Benzene is a hydrocarbon that is commonly used as a commercial solvent. However, it is carcinogenic; i.e., accumulations in the body can cause cancer. What is the vapor pressure of benzene at 58◦C? The normal boiling point of benzene is 80.0◦C and its molar heat of vaporization is 30.8 kJ/mol.
Answer in units of torr
Use the Clausius-Clapeyron equation. One piece of information you must know is that at the boiling point of benzene the vapor pressure is 760 mm.
To find the vapor pressure of benzene at 58°C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at one temperature to its boiling point and molar heat of vaporization:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where:
P2 = vapor pressure at the desired temperature (58°C)
P1 = vapor pressure at the boiling point (80.0°C)
ΔHvap = molar heat of vaporization (30.8 kJ/mol)
R = ideal gas constant (0.0821 L*atm/(K*mol))
T2 = temperature at which vapor pressure is desired (58°C + 273.15 = 331.15 K)
T1 = boiling point temperature (80.0°C + 273.15 = 353.15 K)
Let's substitute the given values into the equation:
ln(P2/760) = (-30.8 kJ/mol / 0.0821 L*atm/(K*mol)) * (1/331.15 K - 1/353.15 K)
Now, let's solve for ln(P2/760):
ln(P2/760) = (-30.8 kJ/mol / 0.0821 L*atm/(K*mol)) * (0.0030188 - 0.0028314)
ln(P2/760) ≈ (-374.94) * (0.0001874)
ln(P2/760) ≈ -0.07022
To find the vapor pressure (P2), we need to take the exponent of both sides of the equation:
P2/760 = e^(-0.07022)
Now, let's solve for P2:
P2 ≈ 760 * e^(-0.07022)
P2 ≈ 760 * 0.931077
P2 ≈ 707.02 torr
Therefore, the vapor pressure of benzene at 58°C is approximately 707.02 torr.