Question

Benzene ( \(l\) ) has a vapor pressure of 0.1269 bar at \(298.15 \mathrm{K}\) and an enthalpy of vaporization of \(30.72 \mathrm{kJ} \mathrm{mol}^{-1}\). The \(C_{P, m}\) of the vapor and liquid phases at that temperature are 82.4 and \(136.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\), respectively. Calculate the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) at \(340.0 \mathrm{K}\) assuming a. that the enthalpy of vaporization does not change with temperature. b. that the enthalpy of vaporization at temperature \(T\) can be calculated from the equation \(\Delta H_{\text {vaporization}}(T)=\) \(\Delta H_{\text {vaporiation}}\left(T_{0}\right)+\Delta C_{P}\left(T-T_{0}\right)\) assuming that \(\Delta C_{P}\) does not change with temperature.

Short answer

The vapor pressure of benzene at \(340.0\, \mathrm{K}\) is: - For part (a), assuming the enthalpy of vaporization remains constant: \(0.6175\, \mathrm{bar}\) - For part (b), considering the change in enthalpy of vaporization with temperature: \(0.7775\, \mathrm{bar}\)

Step-by-step solution

Write down the known parameters

We are given the following information for benzene: - Vapor pressure at 298.15 K, \(P_1 = 0.1269\, \mathrm{bar}\) - Enthalpy of vaporization, \(\Delta H_\mathrm{vaporization} = 30.72\, \mathrm{kJ\, mol^{-1}}\) - Molar heat capacity of the vapor phase, \(C_{P, m, vapor} = 82.4\, \mathrm{J\,K^{-1}\,mol^{-1}}\) - Molar heat capacity of the liquid phase, \(C_{P, m, liquid} = 136.0\, \mathrm{J\,K^{-1}\,mol^{-1}}\) - Temperature at which we need to calculate the vapor pressure, \(T_2 = 340.0\, \mathrm{K}\)

Use Clausius-Clapeyron equation for part (a)

For part (a), we assume the enthalpy of vaporization does not change with temperature. To calculate the vapor pressure at the new temperature, we can use the Clausius-Clapeyron equation: \[\ln(\frac{P_2}{P_1}) = -\frac{\Delta H_\mathrm{vaporization}}{R}(\frac{1}{T_2} - \frac{1}{T_1})\] where \(P_1\) is the vapor pressure at the initial temperature \(T_1 = 298.15\, \mathrm{K}\) and \(P_2\) is the vapor pressure at the new temperature \(T_2 = 340.0\, \mathrm{K}\). R is the universal gas constant (8.314 J/(mol K)).

Calculate vapor pressure for part (a)

Solve the Clausius-Clapeyron equation from step 2 for \(P_2\): \[P_2 = P_1 \cdot \exp{(-\frac{\Delta H_\mathrm{vaporization}}{R}(\frac{1}{T_2} - \frac{1}{T_1}))}\] Plugging in the given values: \[P_2 = 0.1269 \cdot \exp{(-\frac{30.72 \times 10^3\, \mathrm{J\,mol^{-1}}}{8.314\, \mathrm{J\,K^{-1}\,mol^{-1}})(\frac{1}{340.0\, \mathrm{K}} - \frac{1}{298.15\, \mathrm{K}}))}\] Calculating the vapor pressure at 340 K: \[P_2 \approx 0.6175\, \mathrm{bar}\]

Calculate the change in enthalpy of vaporization for part (b)

For part (b), we need to consider the change in enthalpy of vaporization with temperature. We're given the formula: \[\Delta H_\mathrm{vaporization}(T) = \Delta H_\mathrm{vaporization}(T_0) + \Delta C_{P}(T-T_0)\] where \(\Delta H_\mathrm{vaporization}(T_0)\) is the enthalpy of vaporization at the reference temperature and \(\Delta C_P = C_{P, m, vapor} - C_{P, m, liquid}\). First, calculate \(\Delta C_P\): \[\Delta C_P = 82.4 - 136.0\, \mathrm{J\,K^{-1}\,mol^{-1}} = -53.6\, \mathrm{J\,K^{-1}\,mol^{-1}}\] Now, calculate the enthalpy of vaporization at the new temperature: \[\Delta H_\mathrm{vaporization}(340) = 30.72\times 10^3 + (-53.6)(340 - 298.15)\, \mathrm{J\,mol^{-1}} \approx 28.35\times 10^3\, \mathrm{J\,mol^{-1}}\]

