Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.600 M , [B] = 0.650 M , and [C] = 0.500 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.470 M and [C] = 0.630 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

Answer:

Given initial concentrations of [A]=0.6 M, [B]=0.65 M, [C]=0.5 M.

At equilibrium [A]=0.47 M, and [C]=0.63 M.

The equilibrium reaction is

A + 2B <------> C

Initial. 0.6 0.65 . 0.5

Change . - x . -2x . +x

Equilibrium . 0.6-x. 0.65-2x . 0.5+x

Therefore [A]=0.47 =0.6-x

x=0.6-0.47=0.13

And also [C]=0.5+x=0.63

x=0.63-0.5=0.13

Therefore [B]=0.65-2x=0.65-(2x0.13)=0.39 M.

Equilibrium constant, Kc=[C]/[A][B]^2

Kc=(0.63)/(0.47)(0.39^2)

Kc=9.63.

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