JEE Main & Advanced
Chemistry
Equilibrium / साम्यावस्था
Question Bank
Self Evaluation Test - Equilibrium
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 \[{{K}_{C}}=4\]