Question

The equilibrium constant $K_C$ for the reaction,
$A\left(g\right)+2B\left(g\right)\rightleftharpoons 3C\left(g\right)$ is $2\times {10}^{-3}$
What would be the equilibrium partial pressure of gas $C$ if initial pressure of gas $A$ and $B$ are $1$ and $2$.

Answer 1

$\begin{array}{c}{K}_C=\frac{{\left[C\right]}^3}{{\left[B\right]}^2\left[A\right]}\Rightarrow \frac{{\left[3x\right]}^3}{\left[1-x\right]{\left[2-2x\right]}^2}\Rightarrow 2\times {10}^{-3}\\ \Rightarrow \frac{27x^3}{\left(1-x\right)4{\left(1-x\right)}^2}=2\times {10}^{-3}\Rightarrow \frac{x^3}{{\left(1-x\right)}^3}=\frac{2\times {10}^{-3}\times 4}{27}\\ \frac{x}{1-x}={\left(\frac{8\times {10}^{-3}}{27}\right)}^{113}\Rightarrow \frac{2}{3}\times {10}^{-1}\\ x=\left(1-x\right)\times \frac{0.2}{3}\Rightarrow 3x=0.2-0.2x\\ 3.2x=0.2\\ x=\frac{0.2}{3.2}\Rightarrow \frac{1}{16}\Rightarrow 0.0625\\ A=1-0.0625\Rightarrow 0.9375atm\\ B=2-2\times 0.0625\Rightarrow 1.875atm\\ C=3\times 0.0625\Rightarrow 0.1875atm\end{array}$