The system N2O4 <--->2NO2 maintained in a closed vessel at 60º C & a pressure of 5 atm has an average (i.e. observed) molecular weight of 69, calculate Kp.
At what pressure at the same temperature would the observed molecular weight be (230/3)
I believe that is 15 atm for Ptotal and not 1.5.
I would do this.
Let x = fraction of gas that is NO2
The 1-x = fraction that is N2O4
molar mass NO2 = 46
molar mass N2O4 = 92
46(x) + (92(1-x) = 69
Solve for x and I get 0.5 which is the mole fraction of these gases.
Then pNO2 = XNO2*Ptotal
and pN2O4 = XN2O4*Ptotal
Finally, Kp = p^2NO2/pN2O4
I think Kp = 2.5 but you need to confirm that.
Part 2 is similar.
46(x) + 92(1-x) = 230/3 = 76.67
x = about 0.333 = fraction NO2 = XNO2
1-x = about 0.666 = XN2O4
Then pNO2 = XNO2*Ptotal
and pN2O4 = XN2O4*Ptotal
Substitute pNO2 and pN2O4 into Kp and solve for Ptotal
Kp = p^2NO2/pN2O4 = (0.333P)^2/(0.667P)
and I solved for Ptotal as 1.5 atm. You should confirm that.
Answer is 15 atm lol
To calculate Kp, we can use the ideal gas law equation:
PV = nRT
Where:
- P is the pressure in atm
- V is the volume in liters (assumed to be constant in this case)
- n is the number of moles
- R is the ideal gas constant (0.0821 L·atm/mol·K)
- T is the temperature in Kelvin
First, let's calculate the number of moles in the given system:
For N2O4:
1 mole of N2O4 consists of 2 moles of NO2.
Since the molecular weight of N2O4 is 92 g/mol, the mass of N2O4 in the system is 69 g.
Using the formula:
moles = mass / molecular weight
moles of N2O4 = 69 g / (92 g/mol) = 0.75 mol
moles of NO2 = 2 * 0.75 mol = 1.5 mol
Now, let's find the volume of the closed vessel. Since the volume is not given, we can assume it to be 1 liter (since it's not mentioned whether the volume changes or not).
Using the ideal gas law equation, we can rearrange it to solve for P:
P = nRT / V
P = (1.5 mol) * (0.0821 L·atm/mol·K) * (333 K) / 1 L
P = 40 atm
So, at a temperature of 60 ºC (333 K) and pressure of 5 atm, the system N2O4 <--> 2NO2 has an observed pressure of 40 atm.
To find the pressure at the same temperature for an observed molecular weight of (230/3), we need to calculate the moles of N2O4 and NO2 for this new molecular weight.
Using the formula (moles = mass / molecular weight), moles of N2O4 = (69 g) / (92 g/mol) = 0.75 mol
Let x represent the number of moles of NO2.
Since the molecular weight of NO2 is 46 g/mol, the total mass of NO2 is (230/3) g.
We can set up the following equation:
(46 g/mol) * x = (230/3) g
Solving for x:
x = (230/3) g / (46 g/mol)
x = 5 mol
So, at the same temperature of 60 ºC (333 K), the pressure where the observed molecular weight is (230/3) is given by:
P = nRT / V
P = (5 mol) * (0.0821 L·atm/mol·K) * (333 K) / 1 L
P = 13.495 atm
Therefore, at the same temperature, the pressure required for the observed molecular weight to be (230/3) is approximately 13.495 atm.