equilibrium Question 780
Question: In reaction $ A+2B\rightarrow 2C+D $ , initial concentration of B was 1.5 times of [A], but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is:
Options:
A) 8
B) 4
C) 12
D) 6
Show Answer
Answer:
Correct Answer: B
Solution:
Hence $ K_{C}=\frac{{{(2x)}^{2}}\times x}{(a-x){{(1.5a-2x)}^{2}}} $ Given, at equilibrium
$ \therefore ~( a-x )=( 1.5a-2x ) $
$ \therefore ~a=2x $ On solving $ K_{C}=4 $
