Question

A mixture initially contains A, B, and C in the following concentrations:

[A] = 0.650 M , [B] = 1.15 M , and [C] = 0.300 M .

The following reaction occurs and equilibrium is established:

A+2B⇌C

At equilibrium, [A] = 0.550 M and [C] = 0.400 M . Calculate the value of the equilibrium constant, Kc.

Answer #1

construct ICE table

A + 2B <====> C

I 0.65 1.15 0.3

C -x -2x +x

E 0.65-x 1.15-2x 0.3-x

equilibrium expression Kc = [C] / [A][B]^{2}

he has given the equilibrium concentration of A and C use one of its value and find out the x

from ICE table [A] = 0.65-x = 0.550

x = 0.65 - 0.55 = 0.1

equilibrium concentration of B

[B ] = 1.15-2x = 1.15 - 2(0.1) = 1.15 - 0.2 = 0.95

now we have all the equilibrium concentration substitute in the above equation

Kc = [0.4] / [0.55][0.95]^{2}

Kc = 0.4 / 0.4963

Kc = 0.806

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