Chemistry - Chemical Equilibrium Question with Solution | TestHub

For the given reaction, $$2A(g)+B_2(g)\rightleftharpoons 2A{B}_2(g)$$ the equilibrium constant, $$K_p=\frac{p_{A{B}_2}^2}{p_A^2\cdot p_{B_2}}=\mathrm{16...}(i)$$ For the other given reaction, $$A{B}_2(g)\rightleftharpoons A(g)+\frac{1}{2}{B}_2(g)$$ The equilibrium constant, $$K_p^{\prime }=\frac{p_A\cdot p_{B_2}^{1/2}}{p_{A{B}_2}}...(ii)$$ On squaring Eq. (ii), we obtain, $${\left(K^{\prime }{}_p\right)}^2=\frac{p_A^2\cdot p_{B_2}}{p_{A{B}_2}^2}...(iii)$$ Now, from Eq. (i) and (iii), we obtain, $$\begin{array}{l}{K}_p\cdot {\left(K_p^{\prime }\right)}^2=1\Rightarrow 16\cdot {\left(K_p^{\prime }\right)}^2=1\\ \left(\because K_p=16.0\right)\\ \Rightarrow \left(K_p^{\prime }\right)={\left(\frac{1}{16}\right)}^{1/2}\Rightarrow K_p^{\prime }=\frac{1}{4}=0.25\end{array}$$