Problem 21 - Chemistry and Chemical Reactivity

Question

Equilibrium vapor pressures of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) at various temperatures are given in the table. $$\begin{array}{cc} \text { Temperature }\left(^{\circ} \mathrm{C}\right) & \text { Vapor Pressure }(\mathrm{mm} \mathrm{Hg}) \\ \hline 7.6 & 40 . \\ 26.1 & 100 \\ 60.6 & 400 \\\ 80.1 & 760 \\ \hline \end{array}$$ (a) What is the normal boiling point of benzene? (b) Plot these data so that you have a graph resembling the one in Figure \(12.17 .\) At what temperature does the liquid have an equilibrium vapor pressure of \(250 \mathrm{mm}\) Hg? At what temperature is the vapor pressure \(650 \mathrm{mm}\) Hg? (c) Calculate the molar enthalpy of vaporization for benzene using the the Clausius-Clapeyron equation.

Step-by-Step Solution

Answer

(a) Normal boiling point is 80.1°C. (b) At 250 mm Hg, temperature is about 55°C; at 650 mm Hg, about 75°C. (c) Molar enthalpy of vaporization is about 33,000 J/mol.

Step 1: Understanding Normal Boiling Point

The normal boiling point of a substance is the temperature at which its vapor pressure equals atmospheric pressure, typically 1 atm or 760 mm Hg. According to the given data, the vapor pressure of benzene is 760 mm Hg at 80.1°C. Therefore, the normal boiling point of benzene is 80.1°C.

Step 2: Plotting the Vapor Pressure vs. Temperature Graph

To plot these data points, make a graph with temperature on the x-axis and vapor pressure on the y-axis. Using the given data points, plot the pressure values (40, 100, 250, 400, 650, 760 mm Hg) against their respective temperatures (7.6, 26.1, 60.6, and 80.1°C). Once plotted, use interpolation to estimate the temperatures corresponding to pressures of 250 mm Hg and 650 mm Hg.

Step 3: Estimating Temperature for 250 mm Hg and 650 mm Hg

Based on your plotted graph, locate the 250 mm Hg mark on the vapor pressure axis, draw a horizontal line to intersect the curve, and from that intersection, draw a vertical line to the temperature axis. Repeat this for the 650 mm Hg mark. Using the graph, estimate the temperature for 250 mm Hg, which should be slightly above 60.6°C, and for 650 mm Hg, slightly below 80.1°C, but make precise readings from the graph.

Step 4: Applying the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the change in vapor pressure with the change in temperature as follows: \[\ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]Here, \(P_1 = 100\) mm Hg at \(T_1 = 26.1 + 273.15\) K and \(P_2 = 400\) mm Hg at \(T_2 = 60.6 + 273.15\) K. The ideal gas constant \(R\) is 8.314 J/mol\(\cdot\)K. Solve for \(\Delta H_{vap}\) using these values.

Step 5: Solving for the Molar Enthalpy of Vaporization

Convert temperatures to Kelvin: \(T_1 = 299.25\) K, \(T_2 = 333.75\) K. Calculate the natural log: \[\ln \left(\frac{400}{100}\right) = \ln(4) = 1.3863\]Substitute in the Clausius-Clapeyron equation:\[1.3863 = -\frac{\Delta H_{vap}}{8.314} \left(\frac{1}{333.75} - \frac{1}{299.25}\right)\]Calculate the inverse differences and solve for \(\Delta H_{vap}\). Evaluate to find \(\Delta H_{vap} \approx 33,000 \) J/mol.

Key Concepts

Vapor Pressure

Vapor pressure is an important concept in chemistry, particularly when studying the phase changes of substances. It refers to the pressure exerted by a vapor in equilibrium with its liquid (or solid) phase within a closed system. In simple terms, it's the tendency of particles to escape from a liquid into the air or a vapor state.

Several factors influence vapor pressure:

• Temperature: Higher temperatures increase the kinetic energy of molecules, leading to an increased vapor pressure as more molecules have sufficient energy to escape the liquid.
• Nature of the Liquid: Volatile liquids (like benzene) have high vapor pressures because their molecules readily escape into the gas phase.

Understanding vapor pressure helps predict how a liquid behaves under different temperatures, and it's crucial when determining boiling points, as seen in the benzene example. At the boiling point, vapor pressure equals atmospheric pressure, allowing bubbles to form and liquid to boil.

Molar Enthalpy of Vaporization

The molar enthalpy of vaporization (\(\Delta H_{vap}\)) is the amount of heat energy required to vaporize one mole of a liquid at constant pressure. It's a key factor in understanding how much energy is needed for a phase change from liquid to gas.

To find this value, you can use the Clausius-Clapeyron equation, which connects vapor pressure and temperature to \(\Delta H_{vap}\):
\[\ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]

• \(P_1\) and \(P_2\): vapor pressures at temperatures \(T_1\) and \(T_2\), respectively.
• \(R\): The ideal gas constant, 8.314 J/mol·K.

This equation allows you to calculate the energy needed to convert liquid benzene into vapor, providing essential insight into its thermal properties and behaviors during phase transitions.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the surrounding atmospheric pressure. At this point, the liquid becomes gas, leading to phenomena such as bubbling. Benzene, for instance, has a normal boiling point of 80.1°C because at this temperature, its vapor pressure reaches 760 mm Hg, equivalent to one atmosphere.

Several factors can cause variations in boiling points:

• Elevation: Higher altitudes lower atmospheric pressure, resulting in lower boiling points.
• Impurities: Adding substances can raise or lower a liquid's boiling point.

Understanding the boiling point is important for practical applications such as cooking, where atmospheric conditions significantly affect outcomes. In scientific studies, knowing a compound's boiling point helps determine its volatility and stability under different environmental conditions.

Temperature-Pressure Relationship

The temperature-pressure relationship is a fundamental aspect of thermodynamics and is particularly relevant when studying vapor pressure and boiling points. This relationship shows that as temperature increases, vapor pressure tends to increase as well.

Key points regarding this relationship include:

• Direct Relationship: With increased temperature, the kinetic energy of liquid molecules rises, leading to a higher vapor pressure.
• Clausius-Clapeyron Equation: This mathematical expression links the change in temperature and pressure, providing a method to calculate properties like \(\Delta H_{vap}\).

Graphically, you can plot temperature versus vapor pressure to visualize this connection. In the benzene example, the line's slope in a logarithmic plot indicates how pressure varies with temperature changes. Recognizing and manipulating this relationship allows scientists and engineers to predict how substances will behave under different thermal conditions, crucial for industrial processes and safety assessments.