The vapor pressure of dichloromethane, CH_2Cl_2, at 0 degrees celcius is 134 mmHg. The normal boiling point of dichloromethane is 40 degrees celcius.
Calculate its molar heat of vaporization (in kJ/mol).
Use the Clausius-Clapeyron equation. You have the one pressure point with temperature, the other one you may use the boiling point where the vapor pressure will be 760 torr. Note the correct spelling of celsius.
30.691 Kjule/mole
To calculate the molar heat of vaporization (ΔHvap) of dichloromethane, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
where:
P1 = initial vapor pressure (in Pa)
P2 = final vapor pressure (in Pa)
ΔHvap = molar heat of vaporization (in J/mol)
R = gas constant (8.314 J/(mol*K))
T1 = initial temperature (in K)
T2 = final temperature (in K)
First, we need to convert the vapor pressure from mmHg to Pa. Since 1 mmHg = 133.322 Pa, the vapor pressure of dichloromethane at 0°C is:
P1 = 134 mmHg * 133.322 Pa/mmHg = 17896.02 Pa
Next, we convert the temperatures to Kelvin:
T1 = 0°C + 273.15 K = 273.15 K
T2 = 40°C + 273.15 K = 313.15 K
Now we can solve for ΔHvap. Rearranging the equation, we have:
ΔHvap = -(R * ln(P2/P1)) / (1/T2 - 1/T1)
We can plug in the known values:
P2 = atmospheric pressure at the normal boiling point of dichloromethane (which is 760 mmHg or 101325 Pa)
ΔHvap = -(8.314 J/(mol*K) * ln(101325 Pa / 17896.02 Pa)) / (1/313.15 K - 1/273.15 K)
Calculating this value will give us the molar heat of vaporization of dichloromethane in J/mol. To convert it to kJ/mol, we divide by 1000:
ΔHvap (kJ/mol) = ΔHvap (J/mol) / 1000
By following these steps and performing the calculations, you will find the molar heat of vaporization of dichloromethane at its normal boiling point.