Chemistry - Chemical Equilibrium Question with Solution | TestHub

$$2{\mathrm{AB}}_2\rightleftharpoons 2\mathrm{AB}+{\mathrm{B}}_2$$

Initial:

$$\begin{array}{lcc}A{B}_2& AB& B_2\\ 2& 0& 0\end{array}$$

Final:

$$\begin{array}{lcc}A{B}_2& AB& B_2\\ 2-2x& 2x& x\end{array}$$

The total number of moles is

$$(2-2x)+2x+x=2+x$$

The partial pressures are:

$$P_{{\mathrm{AB}}_2}=\frac{2-2x}{2+x}\times P$$

$$P_{\mathrm{AB}}=\frac{2x}{2+x}\times P$$

$$P_{{\mathrm{B}}_2}=\frac{x}{2+x}\times P$$

The equilibrium constant $K_P$ is given by:

$$K_P=\frac{{\left(P_{\mathrm{AB}}\right)}^2\left(P_{{\mathrm{B}}_2}\right)}{{\left(P_{{\mathrm{AB}}_2}\right)}^2}=\frac{{\left(\frac{2x}{2+x}\right)}^2{P}^2\times \left(\frac{x}{2+x}\right)P}{{\left(\frac{2-2x}{2+x}\right)}^2\times P^2}=\frac{x^3\times P}{(2+x)(1-x{)}^2}$$