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3B eqilibrium 3A. K1 2C eqilibrium 2B. K2 C eqilibrium D. K3 K for. D equilibrium A=?


nitin.verma 13th Feb, 2021
Answer (1)
ARYAN SAGAR 13th Feb, 2021

Dear student,

Equilibrium Constant ( k ) = [products]^x / [reactants]^y  { where x and y are stoichiometric coefficients of products and reactants respectively}.

Now, coming onto the question :-

K1 = ([A]/[B])^3

So, [B] = [A]/(K1)^1/3

K2 = ([B]/[C])^2

So, [C] = [B]/(K2)^1/2 = [A]/((k1)^1/3 (k2)^1/2)

Now, K3 = [D]/[C]

So, [D] = [C]k3 = [A] K3/ ((k1)^1/3 (k2)^1/2)

Now, as D is in equilibrium with A ,

So, K = [D]/[A]

K= [A] K3/ [A] ((k1)^1/3 (k2)^1/2)

So, the equilibrium constant for this reaction is:-

K = K3/ ((k1)^1/3 (k2)^1/2)

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