Neat pure ethanol C2H5OH has an enthalpy of vaporization of 39.3 Kj/mol, and a vapor pressure of 0.308 atm at 50 C. What is the boiling point, at atmospheric pressure of 1 atm, of a solution consisting of 35 G of phenol (a non-electrolyte hydrocarbon with a molecular formula C6H6O) in 250 g of ethanol?
Do you have an answer? Is there some reason you can't use the colligative property of solutions (at least I don't see Kb listed nor a normal boiling point for ethanol so they may expect you to go another route.) Here is the best I've been able to come up with but I've not done one like this before so take the information with a grain of salt.
mols phenol = grams/molar mass.
mols ethanol = g/molar mass.
Xethanol = ?
pethanol at normal boiling point = Xethanol*760 and call that p1
Then substitute into the Clausius-Clapeyron equation. I used p2 = 760 and p1 from above.
T1 = 78.4 + 273.15 (The 78.4 is not listed in the problem so this may be off limits.)
Solve for T2. I get very close to 80 C. Interestingly, if I look up the Kb for elevation of boiling point for ethanol and go through that calculation, I also end up with a new boiling point of very close to 80 C.
To find the boiling point of the solution, we need to consider the effect of the presence of phenol on the vapor pressure of ethanol.
Step 1: Convert the mass of phenol to moles.
The molar mass of phenol (C6H6O) is:
Mass of C: 6 * 12.01 g/mol = 72.06 g/mol
Mass of H: 6 * 1.01 g/mol = 6.06 g/mol
Mass of O: 1 * 16.00 g/mol = 16.00 g/mol
Total molar mass of phenol = 72.06 + 6.06 + 16.00 = 94.12 g/mol
To convert grams to moles, use the formula:
moles = mass (g) / molar mass (g/mol)
moles of phenol = 35 g / 94.12 g/mol = 0.372 mol
Step 2: Calculate the total moles of the solution.
The moles of ethanol can be calculated using the formula:
moles = mass (g) / molar mass (g/mol)
moles of ethanol = 250 g / (46.07 g/mol) = 5.43 mol
The total moles in the solution = moles of phenol + moles of ethanol
= 0.372 mol + 5.43 mol = 5.802 mol
Step 3: Calculate the mole fraction of ethanol.
The mole fraction of a component is the ratio of moles of that component to the total moles.
mole fraction of ethanol = moles of ethanol / total moles
= 5.43 mol / 5.802 mol
= 0.936
Step 4: Calculate the vapor pressure of ethanol in the solution.
Use the equation:
ln(P1/P2) = ΔHvap/R * (1/T2 - 1/T1)
where:
P1 = vapor pressure of pure ethanol
P2 = vapor pressure of the solution
ΔHvap = enthalpy of vaporization of ethanol (39.3 kJ/mol)
R = gas constant (8.314 J/(mol * K))
T1 = boiling point of pure ethanol
T2 = boiling point of the solution
Rearranging the equation gives:
T2 = 1 / (1/T1 - (ΔHvap/R) * ln(P2/P1))
At atmospheric pressure (1 atm), the vapor pressure of pure ethanol is equal to 1 atm.
Let's assume the boiling point of pure ethanol is T1.
Step 5: Calculate the boiling point of the solution.
T2 = 1 / (1/T1 - (ΔHvap/R) * ln(P2/P1))
Substituting the given values:
ΔHvap = 39.3 kJ/mol = 39.3 * 10^3 J/mol
R = 8.314 J/(mol * K)
P2 = 1 atm
P1 = 0.308 atm
T2 = boiling point of the solution (to be calculated)
Let's calculate T2:
T2 = 1 / (1/T1 - (39.3 * 10^3 J/mol / (8.314 J/(mol * K)) * ln(1/0.308))
Using the given conditions, T2 can be calculated.
To determine the boiling point of the solution consisting of phenol and ethanol, we need to calculate the vapor pressure of the solution at 1 atm and then find the boiling point corresponding to that vapor pressure.
First, let's calculate the mole fraction of ethanol and phenol in the solution:
Mole fraction (X) = (moles of component) / (total moles of all components)
To find the moles of ethanol (C2H5OH), we can use the given mass and molar mass:
Moles of ethanol = (mass of ethanol) / (molar mass of ethanol)
= (250 g) / (46.07 g/mol) [Molar mass of ethanol = 46.07 g/mol]
= 5.426 mol
Similarly, to find the moles of phenol (C6H6O):
Moles of phenol = (mass of phenol) / (molar mass of phenol)
= (35 g) / (94.11 g/mol) [Molar mass of phenol = 94.11 g/mol]
= 0.372 mol
Next, calculate the total moles of the solution:
Total moles of solution = moles of ethanol + moles of phenol
= 5.426 mol + 0.372 mol
= 5.798 mol
Now, we can calculate the mole fraction of ethanol and phenol in the solution:
Mole fraction of ethanol = (moles of ethanol) / (total moles of solution)
= 5.426 mol / 5.798 mol
= 0.936
Mole fraction of phenol = (moles of phenol) / (total moles of solution)
= 0.372 mol / 5.798 mol
= 0.064
Using Raoult's Law, which states that the vapor pressure of a component in a solution is equal to the product of the mole fraction of that component and its vapor pressure in the pure state, we can calculate the vapor pressure of the solution.
Vapor pressure of solution = (mole fraction of ethanol) * (vapor pressure of ethanol) + (mole fraction of phenol) * (vapor pressure of phenol)
The vapor pressure of ethanol is given as 0.308 atm.
The vapor pressure of phenol can be found using its vapor pressure at boiling point (assuming it behaves ideally):
Vapor pressure of phenol = 1 atm
Now, calculate the vapor pressure of the solution:
Vapor pressure of solution = (0.936) * (0.308 atm) + (0.064) * (1 atm)
= 0.287 atm + 0.064 atm
= 0.351 atm
Finally, to find the boiling point of the solution at atmospheric pressure of 1 atm, we consult a vapor pressure versus temperature chart for ethanol.
From the chart, we find that at 1 atm, the boiling point of ethanol is approximately 78.3°C.
Therefore, the boiling point of the solution consisting of 35 g of phenol in 250 g of ethanol is around 78.3°C.