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(C) Given reactions:$(1) \ 2PQ ightleftharpoons P_2 + Q_2\,;\,K_1 = 2.5 	imes 10^5$$(2) \ PQ + \frac{1}{2}R_2 ightleftharpoons PQR\,;\,K_2 = 5

Vedclass · Accepted Answer

$1 	imes 10^{-5}$

Vedclass · Answer

(C) Given reactions:$(1) \ 2PQ ightleftharpoons P_2 + Q_2\,;\,K_1 = 2.5 	imes 10^5$$(2) \ PQ + \frac{1}{2}R_2 ightleftharpoons PQR\,;\,K_2 = 5 	imes 10^{-3}$To obtain the target reaction $\frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 ightleftharpoons PQR$,we manipulate the given reactions:Reverse reaction $(1)$ and divide by $2$:$\frac{1}{2}P_2 + \frac{1}{2}Q_2 ightleftharpoons PQ\,;\,K_3 = \sqrt{\frac{1}{K_1}} = \sqrt{\frac{1}{2.5 	imes 10^5}} = \sqrt{4 	imes 10^{-6}} = 2 	imes 10^{-3}$Add this new reaction to reaction $(2)$:$(\frac{1}{2}P_2 + \frac{1}{2}Q_2 ightleftharpoons PQ) + (PQ + \frac{1}{2}R_2 ightleftharpoons PQR) \implies \frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 ightleftharpoons PQR$The equilibrium constant $K_{eq}$ for the target reaction is:$K_{eq} = K_3 	imes K_2 = (2 	imes 10^{-3}) 	imes (5 	imes 10^{-3}) = 10 	imes 10^{-6} = 1 	imes 10^{-5}$

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