The boiling point of ethanol is 78.5C and delta H vap = 40.5kJ/mol. Calculate the vapor pressure (in torr) of ethanol at 0.0C ?
Use the Clausius-Clapeyron equation. The vapor pressure of ANY substance at its boiling point is atmospheric pressure, so if we assume standard conditions, the vp of ethanol at its boiling point will be 760 mm Hg. Don't forget to use Kelvin for temperature.
To calculate the vapor pressure of ethanol at 0.0°C, we can make use of the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its boiling point and enthalpy of vaporization.
The equation is as follows:
ln(P2/P1) = [ΔHvap/R] * [(1/T1) - (1/T2)]
Where:
P1 is the vapor pressure at the first temperature (in this case, 78.5°C)
P2 is the vapor pressure at the second temperature (which we want to solve for)
ΔHvap is the enthalpy of vaporization (40.5 kJ/mol)
R is the gas constant (8.314 J/(mol K))
T1 and T2 are the temperatures in Kelvin at P1 and P2, respectively.
First, let's convert the given temperatures from Celsius to Kelvin:
T1 (boiling point) = 78.5°C + 273.15 = 351.65 K
T2 (0.0°C) = 0.0°C + 273.15 = 273.15 K
Now we can substitute the values into the Clausius-Clapeyron equation:
ln(P2/1 atm) = [(40.5 kJ/mol) / (8.314 J/(mol K))] * [(1/351.65 K) - (1/273.15 K)]
We divide ΔHvap by 8.314 to convert kJ to J and then solve for the natural logarithm of P2 divided by 1 atm (since vapor pressure is often reported relative to atmospheric pressure).
Next, rearrange the equation to solve for P2:
P2/1 atm = e^([(40.5 kJ/mol) / (8.314 J/(mol K))] * [(1/351.65 K) - (1/273.15 K)])
P2 = 1 atm * e^([(40.5 kJ/mol) / (8.314 J/(mol K))] * [(1/351.65 K) - (1/273.15 K)])
Finally, substitute the values into the equation and solve:
P2 = 1 atm * e^([0.00487] * [0.00306134])
P2 = 1 atm * e^(0.000014903)
Using a scientific calculator or an online calculator, calculate e^0.000014903, which gives approximately 1.000014919.
P2 = 1 atm * 1.000014919
Now convert from atmospheres to torr:
1 atm = 760 torr
P2 = 760 torr * 1.000014919
P2 ≈ 760.0101 torr
Therefore, the vapor pressure of ethanol at 0.0°C is approximately 760.0101 torr.