Calculate the ÄHvap , in kJ/mol, for ethanol, given that the vapor pressure at 45.0°C is 155.0 torr. The normal boiling point of ethanol is 78°C. Express your answer to two decimal places.
Use the Clausius-Clapeyron equation.
p1 = 155 torr
T1 = 45.0C
p2 = 760 torr
T2 = 78 C
change dHvap to J/mol
To calculate the enthalpy of vaporization (ΔHvap) for ethanol, we can use the Clausius-Clapeyron equation. The Clausius-Clapeyron equation relates the vapor pressure of a substance at one temperature to its vapor pressure at another temperature and the enthalpy of vaporization.
The Clausius-Clapeyron equation is given as:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
R = ideal gas constant, 8.314 J/(mol·K)
T1 = temperature in Kelvin
T2 = temperature in Kelvin
First, we need to convert the given temperatures to Kelvin.
45.0°C = 45.0 + 273.15 = 318.15K
78°C = 78.0 + 273.15 = 351.15K
Next, let's plug in the given values into the Clausius-Clapeyron equation:
ln(155.0 torr/P1) = -ΔHvap/(8.314 J/(mol·K)) * (1/351.15K - 1/318.15K)
Now, we can solve for ΔHvap. Rearrange the equation:
ΔHvap = -R * (T2 - T1) / (ln(P2/P1))
Substitute the values:
ΔHvap = -(8.314 J/(mol·K)) * (351.15K - 318.15K) / ln(155.0 torr/P1)
Note: We have an unknown variable in the equation, which is P1. However, since the problem does not provide P1, we can assume that the vapor pressure is very low compared to P2, so we can approximate P1 as zero. Hence, ln(155.0 torr/P1) can be approximated as ln(155.0 torr/0) = ln(infinity) = infinity.
Therefore, the equation simplifies to:
ΔHvap ≈ -(8.314 J/(mol·K)) * (351.15K - 318.15K) / infinity
As we divide by infinity, we can approximate the final equation as:
ΔHvap ≈ 0
Hence, the enthalpy of vaporization for ethanol is approximately 0 kJ/mol.