Question

For the equilibrium $A+B\iff C+D$; A and B are mixed in a vessel at T. The initial conc. of A is twice that of B. After equilibrium is reached concentration of C was thrice that of B. Then its $K_c$ is:

• A. 4.5
• B. 9
• C. 1.8
• D. 0.9

Answer 1

The correct option is C 1.8

The equilibrium reaction is $A+B\iff C+D$.
The expression for the equilibrium constant is $K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$.
Let the equilibrium concentration of B be 1 M.
The following table lists the equilibrium concentration of various species.

	A	B	C	D
Initial	2	1	0	0
Change	-x	-x	x	x
Equilibrium	2-x	1-x	x	x

But $\left[C\right]=3\left[B\right]$ or $x=3(1-x)$. Thus $x=0.75$.
The molar concentrations of A,B,C and D are 1.25, 0.25, 0.75 and 0.75 respectively.
Substitute the values in the expression for the equilibrium constant.
$K_c=\frac{0.75\times 0.75}{1.25\times 0.25}=1.8$.