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(B) Given reactions:$(i) \ 2 P_{(g)} + 3 Cl_{2(g)} \rightleftharpoons 2 PCl_{3(g)} \quad K_1$$(ii) \ PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(

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$K_1 K_2^2$

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(B) Given reactions:$(i) \ 2 P_{(g)} + 3 Cl_{2(g)} ightleftharpoons 2 PCl_{3(g)} \quad K_1$$(ii) \ PCl_{3(g)} + Cl_{2(g)} ightleftharpoons PCl_{5(g)} \quad K_2$Target reaction:$(iii) \ 2 P_{(g)} + 5 Cl_{2(g)} ightleftharpoons 2 PCl_{5(g)}$To obtain reaction $(iii)$,we multiply reaction $(ii)$ by $2$ and add it to reaction $(i)$:$2 	imes [PCl_{3(g)} + Cl_{2(g)} ightleftharpoons PCl_{5(g)}] \implies 2 PCl_{3(g)} + 2 Cl_{2(g)} ightleftharpoons 2 PCl_{5(g)} \quad K' = K_2^2$Adding $(i)$ and the modified $(ii)$:$(2 P_{(g)} + 3 Cl_{2(g)}) + (2 PCl_{3(g)} + 2 Cl_{2(g)}) ightleftharpoons 2 PCl_{3(g)} + 2 PCl_{5(g)}$Canceling common terms gives:$2 P_{(g)} + 5 Cl_{2(g)} ightleftharpoons 2 PCl_{5(g)}$The equilibrium constant $K$ for the final reaction is the product of the constants:$K = K_1 	imes K_2^2 = K_1 K_2^2$

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