3B eqilibrium 3A. K1 2C eqilibrium 2B. K2 C eqilibrium D. K3 K for. D equilibrium A=?
Answer (1)
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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