Question

The equilibrium constant for the reaction SO2(g)+NO2(g)⇌SO3(g)+NO(g) is 3.0. Find the amount of NO2 that must be added to 2.3 mol of SO2 in order to form 1.1 mol of SO3 at equilibrium.

Answer 1

One other thing that you must state for this problem is the VOLUME in which this occurs. I will assume 1.00 liter to solve it

SO2(g) + NO2(g) = SO3(g) + NO(g)

If 1.1 mol of SO3 forms at equilibrium, then 1.1 mol of NO must also be formed.

Also, if the initial concentration of SO2 is 2.3 mol then at equilibrium its moles must be reduced to (2.3 mol - 1.1 mol = 1.2 mol).

Let the volume = 1 L

So, equilibrium concentrations are;

[SO2] = 1.2 mol /L = 1.2 M

[SO3] = 1.1 M

[NO] = 1.1 M

Let "x" represent the initial concentration of NO2 introduced to the SO2. The amount that will remain at equilibriums will be (x - 1.1) M. Now set up the equilibrium expression.

K = [SO3][NO] / [SO2][NO2]

3.1 = [ (1.1)(1.1) ] / [ (1.2) (x - 1.1) ]

3.1 = (1.21) / (1.2x - 1.32)

1.2x - 1.32 = 1.21 / 3.1

1.2x - 1.32 = 0.39

1.2x = 0.39 + 1.32

1.2x = 1.71

x = 1.71 / 1.2

x = 1.43 M

Answer: 1.43 M or 1.43 moles/Liter must be introduced (3 significant figures).