Question

For the reaction $A\left(g\right)+2B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right);$ $K_c={10}^{12}volme.mol{e}^{-1}$. If the initial moles of $A,B,C$ and $D$ are $0.5,1,0.5$ and $3.5$ moles respectively in a one litre vessel. The equilibrium concentration of $B$ is:

• A. ${10}^{-4}$
• B. $2\times {10}^{-4}$
• C. $4\times {10}^{-4}$
• D. $8\times {10}^{-4}$

Answer 1

Answer: (B)

$2\times {10}^{-4}$

The equilibrium reaction is $A\left(g\right)+2B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right).$

The initial moles of $A,B,C$ and $D$ are $0.5,1,0.5$ and $3.5$ moles respectively.

The equilibrium moles of $A,B,C$ and $D$ are $0.5-x,1-2x,0.5+x$ and $3.5+x$ moles respectively.

The expression for the equilibrium constant is $K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right][B{]}^2}$.

Substitute values in the above expression.

$K_c=\frac{(0.5+x)(3.5+x)}{(0.5-x)(1-2x{)}^2}$

Hence, $x=0.4999$ and $\left[B\right]=(1-2x)=2\times {10}^{-4}$.