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 $



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