Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.450 M , [B] = 0.600 M , and [C] = 0.350 M . The following reaction occurs and equilibrium is established:

A+2B⇌C

At equilibrium, [A] = 0.290 M and [C] = 0.510 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

The given chemical transformation,

              A+2B⇌C

Kc for this reaction is given as,

Kc = [C] / {[A][B]²}…………..(1)

Where [A], [B], [C] are equilibrium concentrations.

We know the equilibrium concentration of A and C species but of B we have to find out.

a)Finding of equilibrium concentration of B.

From the stoichiometry or reaction its clear that 1 mole of B react with 2 moles of B to give the 1 mole product C.

This means, Concentration of B consumed = 2 x concentration of A consumed.

Let us find out concentration of A consumed,

Initial concentration of A = 0.450 M

Equilibrium concentration of A = 0.290 M

Hence concentration of A consumed = 0.450 – 0.290 = 0.160 M

And hence concentration of B consumed = 2 x 0.16 = 0.320 M

Initial concentration of B = 0.600 M

Hence equilibrium concentration of B ([B]) = Initial concentration – Concentration consumed

[B] = 0.600 – 0.320 = 0.28 M

[B] = 0.28 M…….. i.e. equilibrium concentration of B

b) calculation of Kc for given reaction,

We have, [A]= 0.290 M, [B]= 0.280, [C]= 0.510 M

Let us put all these values in eq.1

Kc = (0.510) / (0.290)(0.280)²

Kc = (0.510) / (0.290)(0.0784)

Kc = 22.431

Equilibrium constant Kc for given reaction is 22.431.

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