Enough SO3 is added to an evacuated container so its initial pressure is 4.00 atm. Calculate the mole fraction of O2 in the equilibrium mixture if the total pressure at equilibrium is 5.00 atm.

To calculate the mole fraction of O2 in the equilibrium mixture, we need to know the number of moles of O2 and the total number of moles in the system.

To get started, let's assume that x moles of O2 are present in the equilibrium mixture. Since the total pressure at equilibrium is 5.00 atm, and SO3 is a stoichiometric reactant with O2 in a 2:1 ratio, the partial pressure of O2 is 2x (twice the mole fraction of O2).

Now, let's consider the initial pressure in the container, which is 4.00 atm. Since SO3 is the only component initially present, the initial partial pressure of SO3 is 4.00 atm. As the reaction proceeds to equilibrium, some of the SO3 will react to produce O2, resulting in a decrease in the partial pressure of SO3.

According to the reaction stoichiometry, for every 2 moles of SO3 that react, 1 mole of O2 is produced. Therefore, the change in the partial pressure of SO3 will be twice the change in the partial pressure of O2. Let's denote this change as ∆P.

Since the initial partial pressure of O2 is 0 (because there is no O2 initially present), we can write the equilibrium partial pressure of O2 as 2x - ∆P.

Now, let's consider the reaction as it approaches equilibrium. The total pressure at equilibrium is given as 5.00 atm, which is the sum of the partial pressures of SO3 and O2. Therefore, we can write the equation:

(4.00 - ∆P) + (2x - ∆P) = 5.00

Simplifying, we get:

6.00 - 2∆P + 2x = 5.00

Rearranging the equation:

2x - 2∆P = -1.00

Since we are interested in the mole fraction of O2, let's express the moles of O2 and the total moles in terms of x:

moles of O2 = x
total moles = x + moles of SO3

The mole fraction of O2 (denoted as XO2) is given by the ratio of moles of O2 to the total moles:

XO2 = moles of O2 / total moles

Substituting the values of moles of O2 and total moles:

XO2 = x / (x + moles of SO3)

We know that the mole ratio of SO3 to O2 is 2:1, so the moles of SO3 is twice the moles of O2:

moles of SO3 = 2 * x

Substituting the value of moles of SO3:

XO2 = x / (x + 2 * x)

Simplifying:

XO2 = x / (3 * x)

Now, let's substitute the expression for XO2 in terms of ∆P into the equation we derived earlier:

2x - 2∆P = -1.00

Substituting the value of X in terms of x:

2 * (XO2 * total moles) - 2 * ∆P = -1.00

Substituting the value of total moles in terms of x:

2 * (XO2 * (x + 2 * x)) - 2 * ∆P = -1.00

Simplifying:

2 * (x / (3 * x)) * (3 * x) - 2 * ∆P = -1.00

Cancelling out the common terms:

2 - 2 * ∆P = -1.00

Solving for ∆P:

∆P = (2 + 1.00) / 2

∆P = 1.50

Now, let's substitute the value of ∆P back into the equation for XO2:

XO2 = x / (x + 2 * x)

XO2 = x / (3 * x)

XO2 = 1 / 3

Therefore, the mole fraction of O2 in the equilibrium mixture is 1/3.