What is the delta Hvap of a liquid that has a vapor pressure of 612 torr at 77.6 degrees C and a boiling point of 94.9 degrees C at 1 atm?
To calculate the delta Hvap (enthalpy of vaporization) of a liquid, you need to use the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance at two different temperatures to its enthalpy of vaporization.
The equation is:
ln(P2/P1) = (delta Hvap/R) * (1/T1 - 1/T2)
where:
- P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively,
- R is the ideal gas constant (8.314 J/(mol·K) or 0.0821 L·atm/(mol·K)),
- delta Hvap is the enthalpy of vaporization,
- T1 and T2 are the temperatures in Kelvin.
To solve the equation, we need to convert the temperatures from Celsius to Kelvin:
T1 = 77.6 + 273.15 = 350.75 K
T2 = 94.9 + 273.15 = 368.05 K
Given:
P1 = 612 torr
P2 = 1 atm
Now we can substitute the values into the equation and solve for delta Hvap:
ln((1 atm)/(612 torr)) = (delta Hvap/0.0821) * (1/350.75 - 1/368.05)
First, convert the atmospheric pressure from torr to atm:
1 atm = 760 torr
ln(760/612) = (delta Hvap/0.0821) * (1/350.75 - 1/368.05)
Note: ln is the natural logarithm.
Now, take the natural logarithm of (760/612):
ln(760/612) ≈ 0.221
Substituting this into the equation:
0.221 = (delta Hvap/0.0821) * (1/350.75 - 1/368.05)
Now, simplify the equation:
0.221 = (delta Hvap/0.0821) * (-0.00483)
Rearrange the equation to solve for delta Hvap:
delta Hvap/0.0821 = 0.221 / -0.00483
delta Hvap ≈ (-0.00483 * 0.221) / 0.0821
delta Hvap ≈ -0.0104 kJ/mol
Therefore, the delta Hvap of the liquid is approximately -0.0104 kJ/mol. Note that the negative sign indicates that the process is exothermic.