Problem 12 - Chemistry and Chemical Reactivity

Question

The enthalpy of vaporization of liquid mercury is \(59.11 \mathrm{kJ} / \mathrm{mol} .\) What quantity of energy as heat is required to vaporize \(0.500 \mathrm{mL}\) of mercury at \(357^{\circ} \mathrm{C},\) its normal boiling point? The density of mercury is \(13.6 \mathrm{g} / \mathrm{mL}.\)

Step-by-Step Solution

Answer

Approximately 2.002 kJ of energy is required to vaporize the mercury.

Step 1: Calculate the mass of mercury

First, we need to determine the mass of mercury. Use the formula for mass: \( \text{mass} = \text{volume} \times \text{density} \). Therefore, the mass of mercury is: \[ 0.500 \text{ mL} \times 13.6 \text{ g/mL} = 6.8 \text{ g} \].

Step 2: Convert mass to moles

Next, convert the mass of mercury to moles using the molar mass of mercury, which is approximately 200.59 g/mol. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Therefore, the moles of mercury are: \[ \frac{6.8 \text{ g}}{200.59 \text{ g/mol}} \approx 0.0339 \text{ mol} \].

Step 3: Calculate energy required for vaporization

Use the enthalpy of vaporization to calculate the energy required to vaporize the mercury. The formula is: \( Q = n \times \Delta H_{vap} \), where \( n \) is the moles of mercury and \( \Delta H_{vap} \) is the enthalpy of vaporization. Thus, \( Q = 0.0339 \text{ mol} \times 59.11 \text{ kJ/mol} \approx 2.002 \text{ kJ} \).

Key Concepts

Mercury

Mercury is a unique element often recognized by its liquid state at room temperature. Unlike most metals, mercury is found in its liquid form, making it quite distinct.
It's also known for its shiny, silver color and its ability to flow, much like water. In this exercise, we explore the process involved in turning liquid mercury into vapor by applying heat. This process is known as vaporization, and it's essential to understand how much energy is needed to bring about this change. If you ever wondered why it takes energy to change the state of a substance, it's because you need to overcome the forces that hold the particles together in the liquid state. For mercury, those forces are relatively weak compared to solids but still significant enough to require attention when calculating energy requirements.

Energy Calculation

When it comes to determining the amount of energy needed to vaporize mercury, we rely on the concept of enthalpy of vaporization. This is expressed in terms of energy per mole, usually in kilojoules per mole (kJ/mol). In our case, mercury requires 59.11 kJ/mol to transition from liquid to gas at its boiling point.
To compute the required energy:

• First, determine the mass of mercury, using its volume and its known density.
• Then, convert this mass into moles, using mercury's molar mass (200.59 g/mol).
• Finally, apply the formula for energy calculation: \[ Q = n \times \Delta H_{vap}, \] where \( n \) represents the moles, and \( \Delta H_{vap} \) is the enthalpy of vaporization.

This simple calculation shows how enthalpy, a fundamental thermodynamic quantity, helps us understand and predict energy requirements for phase changes, offering insight into how substances behave under the influence of heat.

Density

Density is a crucial concept when it comes to calculating energy for substances like mercury. It expresses the mass of a substance per unit volume, often in units such as grams per milliliter (g/mL).
In our scenario, mercury has a density of 13.6 g/mL. This means every milliliter of mercury weighs 13.6 grams. Why is this important? The density allows us to find the mass of the mercury we have, which is a necessary step before we can calculate the moles and subsequently the energy required for vaporization. Here's how it works:

• Take the given volume of mercury, multiply by its density to get its mass.
• This mass calculation is fundamental because all further calculations depend on accurately knowing how much mercury you are working with.

By understanding density, we can seamlessly move to subsequent steps in energy calculation, ensuring our answers are both accurate and meaningful.