Question

SO3(g) + NO(g) NO2(g) + SO2(g) Kc was found to be 0.500 at a certain temperature. A reaction mixture is prepared in which 0.375 mol NO2 and 0.375 mol of SO2 are placed in a 4.00 L vessel. After the system reaches equilibrium what will be the equilibrium concentrations of all four gases? How do these equilibrium values compare to the values when 0.750 mol of SO₃ and 0.750 mol of NO are placed in a 8.00 L container and allowed to react.

Answer 1

Kc = [NO2][SO2]/[SO3][NO] = 0.500

molar concentration [NO2] = [SO2] = moles/L = 0.375/4 = 0.094 M

                      NO     +     SO3   ----------->      NO2        +        SO2

initial                -                -                            0.094                 0.094

change           -x              -x                              +x                      +x

equilibrium      x                x                          0.094-x              0.094-x

So we have,

0.5 = (0.094-x)(0.094-x)/x^2

0.5x^2 = 8.836 x 10^-3 - 0.188x + x^2

0.5x^2 - 0.188x + 8.836 x 10^-3 = 0

x = 0.055 M

So the equilibrium concentrations of,

[SO3] = 0.055 M

[NO] = 0.055 M

[NO2] = 0.094 - 0.055 = 0.039 M

[SO2] = 0.039 M

Now If the initial concentrations for,

SO3 = 0.750 mol

NO = 0.750 mol

Molar concentrations are,

[SO3] = 0.750/8 = 0.094 M

[NO] = 0.094 M

let x be the change in cocentration at equilibrium,

0.5 = x^2/(0.094-x)^2

0.5(8.836 x 10^-3 - 0.188x + x^2) = x^2

4.418 x 10^-3 - 0.094x + 0.5x^2 = x^2

0.5x^2 + 0.094x - 4.418 x 10^-3 = 0

x = 0.039 M

So the equilibrium concentrations becomes,

[SO3] = 0.094 - 0.039 = 0.055 M

[NO] = 0.055 M

[SO2] = 0.039 M

[NO2] = 0.039 M