If K_1, K_2, K_3 are equilibrium constants fo | vedclass

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(B) Given reactions:$1) A + D \rightleftharpoons AD, K_1 = \frac{[AD]}{[A][D]}$$2) AD + D \rightleftharpoons AD_2, K_2 = \frac{[AD_2]}{[AD][D]}$$3) AD

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$\log K_1 + \log K_2 + \log K_3 = \log K$

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(B) Given reactions:$1) A + D ightleftharpoons AD, K_1 = \frac{[AD]}{[A][D]}$$2) AD + D ightleftharpoons AD_2, K_2 = \frac{[AD_2]}{[AD][D]}$$3) AD_2 + D ightleftharpoons AD_3, K_3 = \frac{[AD_3]}{[AD_2][D]}$Adding these three equations gives the net reaction:$A + 3D ightleftharpoons AD_3$The equilibrium constant $K$ for the net reaction is the product of the individual equilibrium constants:$K = K_1 	imes K_2 	imes K_3$Taking the logarithm on both sides:$\log K = \log(K_1 	imes K_2 	imes K_3)$$\log K = \log K_1 + \log K_2 + \log K_3$

If $K_1, K_2, K_3$ are equilibrium constants for the formation of $AD, AD_2, AD_3$ respectively as follows: $A + D \rightleftharpoons AD$,$AD + D \rightleftharpoons AD_2$,$AD_2 + D \rightleftharpoons AD_3$. Then the equilibrium constant '$K$' for $A + 3D \rightleftharpoons AD_3$ is related as:

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