question_answer

In reaction \[A+2B\rightleftharpoons 2C+D\], 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:

A) 8                     

B) 4

C) 12                    

D) 6

Correct Answer: B

Solution :

[b]   Hence \[{{K}_{C}}=\frac{{{(2x)}^{2}}\times x}{(a-x){{(1.5a-2x)}^{2}}}\] Given, at equilibrium \[\therefore ~\left( a-x \right)=\left( 1.5a-2x \right)\] \[\therefore ~a=2x\] On solving \[{{K}_{C}}=4\]