Question

For the reaction $A+2B\rightleftharpoons 2C+D$, initial concentration of A is a and that of B is 1.5 times that of A. Concentration of A and D are same at equilibrium. What should be the concentration of B at equilibrium?

• A. $\frac{a}{4}$
• B. $\frac{a}{2}$
• C. $\frac{3a}{4}$
• D. All of the above

Answer 1

Answer: (B)

$\frac{a}{2}$

Let x be the change in the concentration of A to reach equilibrium.

	A	B	C	D
Initial concentration	a	1.5a	0	0
Equilibrium concentration	$a-x$	$1.5a-2x$	$2x$	$x$

Concentration of A and D are same at equilibrium.
$a-x=x$
$a=2x$
$x=0.5$
The equilibrium concentration of B is $1.5a-2x=1.5a-a=\frac{a}{2}$.