Problem 12 - Chemistry and Chemical Reactivity

Question

The enthalpy of vaporization of liquid mercury is \(59.11 \mathrm{kJ} / \mathrm{mol} .\) What quantity of 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

The heat required is approximately 2.00 kJ.

Step 1: Determine the Mass of Mercury

First, use the given density to calculate the mass of mercury. The formula for density is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Given the density of mercury is \( 13.6 \, \mathrm{g/mL} \) and the volume is \( 0.500 \, \mathrm{mL} \), calculate the mass: \[ \text{Mass} = \text{Density} \times \text{Volume} = 13.6 \, \mathrm{g/mL} \times 0.500 \, \mathrm{mL} = 6.8 \, \mathrm{g} \]

Step 2: Convert Mass to Moles

Now, convert the mass of mercury to moles. The molar mass of mercury (Hg) is approximately \( 200.59 \, \mathrm{g/mol} \). Use the formula: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]\[ \text{Moles of Hg} = \frac{6.8 \, \mathrm{g}}{200.59 \, \mathrm{g/mol}} \approx 0.0339 \, \mathrm{mol} \]

Step 3: Calculate Heat Required Using Enthalpy of Vaporization

Finally, calculate the amount of heat required to vaporize the mercury using its enthalpy of vaporization, which is given as \( 59.11 \, \mathrm{kJ/mol} \). Use the formula: \[ \text{Heat} = \text{Enthalpy of Vaporization} \times \text{Moles} \]\[ \text{Heat} = 59.11 \, \mathrm{kJ/mol} \times 0.0339 \, \mathrm{mol} \approx 2.00 \, \mathrm{kJ} \]

Key Concepts

Density Calculation

The concept of density is fundamental when dealing with the mass and volume of a substance. Density is defined as the mass of an object divided by its volume. It provides a measure of how much "stuff" is packed into a given space. To calculate density, you can use the formula: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Here, density helps us determine the mass of mercury given a specific volume.

Given the density of mercury as \( 13.6 \, \mathrm{g/mL} \) and the volume as \( 0.500 \, \mathrm{mL} \), we calculate the mass by rearranging the density formula to solve for mass: \( \text{Mass} = \text{Density} \times \text{Volume} \).

• Multiply \( 13.6 \, \mathrm{g/mL} \) by \( 0.500 \, \mathrm{mL} \)
• You get \( 6.8 \, \mathrm{g} \) as the mass of the mercury.

Understanding how to manipulate the density formula allows you to solve problems involving mass and volume easily.

Molar Mass Conversion

Converting mass to moles is a crucial step in chemistry. It allows you to connect the macroscopic world that we can measure directly, like grams of a substance, to the microscopic number of molecules or atoms. Molar mass functions as the conversion factor in this process.

The molar mass of mercury (Hg) is about \( 200.59 \, \mathrm{g/mol} \). To find the number of moles, we use the relationship: \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \). For the exercise:

• Divide the mass of mercury \( 6.8 \, \mathrm{g} \) by its molar mass \( 200.59 \, \mathrm{g/mol} \)
• This gives approximately \( 0.0339 \, \mathrm{mol} \) of mercury.

This conversion is often a necessary step before performing chemical calculations, such as determining the amount of reactant or product in a reaction.

Heat Calculation

Calculating the heat required to change the state of a substance is essential in thermodynamics. When dealing with vaporization, we use the enthalpy of vaporization, symbolized as \( \Delta H_{\text{vap}} \). This value indicates the amount of heat needed to convert one mole of a liquid to a gas at constant temperature and pressure.

To find the heat needed for vaporizing the mercury, we apply the formula: \( \text{Heat} = \text{Enthalpy of Vaporization} \times \text{Moles} \). Given that the enthalpy of vaporization of mercury is \( 59.11 \, \mathrm{kJ/mol} \) and we have \( 0.0339 \, \mathrm{mol} \) of mercury:

• Multiply \( 59.11 \, \mathrm{kJ/mol} \) by \( 0.0339 \, \mathrm{mol} \)
• This calculates to approximately \( 2.00 \, \mathrm{kJ} \).

This calculation helps in understanding how much energy is required to phase transition substances under standard conditions.