Question

For the reaction $A+2B\rightleftharpoons 2C$ at equilibrium $\left[C\right]=1.4M.[A{]}_o=1M,[B{]}_o=2M,[C{]}_o=3M$. The value of $K_c$ is:

(Given: volume =1 litre, $[{]}_o$ is initial concentration)

• A. 0.084
• B. 8.4
• C. 48
• D. None of these.

Answer 1

Answer: (A)

0.084

The equilibrium reaction is $A+2B\rightleftharpoons 2C.$

The initial concentrations of A, B and C are 1M, 2M, and 3M respectively.

The equilibrium concentrations of A, B and C are $1-xM,2-2xM$ and $3+2xM$ respectively.

The expression for the equilibrium constant is $K_c=\frac{[C{]}^2}{\left[A\right][B{]}^2}=\frac{(3+2x{)}^2}{(1-x)(2-2x{)}^2}.$

But $\left[C\right]=1.4M=3+2x$ or $x=-0.8$

$\therefore \left[A\right]=1-(-0.8)=1.8M;\left[B\right]=2-2(-0.8)=3.6M$

Substitute values in the equilibrium constant expression.

$K_c=\frac{[C{]}^2}{\left[A\right][B{]}^2}=\frac{(3+2x{)}^2}{(1-x)(2-2x{)}^2}=\frac{(1.4{)}^2}{\left(1.8\right)(3.6{)}^2}=\mathrm{0.084.}$