Question

For the reaction SO₂(g) + NO₂(g) == NO(g) + SO₃(g), Kc = 85.0 at 460

Answer 1

Part A

                    SO2(g) + NO2(g) ---> NO(g) + SO3(g)
Initial concentration.        0.05    0.05M       0           0
Change in conc              - x           -x                   +x          +x
At equilibrium                 0.05-x,    0.05-x          x,          x

Kc = [SO3][NO] / [SO2][NO2]
= x X x /(0.05-x)(0.05-x) = 85.0
x^2 / 0.0025 -0.1x +x^2 = 85

x^2 = 0.2125 - 8.5x +85x^2

0.2125 - 8.5x +84x^2 = 0

On solving
And our answers are x= 0.056 or 0.045
If we have x = 0.056,the final concentration of the reactants will become negative, so we discard x = 0.05M, so         x = 0.045

So equilbrium concentration will be:

[SO2] = [NO2] = 0.05-0.045 = 0.005M

[NO] = [SO3] = 0.045M

Part B:

                                    SO2(g) + NO2(g) ---> NO(g) + SO3(g)
Initial concentration.    0.00831    0.00831M       0           0
Change in conc              - x           -x                   +x          +x
At equilibrium                 0.00831-x,    0.00831-x          x,          x

Kc = [SO3][NO] / [SO2][NO2]
= x X x /(0.00831-x)(0.00831-x) = 85.0
x^2 / 0.000069 -0.0166x +x^2 = 85

x^2 = 0.00586 - 1.411x +85x^2

0.00586 - 1.411x +84x^2 = 0

On solving
And our answers are x= 0.00928 or 0.00751
If we have x =0.00928 ,the final concentration of the reactants will become negative, so we discard x = 0.00928M, so         x = 0.00751

So equilbrium concentration will be:

[SO2] = [NO2] = 0.00831-0.00751 = 0.0008M

[NO] = [SO3] = 0.00751M