Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.350 M , [B] = 1.40 M , and [C] = 0.450 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.190 M and [C] = 0.610 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

A + 2B <==> C

Kc = [C] / [A] [B]^2
A + 2B yields C
[ A ]= 0.350 , [B] = 1.40 M , [C] = 0.450 M initial conditions

A + 2B <==> C

-x -2x x ......... amts change
0.350-x 1.40-2x 0.450 + x .......... equilibrium concentration

at equilibrium 0.350- x = 0.190 M
x = 0.16, so we know we can find eq conc of B...
1.40-2x = ?
1.40-2(0.16) = 1.08 M

know take your eq concen of A, B and C and plug them into the Kc equation

Kc = [C] / [A] [B]^2

Kc = ([0.61] / [1.08]² x [0.16] ) = 3.2686