Problem 159 - Chemistry

Question

One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime (CaO) at \(600 .\) C followed by condensation of the mercury vapor. Given the heat of vaporization of mercury \((296 \mathrm{J} / \mathrm{g} \text { ) and the vapor pressure of mercury }\) at \(25.0^{\circ} \mathrm{C}\left(2.56 \times 10^{-3} \text { torr), what is the vapor pressure of the }\right.\) 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 \(\mathrm{-L}\) container?

Step-by-Step Solution

Answer

The vapor pressure of condensed mercury at 300°C is \(2.25 \times 10^{-1}\) torr, and there are approximately \(5.85 \times 10^{20}\) mercury atoms present in the 15.0-L container at that temperature.

Step 1: Calculate the vapor pressure at 300°C using the Clausius-Clapeyron Equation

The Clausius-Clapeyron Equation is: \[\ln \frac{P_2}{P_1} = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] where \(P_1\) is the vapor pressure at temperature \(T_1\), \(P_2\) is the vapor pressure at temperature \(T_2\), \(\Delta H_{vap}\) is the heat of vaporization of the substance, and R is the ideal gas constant. In our case, we have: \(P_1 = 2.56 \times 10^{-3}\) torr (vapor pressure at 25.0°C), \(T_1 = 273.15 + 25 = 298.15\) K (temperature in Kelvin), \(T_2 = 273.15 + 300 = 573.15\) K (temperature in Kelvin), and \(\Delta H_{vap} = 296\) J/g. First, we need to convert the heat of vaporization to J/mol (R's unit needs to match). We have: \[\Delta H_{vap} = 296 \frac{\text{J}}{\text{g}} \times \frac{200.59\text{ g}}{\text{mol}} = 59400 \frac{\text{J}}{\text{mol}}\] Then, let's solve for \(P_2\): \[P_2 = P_1 \times e^{-\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\] Substitute the values and solve for \(P_2\): \[P_2 = (2.56 \times 10^{-3}\text{ torr}) \times e^{-\frac{59400 \mathrm{J} / \mathrm{mol}}{8.314 \mathrm{J} / \mathrm{mol\cdot K}} \left(\frac{1}{573.15\text{ K}} - \frac{1}{298.15\text{ K}}\right)}\] After calculating, we get: \[P_2 = 2.25 \times 10^{-1} \text{ torr}\]

Step 2: Use the Ideal Gas Law to find the number of moles of mercury vapor

Now that we have the vapor pressure at 300°C, we can use the Ideal Gas Law equation to find the number of moles of mercury vapor present in the 15.0-L container. The Ideal Gas Law equation is: \[PV = nRT\] where P is the pressure in pascals, V is the volume in liters, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the pressure from torr to pascals: \[P_2 = 2.25 \times 10^{-1} \text{ torr} \times \frac{101325}{760} \text{Pa} = 30.05 \text{ Pa}\] Now, let's substitute the values into the Ideal Gas Law equation and solve for n: \[n = \frac{PV}{RT} = \frac{30.05 \text{ Pa} \times 15.0 \text{ L}}{8.314 \text{ J/mol⋅K} \times 573.15 \text{ K}}\] After calculating, we get: \[n = 9.72 \times 10^{-4}\text{ mol}\]

Step 3: Calculate the number of mercury atoms

Finally, we need to convert the number of moles to the number of atoms using Avogadro's number: \[N_\text{atoms} = n \times N_A\] where \(N_\text{atoms}\) is the number of mercury atoms, n is the number of moles, and \(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \text{mol}^{-1}\)). Substitute the values and solve for \(N_\text{atoms}\): \[N_\text{atoms} = 9.72 \times 10^{-4} \text{ mol} \times 6.022 \times 10^{23} \text{mol}^{-1}\] After calculating, we get: \[N_\text{atoms} = 5.85 \times 10^{20}\] So, the vapor pressure of condensed mercury in the 15.0-L container at 300°C is \(2.25 \times 10^{-1}\) torr, and there are approximately \(5.85 \times 10^{20}\) mercury atoms present in the container.

Key Concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is a key concept in thermodynamics that allows us to understand how vapor pressure changes with temperature. This equation is particularly useful for substances that undergo phase changes between liquid and vapor. It is expressed as: \[\ln \frac{P_2}{P_1} = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\], where:

• \( P_1 \) is the initial vapor pressure at temperature \( T_1 \),
• \( P_2 \) is the vapor pressure at temperature \( T_2 \),
• \( \Delta H_{vap} \) is the heat of vaporization,
• and \( R \) is the ideal gas constant.

This equation helps us calculate the vapor pressure of mercury as it changes temperature, from 25°C to 300°C in the given problem. The equation derives from integrating the relation between pressure and temperature of a liquid-vapor mixture, emphasizing the exponential relationship between vapor pressure and temperature changes.
When solving for the new vapor pressure (\( P_2 \)), it's important to remember that all temperatures should be converted to Kelvin, and the heat of vaporization should match the units of \( R \). This ensures accuracy in the results obtained using this formula.

Ideal Gas Law

The Ideal Gas Law is essential for understanding the behavior of gases under typical conditions. This law is encapsulated in the equation:\[ PV = nRT \],where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) represents the number of moles,
• \( R \) is the ideal gas constant,
• and \( T \) is the temperature in Kelvin.

In the case of mercury vapor at 300°C in a closed container, this formula is used to determine how many moles of mercury are present. The conversion of pressure from torr to pascals and the use of consistent units is crucial to solve the equation correctly. By plugging in the known values of pressure (converted to pascals from torr), volume (in liters), and temperature (in Kelvin), we can solve for \( n \), which gives us the total number of moles of mercury vapor.
This calculated amount can further be used to find the number of mercury atoms in the vapor, highlighting the strong connection between macroscopic measurements and molecular quantities in chemistry.

Elemental Mercury

Elemental mercury, represented as Hg, is a unique metal with several interesting properties, including its liquid state at room temperature. It is dense, has a high surface tension, and evaporates easily even at low temperatures, making it significant in a variety of applications and scientific studies.
In the preparation of elemental mercury via roasting cinnabar (HgS), mercury is extracted as a vapor and condenses into its liquid form upon cooling. The knowledge about its vapor pressure at different temperatures is critical, especially in industrial processes that involve mercury vapors. Understanding the changes in vapor pressure using the Clausius-Clapeyron equation is important when studying elemental mercury behavior at varying temperatures.
Despite its utility, handling mercury requires care due to its toxic effects. Prolonged exposure to mercury vapor can lead to health issues, making it imperative to understand and evaluate the conditions (such as pressure and temperature) when working with this element, especially at a laboratory or industrial scale.