Question

At a certain temperature the equilibrium constant $K_c$ is $0.25$ for the reaction $A_{2\left(g\right)}+B_{2\left(g\right)}\rightleftharpoons C_{2\left(g\right)}+D_{2\left(g\right)}$. If we take $1$ mole of each of the four gases in a $10$ litre container, what would be equilibrium concentration of $A_{2\left(g\right)}$?

• A. $0.331M$
• B. $0.033M$
• C. $0.133M$
• D. $1.33M$

Answer 1

The correct option is D $1.33M$

Option D is correct

$A_2+B_2\to C_2+D_2$

At t=0 1 1 1 1

t=eq 1-x 1-x 1+x 1+x

$K_C=\frac{[1+x{]}^2}{[1-x{]}^2}$

Given $K_C=0.25$ so

$0.25=\frac{[1+x{]}^2}{[1-x{]}^2}$

$0.25=\frac{1+2x+x^2}{1-2x+x^2}$

$0.25-0.5x+0.25x^2=1+2x+x^2$

$0.75x^2+2.5x+0.75=0$

divide the above equation by 2.5

$0.3x^2+x+0.3=0$

Solving for x

a=0.3, b=1, c=0.3

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-1\pm \sqrt{1-4\times 0.3\times 0.3}}{2\times 0.3}$

$x=-0.33$ & $x=-3$

Equilibrium concentration of $A_2=1-x$

$1+0.33=1.33$ M