Question

$2$ moles of $A$ and $3$ moles of $B$ are mixed and the reaction is carried at ${400}^{o}C$ according to the equation,
$A+B\rightleftharpoons 2C$
the equilibrium constant of the reaction is $4$. Find the number of moles of $C$ at equilibrium.

Answer 1

2 moles of A and 3 moles of B are mixed. x moles of A will react with x moles of B to form 2x moles of C. $2-x$ moles of A and $3-x$ moles of B will remain.

Let V L be the total volume.

$K_c=\frac{[C{]}^2}{\left[A\right]\left[B\right]}$
$1=\frac{[\frac{\text{2x mol}}{\text{V L}}{]}^2}{\left[\frac{\text{2-x mol}}{\text{V L}}\right]\left[\frac{\text{3-x mol}}{\text{V L}}\right]}$
$1=\frac{[2x{]}^2}{[2-x][3-x]}$
$4x^2=[2-x][3-x]$
$4x^2=2[3-x]-x[3-x]$
$4x^2=6-2x-3x+x^2$
$3x^2+5x-6=0$

This is quadratic equation with solution

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-5\pm \sqrt{5^2-4\times 3\times (-6)}}{2\left(3\right)}$
$x=\frac{-5\pm \sqrt{5^2-4\times 3\times (-6)}}{6}$
$x=0.808$ or $x=-2.47$

The value $x=-2.47$ is discarded as number of moles cannot be negative.

The number of moles of C present at equilibrium $=2x=2\left(0.808\right)=1.6$ moles.