Problem 8 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The vapour pressure of tetrachloromethane \(\left(\mathrm{CCl}_{4}\right)\) at \(0^{\circ} \mathrm{C}\) is \(44.0 \mathrm{mbar}\) and at \(50^{\circ} \mathrm{C}\) it is \(422.0 \mathrm{mbar}\). Use these data to estimate the enthalpy change of vaporization of \(\mathrm{CCl}_{4}\) and to estimate its standard boiling point. (Section 17.2 )

Step-by-Step Solution

Answer

The enthalpy of vaporization is approximately 34.1 kJ/mol, and the standard boiling point is about 78°C.

Step 1: Understanding the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the change in vapor pressure with temperature to the enthalpy of vaporization. It is given by:\[\ln \left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]where \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \), respectively, \( \Delta H_{vap} \) is the enthalpy of vaporization, and \( R \) is the ideal gas constant \(8.314 \, J \, mol^{-1} \, K^{-1}\).

Step 2: Plugging in Values

Use the given data to plug into the Clausius-Clapeyron equation. The temperatures should be converted to Kelvin:\( T_1 = 0^{\circ}C + 273.15 = 273.15 \, K \)\( T_2 = 50^{\circ}C + 273.15 = 323.15 \, K \)Substitute into the equation:\[\ln \left(\frac{422.0}{44.0}\right) = \frac{\Delta H_{vap}}{8.314} \left(\frac{1}{273.15} - \frac{1}{323.15}\right)\]

Step 3: Solving the Equation for Enthalpy of Vaporization

Calculate the natural logarithm:\[\ln \left(\frac{422.0}{44.0}\right) = \ln (9.591) = 2.257\]Calculate the right-hand side:\[\left(\frac{1}{273.15} - \frac{1}{323.15}\right) = 0.000550\]Substitute these into the equation:\[2.257 = \frac{\Delta H_{vap}}{8.314} \times 0.000550\]Solve for \( \Delta H_{vap} \):\[\Delta H_{vap} = \frac{2.257 \times 8.314}{0.000550} = 34100 \, J \, mol^{-1}\]The enthalpy change of vaporization is approximately \(34.1 \, kJ \, mol^{-1}\).

Step 4: Estimating the Standard Boiling Point

The standard boiling point is the temperature at which the vapor pressure equals 1 atm (1013 mbar). Start from the same equation\[\ln \left(\frac{1013}{44.0}\right) = \frac{34100}{8.314} \left(\frac{1}{273.15} - \frac{1}{T_b}\right)\]and solve for \( T_b \). Calculate \( \ln(22.977) = 3.134 \).This gives:\[3.134 = \frac{34100}{8.314} \left(\frac{1}{273.15} - \frac{1}{T_b}\right)\]Simplifying further and solving gives:\( T_b \approx 351 \, K \)\, which is equivalent to approximately \(78^{\circ}C\).

Key Concepts

Clausius-Clapeyron equation

The Clausius-Clapeyron equation is essential in thermodynamics for understanding how the vapor pressure of a substance changes with temperature. This equation is particularly useful for substances undergoing phase changes, like the vaporization of a liquid. The equation is expressed as:\[\ln \left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]Here,

• \(P_1\) and \(P_2\) are the vapor pressures at two different temperatures, \(T_1\) and \(T_2\).
• \(\Delta H_{vap}\) represents the enthalpy of vaporization, which is the energy required to convert a mole of liquid into vapor.
• \(R\) is the ideal gas constant, valued at \(8.314 \, J \, mol^{-1} \, K^{-1}\).

In essence, this equation allows you to calculate the enthalpy of vaporization if you know the vapor pressures at two different temperatures. The natural logarithm on the left side of the equation indicates that this relationship is logarithmic, meaning that even small changes in temperature can result in significant changes in vapor pressure.
To solve problems using this equation, it's crucial to ensure that all temperatures are converted to the Kelvin scale. This conversion helps in maintaining consistency and accuracy in calculations.

standard boiling point

The standard boiling point of a substance is defined as the temperature at which its vapor pressure equals the standard atmospheric pressure, which is 1 atmosphere (1013 mbar). For many substances, this is a crucial parameter as it indicates the temperature at which a liquid's vapor pressure will overcome atmospheric pressure, allowing it to boil.
To estimate the standard boiling point using the Clausius-Clapeyron equation, you set \(P_2\) as the standard atmospheric pressure and solve for temperature:\[\ln \left(\frac{1013}{44.0}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{273.15} - \frac{1}{T_b}\right)\]Here, \(T_b\) stands for the boiling temperature you're solving for. You rearrange the equation to isolate \(\frac{1}{T_b}\) on one side and then solve for \(T_b\). This detailed calculation requires careful manipulation of the equation, ensuring each operation is methodically done to prevent errors.
Knowing the standard boiling point has practical applications:

• It's useful in industrial processes where the control of temperature and pressure is essential.
• It helps in understanding a substance's behavior under different atmospheric conditions.

vapor pressure calculation

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. Calculating vapor pressure is important to understand a liquid's volatility, which indicates how quickly it evaporates.
In the context of the Clausius-Clapeyron equation, vapor pressure calculations are used to evaluate how pressure changes with temperature for a given substance. To calculate vapor pressure:

• First, gather experimental data for vapor pressure at different temperatures.
• Convert the temperatures to Kelvin.
• Apply the Clausius-Clapeyron equation to relate the pressure changes with temperature changes.

Calculating vapor pressure aids in predicting how a substance will behave under certain conditions, which is crucial in designing chemical processes. Knowing the vapor pressure at two different temperatures allows you to calculate the enthalpy of vaporization, which is a measure of the energy required to vaporize the substance. Such calculations help in various scientific and engineering applications, such as designing cooling systems, understanding weather patterns, and working in industries that require precise temperature and pressure control.