A sample of PCl5 weighing 2.69 grams is placed in a 1.000 liter flask and vaporized completely at 250 o C.
The final pressure observed at this temperature is 1.000 atmospheres. Note that some, but not all of the PCl5 decomposes according to the equation: PCl5 (g) ---------> PCl3 (g)+Cl2(g)
What are the pressures ( in atm) of PCl5 , PCl3 and Cl2 once the reaction is complete and at equilibrium ?
To determine the pressures of PCl5, PCl3, and Cl2 at equilibrium, we need to use the information given and the ideal gas law equation:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant
T is the temperature
First, we need to find the number of moles of PCl5 initially present in the flask.
Using the molar mass of PCl5 (208.25 g/mol) and the given mass (2.69 g), we can calculate the number of moles:
moles of PCl5 = mass / molar mass = 2.69 g / 208.25 g/mol = 0.0129 mol
Since PCl5 decomposes according to the equation:
PCl5 (g) -> PCl3 (g) + Cl2 (g)
We can see that the moles of PCl5 decrease while the moles of PCl3 and Cl2 increase. Let's denote the number of moles of PCl3 and Cl2 formed as 'x'.
Now, let's calculate the moles of PCl5, PCl3, and Cl2 at equilibrium:
moles of PCl5 = 0.0129 mol - x
moles of PCl3 = x
moles of Cl2 = x
Next, we can use the ideal gas law to find the pressures of each component at equilibrium.
For PCl5:
P1 * V = n1 * R * T (initial condition when PCl5 is not decomposed)
P1 = (n1 * R * T) / V
For PCl5, the initial moles (n1) is 0.0129 mol, the volume (V) is 1.000 L, the temperature (T) is 250 oC (convert to Kelvin: 250 + 273 = 523K), and the gas constant (R) is 0.0821 L.atm/mol.K. Plugging these values into the equation, we get:
P1 = (0.0129 mol * 0.0821 L.atm/mol.K * 523K) / 1.000 L = 0.557 atm
So, the pressure of PCl5 at equilibrium is 0.557 atm.
For PCl3 and Cl2, we can assume they both have the same pressure (x) at equilibrium.
For PCl3:
P2 * V = n2 * R * T (where n2 = x)
P2 = (x * 0.0821 L.atm/mol.K * 523K) / 1.000 L = 42.9x atm
For Cl2:
P3 * V = n3 * R * T (where n3 = x)
P3 = (x * 0.0821 L.atm/mol.K * 523K) / 1.000 L = 42.9x atm
Since the total pressure at equilibrium is 1.000 atm, we can sum up the partial pressures of PCl3 and Cl2 and set it equal to 1:
42.9x + 42.9x = 1.000 atm
85.8x = 1.000 atm
x = 1.000 atm / 85.8
x = 0.0116 atm
Now, we can substitute the value of x into P2 and P3:
P2 = 42.9 * 0.0116 atm ≈ 0.496 atm
P3 = 42.9 * 0.0116 atm ≈ 0.496 atm
Therefore, the pressures of PCl5, PCl3, and Cl2 at equilibrium are approximately:
PCl5: 0.557 atm
PCl3: 0.496 atm
Cl2: 0.496 atm