Chemistry - Chemical Equilibrium Question with Solution | TestHub

$$\begin{array}{c}\text{First}\text{equilibrium:}\text{AB}\text{=}\text{0.3,}{\text{A}}_{\text{2}}\text{=}{\text{B}}_{\text{2}}\text{=}\text{0.2}\text{-}\text{0.15}\text{=}\text{0.05}\text{mol}{\text{K}}_{\text{1}}\text{=}{\text{(0.3)}}^{\text{2}}\\ \text{/[(0.05)(0.05)]}\text{=}\text{36}\text{(}\Delta \text{n}\text{=}\text{0,}\text{so}\text{moles}\text{can}\text{be}\text{used)}\\ \text{AB}\text{falls}\text{from}\text{0.3}\text{to}\text{0.24,}\text{so}\text{0.06}\text{mol}\text{AB}\text{reacts}\text{back,}\text{giving}\text{0.03}\text{mol}\text{each}\text{of}\\ {\text{A}}_{\text{2}}\text{and}{\text{B}}_{\text{2}}\\ {\text{B}}_{\text{2}}\text{=}\text{0.05}\text{+}\text{0.03}\text{=}\text{0.08}\text{mol}\text{(B₂}\text{takes}\text{no}\text{part}\text{in}\text{the}\text{second}\text{reaction)}\\ {\text{K}}_{\text{1}}\text{=}{\text{(0.24)}}^{\text{2}}{\text{/[(A}}_{\text{2}}\text{)(0.08)]}\text{=}\text{36,}\text{so}{\text{A}}_{\text{2}}\text{=}\text{0.0576/2.88}\text{=}\text{0.02}\text{mol}\\ {\text{A}}_{\text{2}}\text{used}\text{by}{\text{C}}_{\text{2}}\text{=}\text{0.05}\text{+}\text{0.03}\text{-}\text{0.02}\text{=}\text{0.06}\text{mol,}\text{so}\text{AC}\text{=}\text{0.12}\text{mol}\\ {\text{C}}_{\text{2}}\text{=}\text{0.1}\text{-}\text{0.06}\text{=}\text{0.04}\text{mol}\\ {\text{K}}_{\text{2}}\text{=}{\text{(0.12)}}^{\text{2}}\text{/[(0.02)(0.04)]}\text{=}\text{0.0144/0.0008}\text{=}\text{18}\\ \text{Answer}\text{is}\text{0018}\end{array}$$