Question

Chloroform, \(\mathrm{CHCl}_{3},\) was once used as an anesthetic. In spy movies it is the liquid put in handkerchiefs to render victims unconscious. Its vapor pressure is \(197 \mathrm{~mm} \mathrm{Hg}\) at \(23^{\circ} \mathrm{C}\) and \(448 \mathrm{~mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). Estimate its (a) heat of vaporization. (b) normal boiling point.

Short answer

Question: Determine the heat of vaporization and the normal boiling point of chloroform (CHCl3). Answer: The heat of vaporization for chloroform is approximately 29.4 kJ/mol, and its normal boiling point is approximately 58.2 °C.

Step-by-step solution

Convert temperatures to Kelvin

To use the Clausius-Clapeyron equation, the temperatures must be in units of Kelvin. Convert the given temperatures from Celsius to Kelvin: $$T_1 = 23^{\circ}\mathrm{C} + 273.15 = 296.15\mathrm{K}$$ $$T_2 = 45^{\circ}\mathrm{C} + 273.15 = 318.15\mathrm{K}$$

Calculate heat of vaporization

Using the Clausius-Clapeyron equation, we can now solve for \(\Delta H_{vap}\): $$\ln\frac{448}{197}=-\frac{\Delta H_{vap}}{8.314}\left(\frac{1}{318.15}-\frac{1}{296.15}\right)$$ Rearrange and solve for \(\Delta H_{vap}\): $$\Delta H_{vap}=-8.314\left(\ln\frac{448}{197}\right)\left(\frac{1}{318.15}-\frac{1}{296.15}\right)^{-1}=29369.1\mathrm{~J/mol}$$ (a) The heat of vaporization for chloroform is approximately \(29.4\mathrm{~kJ/mol}\).

Calculate normal boiling point

The normal boiling point is defined as the temperature at which the vapor pressure is equal to 1 atmosphere (or 760 mmHg). Setting \(P_2 = 760\mathrm{~mm} \mathrm{Hg}\) and using the known vapor pressure and temperature \((P_1, T_1)\) and \(\Delta H_{vap} \approx 29.4\mathrm{~kJ/mol}\), we can rearrange the Clausius-Clapeyron equation to find \(T_2\): $$\ln\frac{760}{197}=-\frac{29369.1}{8.314}\left(\frac{1}{T_2}-\frac{1}{296.15}\right)$$ Rearrange and solve for \(T_2\): $$T_2 = \frac{1}{\frac{1}{296.15}-\frac{8.314}{29369.1}\ln\frac{760}{197}} = 331.39\mathrm{K}$$ Convert the boiling point in Kelvin to Celsius: $$T_2 = 331.39\mathrm{K} - 273.15 = 58.24^{\circ}\mathrm{C}$$ (b) The normal boiling point of chloroform is approximately \(58.2^{\circ}\mathrm{C}\).

Key concepts

Vapor Pressure

When liquid substance is placed in a closed container, it will evaporate until an equilibrium is reached where an equal amount of molecules vaporize and condense at the same rate. The pressure exerted by the vapor in equilibrium with its liquid at a given temperature is known as the vapor pressure.

Vapor pressure is influenced by the temperature; as it increases, more molecules have enough kinetic energy to escape into the gas phase, thus increasing the vapor pressure. This key concept is important in understanding how substances transition from liquid to gas.

For example, chloroform's vapor pressure at two different temperatures provides crucial data that allows us to use the Clausius-Clapeyron equation to investigate its thermodynamic properties, such as the heat of vaporization and the normal boiling point.

Heat of Vaporization

The heat of vaporization, denoted by abla H_{vap}, is the amount of energy required to convert one mole of a liquid at its boiling point into vapor without an increase in temperature. This value is intrinsic to a substance and is essential for calculations regarding phase changes.

In the provided exercise, the Clausius-Clapeyron equation helps determine the heat of vaporization by relating the change in vapor pressure with temperature. Through logarithmic manipulation and by knowing two points of vapor pressure and corresponding temperatures, we can calculate the energy needed for chloroform to transition from its liquid state to a vapor.

This calculation aids in understanding the thermal dynamics of the substance and is critical in industrial applications where temperature and pressure conditions are pivotal.

Normal Boiling Point

The normal boiling point is the temperature at which a liquid's vapor pressure is equal to the standard atmospheric pressure at sea level (1 atm or 760 mmHg). It's a particular case of boiling point that is used as a standard reference.

Using the Clausius-Clapeyron equation, we can find this value by knowing the heat of vaporization and a vapor pressure at a given temperature. By setting the second pressure to 1 atmosphere, and solving for the unknown temperature, we find the normal boiling point of the substance.

For chloroform, given its vapor pressure at one such temperature, we apply the Clausius-Clapeyron equation to estimate its normal boiling point. Knowledge of a substance's normal boiling point is very important, for example, in the field of forensic science where the presence of chloroform might need to be detected under normal atmospheric conditions.