Question

For the reaction $A+B\rightleftharpoons C+D$, the initial concentrations of A and B are equal. The equilibrium concentration of C is two times the equilibrium concentration of A.

The value of the equilibrium constant is :

• A. $1$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

For the reaction $A+B\rightleftharpoons C+D$, the initial concentrations of A and B are equal. The equilibrium concentration of C is two times the equilibrium concentration of A. The value of equilibrium constant is 4.

Let the initial concentrations of A and B be X M each Out of this Y M react to reach equilibrium.

	A	B	C	D
Initial concentration (M)	X	X	0	0
Change (M)	-Y	-Y	Y	Y
Equilibrium concentration (M)	X-Y	X-Y	Y	Y

The equilibrium concentration of C is two times the equilibrium concentration of A.

$Y=2(X-Y)$

$X=1.5Y$

The equilibrium constant expression is $K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{Y\times Y}{(X-Y)\times (X-Y)}=(\frac{Y}{X-Y}{)}^2$

But $X=1.5Y$

Hence, $K=(\frac{Y}{1.5Y-Y}{)}^2=4$