Question

One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime (CaO) at \(600 .{ }^{\circ} \mathrm{C}\) followed by condensation of the mercury vapor. Given the heat of vaporization of mercury ( \(296 \mathrm{~J} / \mathrm{g}\) ) and the vapor pressure of mercury at \(25.0^{\circ} \mathrm{C}\left(2.56 \times 10^{-3}\right.\) torr), what is the vapor pressure of the condensed mercury at \(300 .{ }^{\circ} \mathrm{C}\) ? How many atoms of mercury are present in the mercury vapor at \(300 .{ }^{\circ} \mathrm{C}\) if the reaction is conducted in a closed \(15.0\) -L container?

Short answer

The vapor pressure of the condensed mercury at 300°C is 8.12 torr. There are approximately \(2.05\times10^{21}\) atoms of mercury in the mercury vapor at 300°C in the 15.0-L container.

Step-by-step solution

Write down the Clausius-Clapeyron equation

The Clausius-Clapeyron equation relates the vapor pressure of a substance at two different temperatures: \[ \ln \left(\frac{P_2}{P_1}\right) = \frac{-\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] Where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\), \(\Delta H_{vap}\) is the heat of vaporization, and R is the gas constant (8.314 J/mol·K).

Convert the given data to appropriate units

We are given the heat of vaporization of mercury (\(\Delta H_{vap}\)) in J/g, but we need it in J/mol. Also, we need the temperatures in Kelvin. Convert the given data as follows: 1. Heat of vaporization: 296 J/g Mercury has a molar mass of 200.6 g/mol, so \(\Delta H_{vap} = 296\,\text{J/g} \times 200.6\,\text{g/mol} = 59,377.6\,\text{J/mol}\) 2. Temperatures: \(T_1 = 25.0^\circ\text{C} + 273.15\,\text{K} = 298.15\,\text{K}\) \(T_2 = 300.0^\circ\text{C} + 273.15\,\text{K} = 573.15\,\text{K}\)

Use the Clausius-Clapeyron equation to find the vapor pressure

Now, plug in the given values and solve for \(P_2\): \[ \ln \frac{P_2}{2.56 \times 10^{-3}} = \frac{-59,377.6}{8.314}\left(\frac{1}{573.15} - \frac{1}{298.15}\right) \] Solving for \(P_2\): \[ P_2 = 2.56 \times 10^{-3} \times e^{\frac{-59,377.6}{8.314}\left(\frac{1}{573.15} - \frac{1}{298.15}\right)} = 8.12\,\text{torr} \] So the vapor pressure of the condensed mercury at 300°C is 8.12 torr.

Use the ideal gas law to find the number of moles and atoms of mercury vapor

We have the vapor pressure (\(P\)), temperature (\(T\)) and volume (\(V\)) of the container, and we can use the ideal gas law to find the number of moles of mercury vapor: \[ PV=nRT \] Where \(P\) is the pressure in atm, \(V\) is the volume in L, \(n\) is the number of moles, \(R\) is the gas constant (0.0821 L·atm/mol·K), and \(T\) is the temperature in K. First, convert the vapor pressure from torr to atm: \[ P = 8.12\,\text{torr} \times \frac{1\,\text{atm}}{760\,\text{torr}} = 0.01068\,\text{atm} \] Now, plug in the values: \[ n = \frac{PV}{RT} = \frac{0.01068\,\text{atm} \times 15.0\,\text{L}}{0.0821\,\text{L·atm/mol·K} \times 573.15\,\text{K}} = 3.41 \times 10^{-3}\,\text{mol} \] Finally, convert moles to atoms using Avogadro's number (6.022 x 10²³ atoms/mol): \[ \text{Atoms of Hg} = 3.41 \times 10^{-3}\,\text{mol} \times 6.022\times 10^{23}\,\text{atoms/mol} = 2.05 \times 10^{21}\,\text{atoms} \] The number of atoms of mercury in the mercury vapor at 300°C in the 15.0-L container is approximately \(2.05\times10^{21}\) atoms.

Key concepts

Vapor Pressure

Vapor pressure is a crucial concept to understand when dealing with the properties of liquids and gases. It's the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. This equilibrium occurs because at any temperature, molecules of the liquid or solid phase will have enough energy to escape into the vapor phase.

For any substance, as temperature increases, more molecules have sufficient energy to escape into the vapor phase, thus increasing the vapor pressure. Conversely, cooling the substance decreases its vapor pressure. This behavior is important in understanding phenomena like boiling, where the vapor pressure equals the atmospheric pressure, allowing the liquid to transition into the vapor phase.

• As temperature increases, vapor pressure increases.
• Vapor pressure is also unique for each substance at a given temperature due to differences in intermolecular forces.

Vapor pressure plays a significant role in predicting how substances will behave under different conditions, which is why it is crucial when using the Clausius-Clapeyron equation for calculations.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas using the formula: \[ PV = nRT \]where:

• \( P \): Pressure in atmospheres (atm)
• \( V \): Volume in liters (L)
• \( n \): Number of moles of gas
• \( R \): Universal gas constant (0.0821 L·atm/mol·K)
• \( T \): Temperature in Kelvin (K)

The ideal gas law is based on the assumptions of the ideal gas model: that the gas consists of many tiny particles in constant, random motion, and these particles do not interact except by elastic collisions. While real gases sometimes deviate from this ideal behavior at high pressures or low temperatures, this law is a close approximation in many conditions.

To use the ideal gas law effectively, you should remember to convert all units consistently, particularly temperature into Kelvin and pressure into atmospheres. It allows calculations like determining the number of moles of a gas in a container or finding out how volume changes with pressure.

Heat of Vaporization

The heat of vaporization is defined as the amount of energy required to change 1 gram or 1 mole of a liquid into a gas at constant temperature and pressure. It's a measure of the strength of the intermolecular forces within a liquid; stronger intermolecular forces result in a higher heat of vaporization.
For mercury, the heat of vaporization needed for calculations is often provided in Joules per gram (J/g), but when utilizing the Clausius-Clapeyron equation, you will typically need it in Joules per mole (J/mol). In such cases, you multiply the given J/g value by the molar mass of the substance to get J/mol.

Understanding the heat of vaporization helps in determining how much energy is needed to vaporize a liquid and is significant in applications involving phase changes. In the solution exercise, the heat of vaporization is vital for calculating the vapor pressure at different temperatures using the Clausius-Clapeyron equation.\[\]Remember:

• Heat of vaporization is higher for substances with strong intermolecular attractions.
• It is essential for calculating temperature-dependent phenomena like boiling.