Chapter 10: Problem 93
Carbon tetrachloride, \(\mathrm{CCl}_{4}\), has a vapor pressure of 213 torr at \(40 .{ }^{\circ} \mathrm{C}\) and 836 torr at \(80 .{ }^{\circ} \mathrm{C}\). What is the normal boiling point of \(\mathrm{CCl}_{4}\) ?
Short Answer
Expert verified
The normal boiling point of carbon tetrachloride (CCl4) is approximately \( 76.73 ^{\circ} \mathrm{C} \).
Step by step solution
01
Convert temperatures to Kelvin
First, we need to convert the temperatures from Celsius to Kelvin; we do this by adding 273.15 to each temperature:
\( T_1 = 40 ^{\circ}\mathrm{C} + 273.15 = 313.15 \mathrm{K} \)
\( T_2 = 80 ^{\circ}\mathrm{C} + 273.15 = 353.15 \mathrm{K} \)
02
Plug values into the Clausius-Clapeyron equation and solve for \(\Delta H_{vap}\)
We have \( P_1 = 213 \) torr and \( P_2 = 836 \) torr at temperatures \( T_1 = 313.15 \) K and \( T_2 = 353.15 \) K, respectively. Now we plug these values into the Clausius-Clapeyron equation and solve for \( \Delta H_{vap} \):
\( \ln(\frac{836}{213}) = -\frac{\Delta H_{vap}}{8.314}(\frac{1}{353.15} - \frac{1}{313.15}) \)
Now, we can solve for \( \Delta H_{vap} \):
\( \Delta H_{vap} = 8.314 \times \frac{-\ln(\frac{836}{213})}{(\frac{1}{353.15} - \frac{1}{313.15})} \)
\( \Delta H_{vap} \approx 29.86 \mathrm{kJ/mol} \)
03
Use Clausius-Clapeyron to find the boiling point
Now that we have \( \Delta H_{vap} \), we can find the boiling point, which is the temperature at which the vapor pressure is equal to 760 torr (i.e., normal atmospheric pressure). Let that temperature be \( T_b \). Using the Clausius-Clapeyron equation again:
\( \ln(\frac{760}{213}) = -\frac{29.86 \times 10^3 \mathrm{J/mol}}{8.314\mathrm{J/mol·K}}(\frac{1}{T_b} - \frac{1}{313.15\mathrm{K}}) \)
Now, we solve for the boiling point, \( T_b \):
\( T_b = \frac{1}{ (\frac{8.314 \times \ln(\frac{760}{213})}{-29.86 \times 10^3} + \frac{1}{313.15}) }\)
\( T_b \approx 349.88 \mathrm{K} \)
04
Convert the boiling point to Celsius
Finally, we will convert the boiling point from Kelvin to Celsius:
\( T_b = 349.88 \mathrm{K} - 273.15 \)
\( T_b \approx 76.73 ^{\circ} \mathrm{C} \)
The normal boiling point of carbon tetrachloride (CCl4) is approximately 76.73°C.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Clausius-Clapeyron Equation
The Clausius-Clapeyron equation is a fundamental principle in thermodynamics that describes the relationship between the temperature and pressure of a substance at its phase transition points. Specifically, when a substance transitions between its liquid and gas phases, the equation helps predict how its vapor pressure will vary with temperature. It is represented as:
\[ \text{ln}(P_2/P_1) = -\frac{\triangle H_{vap}}{R} \times \bigg(\frac{1}{T_2} - \frac{1}{T_1}\bigg) \]
Where:\
\[ \text{ln}(P_2/P_1) = -\frac{\triangle H_{vap}}{R} \times \bigg(\frac{1}{T_2} - \frac{1}{T_1}\bigg) \]
Where:\
- \(P_1\) and \(P_2\) are the vapor pressures at two different temperatures \(T_1\) and \(T_2\), respectively.
- \(\triangle H_{vap}\) is the enthalpy of vaporization, which is the energy needed to convert a unit mass of liquid into gas at a constant temperature.
- \(R\) is the universal gas constant.
Demystifying Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. All liquids have a specific vapor pressure, which increases as the temperature rises, leading to more molecules escaping from the liquid to the gas phase.
In simpler terms, vapor pressure is an indicator of a liquid's evaporation rate. A high vapor pressure means the liquid easily evaporates at lower temperatures, whereas a low vapor pressure indicates the liquid is less prone to evaporate. This property plays a crucial role in determining the boiling point of a substance, as boiling occurs when the vapor pressure equals the ambient, or atmospheric, pressure. By understanding vapor pressure, we can infer that as external pressure changes, so does the boiling point of a liquid. The normal boiling point refers to the temperature at which a liquid's vapor pressure equals 1 atmosphere (760 torr or 101.3 kPa).
In simpler terms, vapor pressure is an indicator of a liquid's evaporation rate. A high vapor pressure means the liquid easily evaporates at lower temperatures, whereas a low vapor pressure indicates the liquid is less prone to evaporate. This property plays a crucial role in determining the boiling point of a substance, as boiling occurs when the vapor pressure equals the ambient, or atmospheric, pressure. By understanding vapor pressure, we can infer that as external pressure changes, so does the boiling point of a liquid. The normal boiling point refers to the temperature at which a liquid's vapor pressure equals 1 atmosphere (760 torr or 101.3 kPa).
Enthalpy of Vaporization Explained
Enthalpy of vaporization, denoted as \(\triangle H_{vap}\), is a pivotal concept in physical chemistry and is a measure of the energy required to transform one mole of a liquid into gas at a constant temperature and pressure. It's essentially the energy needed to break intermolecular forces in the liquid phase.
The value of \(\triangle H_{vap}\) is crucial in calculating the normal boiling point using the Clausius-Clapeyron equation, as seen in our carbon tetrachloride example. A high \(\triangle H_{vap}\) means that more energy is needed for the substance to vaporize, which typically suggests stronger intermolecular forces, such as hydrogen bonding, in the liquid. Conversely, a lower \(\triangle H_{vap}\) would indicate weaker intermolecular forces. It's important for students to grasp that \(\triangle H_{vap}\) directly influences the boiling point: the greater the enthalpy of vaporization, the higher the temperature required to boil the liquid.
The value of \(\triangle H_{vap}\) is crucial in calculating the normal boiling point using the Clausius-Clapeyron equation, as seen in our carbon tetrachloride example. A high \(\triangle H_{vap}\) means that more energy is needed for the substance to vaporize, which typically suggests stronger intermolecular forces, such as hydrogen bonding, in the liquid. Conversely, a lower \(\triangle H_{vap}\) would indicate weaker intermolecular forces. It's important for students to grasp that \(\triangle H_{vap}\) directly influences the boiling point: the greater the enthalpy of vaporization, the higher the temperature required to boil the liquid.