Problem 42
Question
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.
Step-by-Step Solution
Verified Answer
The normal boiling point of trichloromethane is approximately 334 K, or 61 °C.
1Step 1: 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 \(\Delta H_{vap}=29200 \,J/mol\).
2Step 2: 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 \(T_2\).
3Step 3: 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.
4Step 4: Convert \(T_2\) to Celsius
Finally, convert the temperature \(T_2\) from kelvin to degrees Celsius by subtracting 273 from it.
Key Concepts
Clausius-Clapeyron equationenthalpy of vaporizationnormal boiling point
Clausius-Clapeyron equation
Understanding the behavior of substances as they change phases is crucial in chemistry. The Clausius-Clapeyron equation is a fundamental principle that describes the relationship between vapor pressure and temperature. It is particularly useful when studying phase transitions like boiling and condensation.
The Clausius-Clapeyron equation is expressed as: \[ \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] In this formula,
The equation allows us to calculate the normal boiling point of a compound if we know its vapor pressure at a different temperature, alongside its enthalpy of vaporization. This method is incredibly useful in predicting how changes in temperature affect vapor pressure, a key factor when determining boiling points.
The Clausius-Clapeyron equation is expressed as: \[ \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] In this formula,
- \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \) respectively.
- \( \Delta H_{vap} \) is the enthalpy of vaporization.
- \( R \) is the ideal gas constant (8.314 J/mol·K).
The equation allows us to calculate the normal boiling point of a compound if we know its vapor pressure at a different temperature, alongside its enthalpy of vaporization. This method is incredibly useful in predicting how changes in temperature affect vapor pressure, a key factor when determining boiling points.
enthalpy of vaporization
Enthalpy of vaporization is the amount of energy required to transform a given quantity of a substance from the liquid phase to the gaseous phase at constant pressure. It is expressed in kilojoules per mole (kJ/mol) and is a vital concept in thermodynamics.
To comprehend enthalpy of vaporization, consider what occurs at the molecular level during phase transition:
To comprehend enthalpy of vaporization, consider what occurs at the molecular level during phase transition:
- Molecules are in a liquid state, closely packed together with stronger intermolecular forces.
- When energy is added, these forces become weaker, and molecules move apart, transitioning into a gaseous state.
- More energy is required for substances with higher enthalpy of vaporization because their intermolecular forces are stronger.
normal boiling point
The normal boiling point of a substance is the temperature at which its vapor pressure equals the atmospheric pressure of 1 atm (or 101.3 kPa at sea level). At this temperature, the substance transitions from liquid to gas, and it is a distinct and measurable property of each compound.
When considering the normal boiling point, it's important to note:
When considering the normal boiling point, it's important to note:
- It depends on atmospheric conditions and can vary with altitude or external pressure.
- Substances with higher molecular mass or stronger intermolecular forces usually have higher boiling points.
Other exercises in this chapter
Problem 39
Cyclohexanol has a vapor pressure of \(10.0 \mathrm{mmHg}\) at \(56.0^{\circ} \mathrm{C}\) and \(100.0 \mathrm{mmHg}\) at \(103.7^{\circ} \mathrm{C} .\) Calcula
View solution Problem 41
The normal boiling point of acetone, an important laboratory and industrial solvent, is \(56.2^{\circ} \mathrm{C}\) and its \(\Delta H_{\text {vap }}\) is \(25.
View solution Problem 46
Can \(\mathrm{SO}_{2}\) be maintained as a liquid under a pressure of 100 atm at \(0^{\circ} \mathrm{C} ?\) Can liquid methane be obtained under the same condit
View solution Problem 47
The normal melting point of copper is \(1357 \mathrm{K}\), and \(\Delta \mathrm{H}_{\text {fus }}\) of \(\mathrm{Cu}\) is \(13.05 \mathrm{kJ} \mathrm{mol}^{-1}\
View solution