Question

Sulfur dioxide reacts with oxygen to produce sulfur trioxide in a gas-phase reaction.
(a) A 1.500 L flask was filled with 4.00 x 10^-2 moles of sulfur dioxide and 2.00 x 10^-2 moles of oxygen, and the reaction is initiated at 900K. At equilibrium, the flask contained 2.96 x 10^-2 moles of sulfure trioxide. What is the molarity of each substance in the flask at equilibrium?

(b) Find K. Suppose that the 1.500 L flask was initially filled with 4.00 x 10^-2 moles of sulfur trioxide. Under the same conditions, how many moles of each reactant will be present in the flask when the reaction reaches equilibrium?

Answer 1

a)

Molarity of SO2 = 4.0 x 10^-2 / 1.50 = 0.0267 M

molarity of O2 = 2.0 x 10^-2 / 1.5 = 0.0133 M

At equilibrium :

concentration of SO3 = 2.96 x 10^-2 / 1.5 = 0.01973 M

2 SO2 +     O2     ------------->   2 SO3

0.0267       0.0133                        0

0.0267 - 2x 0.0133 - x    2x

At equilibrium :

concnentration of SO3 = 0.01973 M

2 x = 0.01973 M

x = 9.87 x 10^-3

[SO2] = 0.0267 - 2x = 0.0267 - 2 x 9.87 x 10^-3

[SO3] = 0.0197 M

[SO2] = 6.97 x 10^-3 M

[O2] = 3.43 x 10^-3 M

b)

K = [SO3]^2 / [SO2]^2[O2]

    = (0.01973 )^2 / (6.97 x 10^-3)^2 x 3.43 x 10^-3

K = 2336

2 SO2 +     O2     ------------->   2 SO3

     0           0                     0.0267

2x    x       0.0267 - 2x

K = (0.0267 - 2x)^2 / (2x)^2 x

2336 = (0.0267 - 2x)^2 / 4x^3

x = 3.47 x 10^-3

[SO3] = 0.0198 M

[O2] = 3.47 x 10^-3 M

[SO2] = 6.94 x 10^-3 M