a sample of pcl5 weighing 2.69 placed in 1.0 l flask and completely vaporized at a tem of 250 c the pressur at this tempreture was 1.0 atm if pcl5 dissociated are partual pressur of pcl5 pcl3 and cl2 at condition
Your post isn't very clear. Words omitted etc. don't help decipher the problem. BTW, it's chemistry.
2.69 what? I'll assume 2.69 grams.
mols PCl5 = g/molar mass = 2.69/about 209 = approx 0.013 but you need to be more accurate with the 209 as well as all other calculations that follow that.number, Convert mols to Pressure PCl5 using PV = nRT and P = approx 0.55.
...........PCl5 ==> PCl3 + Cl2
I............0.55............0.........0
C............-p.............p........p
E.........0.55-p...........p.........p
Then Ptotal = 1 = 0.55-p + p + p
Solve for p and 0.55-p. Remember to go through this yourself for better accuracy.
To determine the partial pressures of PCl5, PCl3, and Cl2 at the given conditions, we need to use the ideal gas law and the stoichiometry of the reaction.
The balanced equation for the dissociation of PCl5 is:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Given:
- Mass of PCl5 = 2.69 g
- Volume of the flask = 1.0 L
- Temperature = 250 °C = 523.15 K
- Pressure at this temperature = 1.0 atm
Step 1: Calculate the number of moles of PCl5 using the given mass.
Molar mass of PCl5 = 208.25 g/mol
Number of moles of PCl5 = Mass / Molar mass = 2.69 g / 208.25 g/mol
Step 2: Use the ideal gas law to calculate the total number of moles in the flask.
Ideal gas law: PV = nRT
P = Pressure = 1.0 atm
V = Volume = 1.0 L
R = Ideal gas constant = 0.0821 L·atm/(mol·K)
T = Temperature in Kelvin = 523.15 K
Rearranging the equation, we can solve for n (number of moles):
n = (P * V) / (R * T)
Step 3: Determine the moles of PCl3 and Cl2 produced based on the stoichiometry of the reaction.
From the balanced equation, we know that 1 mole of PCl5 dissociates to produce 1 mole of PCl3 and 1 mole of Cl2.
So, the number of moles of PCl3 and Cl2 is equal to the number of moles of PCl5.
Step 4: Calculate the partial pressures of each component.
Partial pressure = (Number of moles / Total number of moles) * Total pressure
For PCl5:
Partial pressure of PCl5 = (moles of PCl5 / total moles) * pressure
For PCl3:
Partial pressure of PCl3 = (moles of PCl3 / total moles) * pressure
For Cl2:
Partial pressure of Cl2 = (moles of Cl2 / total moles) * pressure
Now you can substitute the values you calculated in steps 1, 2, and 3 to find the partial pressures of PCl5, PCl3, and Cl2 at the given conditions.