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 use the ideal gas law and the concept of partial pressures.
1. First, let's understand the chemical equation for the reaction involving SO3:
2 SO3(g) ⇌ 2 SO2(g) + O2(g)
From this equation, we can see that 2 moles of SO3 produce 1 mole of O2 in the equilibrium mixture.
2. We need to determine the number of moles of SO3 present initially. To do this, we'll use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
In this case, the pressure (P) is 4.00 atm, and we assume the volume (V) is constant. Thus, we can rearrange the ideal gas law equation and solve for the number of moles of SO3 (n):
n = PV / RT
3. Now, we need to determine the total number of moles of gas present at equilibrium. From the chemical equation, we know that for every 2 moles of SO3, we have 1 mole of O2. Therefore, if 'x' moles of SO3 reacted, we would have 'x/2' moles of O2.
4. The total pressure at equilibrium is given as 5.00 atm. According to Dalton's law of partial pressures, the sum of the individual partial pressures of all the gases in the mixture equals the total pressure.
In this case, we have two gases in the equilibrium mixture: SO3 and O2. So, we can set up an equation using the partial pressures:
P_total = P_SO3 + P_O2
where P_total is 5.00 atm and P_SO3 is the partial pressure of SO3. Considering the stoichiometry of the reaction, we can write:
P_total = P_SO3 + (1/2) * P_O2
5. We can substitute the values into the equation and solve for the partial pressure of O2:
5.00 atm = 4.00 atm + (1/2) * P_O2
P_O2 = 2 * (5.00 - 4.00) atm
P_O2 = 2.00 atm
6. Finally, we can calculate the mole fraction of O2 (X_O2) using the definition of mole fraction:
X_O2 = n_O2 / n_total
Since we know that the mole ratio of SO3 to O2 is 2:1, the number of moles of O2 is half of the number of moles of SO3:
n_O2 = n_SO3 / 2
The total number of moles is the sum of the moles of SO3 and O2:
n_total = n_SO3 + n_O2
Now we can substitute the values to calculate the mole fraction of O2:
X_O2 = (n_O2 / n_total) = (n_SO3 / 2) / (n_SO3 + (n_SO3 / 2))
Simplifying this equation, we get:
X_O2 = n_SO3 / (3 * n_SO3 / 2) = 2 / 3
Therefore, the mole fraction of O2 in the equilibrium mixture is 2/3 or approximately 0.67.