Use the Clausius-Clapeyron equation for part (b)

We will now use the updated enthalpy of vaporization at the new temperature in the Clausius-Clapeyron equation: \[P_2 = P_1 \cdot \exp{(-\frac{\Delta H_\mathrm{vaporization}(340)}{R}(\frac{1}{T_2} - \frac{1}{T_1}))}\] Plugging in the given values: \[P_2 = 0.1269 \cdot \exp{(-\frac{28.35\times 10^3\, \mathrm{J\,mol^{-1}}}{8.314\, \mathrm{J\,K^{-1}\,mol^{-1}})(\frac{1}{340\, \mathrm{K}} - \frac{1}{298.15\, \mathrm{K}}))}\] Calculating the vapor pressure at 340 K: \[P_2 \approx 0.7775\, \mathrm{bar}\]

Conclusion

The vapor pressure of benzene at 340.0 K is: - For part (a), assuming the enthalpy of vaporization remains constant: \(0.6175\, \mathrm{bar}\) - For part (b), considering the change in enthalpy of vaporization with temperature: \(0.7775\, \mathrm{bar}\)

Key concepts

Enthalpy of Vaporization

Enthalpy of vaporization is a crucial concept in understanding phase transitions, specifically from the liquid phase to the vapor phase. It represents the amount of energy required to convert one mole of a liquid into a gas at a constant temperature and pressure. For instance, in the given problem with benzene, the enthalpy of vaporization is provided as 30.72 kJ/mol. This means that this energy quantity is needed to break the intermolecular forces holding benzene molecules together in the liquid state.

The enthalpy of vaporization can change with temperature and pressure because the strength of intermolecular forces, and therefore the energy needed to overcome them, can vary. In exercises like the one provided, there may be assumptions made, such as keeping enthalpy constant, which simplifies calculations using the Clausius-Clapeyron equation. Alternatively, changes in enthalpy with temperature can be included using specific formulas to achieve greater accuracy. This relationship becomes apparent in part (b) of our original exercise.

Vapor Pressure

Vapor pressure refers to the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. It's an essential factor in determining how easily a liquid will evaporate. For benzene at 298.15 K, the vapor pressure is 0.1269 bar. As temperature increases, more molecules have enough energy to escape the liquid phase, increasing vapor pressure.

To calculate vapor pressure at a new temperature, like 340.0 K in the given exercise, the Clausius-Clapeyron equation comes into play. This equation connects the vapor pressure change with variations in temperature and enthalpy of vaporization, providing a method to estimate vapor pressure at different temperatures. Understanding how vapor pressure changes with temperature is crucial in diverse applications, from meteorology to chemical engineering, as it impacts things like boiling points and condensation processes.

Molar Heat Capacity

Molar heat capacity indicates the amount of heat needed to change the temperature of one mole of a substance by one degree Kelvin. It manifests differently in vapor and liquid phases. In the context of benzene, we have two heat capacity values: for the vapor phase, it is 82.4 J/K/mol and, for the liquid, 136.0 J/K/mol.

Differences in molar heat capacities for vapor and liquid phases (\\(\Delta C_{P}\)), as seen in the problem, are essential when calculating how enthalpy of vaporization changes with temperature. A substance's heat capacity influences both its thermal stability and response to energy input, serving as a decisive parameter in many thermodynamic calculations.

• This affects boiling point determination,
• Efficiency of heating and cooling processes,
• And energy requirements to induce or reverse phase transitions.

Grasping the significance of molar heat capacity will aid in properly handling phase change calculations and understanding thermal properties of substances.