The vapor pressure of trichloromethane (chloroform) is 40.0 Torr at \(-7.1^{\circ} \mathrm{C}\). Its enthalpy of vaporization is \(29.2 \mathrm{kJ} \mathrm{mol}^{-1} .\) Calculate its normal boiling point.

Short Answer

Expert verified
The normal boiling point of trichloromethane is approximately 334 K, or 61 °C.

Step by step solution

01

Convert units

First, convert the given units to be consistent with the SI units. Convert the given pressure from Torr to atmospheres by dividing 40.0 Torr by 760. Also, convert the enthalpy from kJ to J by multiplying 29.2 kJ by 1000. Finally, convert the temperature from Celsius to Kelvin by adding 273 to -7.1. Then we have: \(P_1=0.0526 \,atm\), \(T_1=266 K\), and \(\Delta H_{vap}=29200 \,J/mol\).
02

Use Clausius-Clapeyron equation

Next, use the Clausius-Clapeyron equation to link the variables: \(\ln \left(\frac{P_2}{P_1}\right)=-\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\). Since the normal boiling point of a substance is defined as the temperature at which its vapor pressure is 1 atm, we have \(P_2=1 \,atm\). Now plug all values into this equation and solve for \(T_2\).
03

Solve for \(T_2\)

Calculate \(T_2\) by rearranging the formula and isolating \(T_2\): \(T_2= \left(\frac{\Delta H_{vap}}{R \cdot \ln(P_2 / P_1)} + \frac{1}{T_1} \right)^{-1}\). Plug the known values into this equation and solve to get the value of \(T_2\) in K.
04

Convert \(T_2\) to Celsius

Finally, convert the temperature \(T_2\) from kelvin to degrees Celsius by subtracting 273 from it.

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