Problem 18 - Chemistry The Molecular Science

Question

Mercury is highly toxic. Although it is a liquid at room temperature, it has a high vapor pressure and a low vaporization enthalpy ( \(294 \mathrm{~J} / \mathrm{g}\) ). Calculate the heat energy transfer required to vaporize \(0.500 \mathrm{~mL}\) mercury at \(357^{\circ} \mathrm{C}\), its normal boiling point. The density of \(\operatorname{Hg}(\ell)\) is \(13.6 \mathrm{~g} / \mathrm{mL}\). Compare your result with the energy transfer needed to vaporize \(0.500 \mathrm{~mL}\) water. The molar vaporization enthalpy of \(\mathrm{H}_{2} \mathrm{O}\) is \(40.7 \mathrm{~kJ} / \mathrm{mol}\).

Step-by-Step Solution

Answer

1999.2 J is needed for mercury, more than the 1130 J for water.

Step 1: Calculate mass of mercury

To find the mass of mercury, use the formula: \[\text{mass} = \text{density} \times \text{volume} \] The density of mercury is given as \(13.6\, \text{g/mL}\) and the volume as \(0.500\, \text{mL}\). Thus, the mass is: \[\text{mass} = 13.6\, \text{g/mL} \times 0.500\, \text{mL} = 6.8\, \text{g} \]

Step 2: Calculate heat energy for mercury

To calculate the heat energy required to vaporize mercury, use the formula: \[ q = \text{mass} \times \text{enthalpy change} \] With the mass of mercury as \(6.8\, \text{g}\) and the vaporization enthalpy as \(294\, \text{J/g}\), the heat energy is: \[ q = 6.8\, \text{g} \times 294\, \text{J/g} = 1999.2\, \text{J} \]

Step 3: Calculate moles of water

To calculate the moles of water, first find the mass of water using its density (approximately \(1.00\, \text{g/mL}\)) and the given volume \(0.500\, \text{mL}\):\[ \text{mass of water} = 1.00\, \text{g/mL} \times 0.500\, \text{mL} = 0.500\, \text{g} \]Then, use the molar mass of water \(18.02\, \text{g/mol}\) to find the moles: \[ \text{moles of water} = \frac{0.500\, \text{g}}{18.02\, \text{g/mol}} \approx 0.0278\, \text{mol} \]

Step 4: Calculate heat energy for water

Using the moles of water calculated and the molar vaporization enthalpy, the heat energy required is: \[ q = \text{moles} \times \text{molar enthalpy change} \] With the molar enthalpy \(40.7\, \text{kJ/mol}\):\[ q = 0.0278\, \text{mol} \times 40.7\, \text{kJ/mol} = 1.13\, \text{kJ} \] Convert to Joules for comparison: \[ q = 1.13\, \text{kJ} \times 1000\, \text{J/kJ} = 1130\, \text{J} \]

Step 5: Compare energy requirements

The energy required to vaporize \(0.500\, \text{mL}\) of mercury is \(1999.2\, \text{J}\), compared to the \(1130\, \text{J}\) needed for the same volume of water. Hence, more energy is required for mercury.

Key Concepts

Understanding Enthalpy Change

Enthalpy change, often represented as \( \Delta H \), is the amount of heat absorbed or released during a chemical process at constant pressure. It's essential in determining how much energy is required for phase changes, such as vaporization. In the exercise, mercury and water have distinct enthalpic values, reflecting the energy needed for their vaporization.

For mercury, the vaporization enthalpy is given as 294 J/g, meaning each gram of mercury needs 294 Joules of energy to transition from liquid to gas. Understanding this helps in calculating the total energy transfer required.

For water, the molar vaporization enthalpy is 40.7 kJ/mol. This high value suggests significant energy is involved, although per unit volume, mercury requires more energy, as shown in the comparison.

Density Calculations Made Simple

Density is a physical property defining mass per unit volume, typically expressed in g/mL or g/cm³. Calculations involving density, like those in the exercise, help convert volume to mass when needed.

For mercury, with a density of 13.6 g/mL, multiplying by the volume gives the mass:

• Mass of mercury = Density × Volume = 13.6 g/mL × 0.500 mL = 6.8 g

In comparison, water has a density of about 1.00 g/mL, which makes the calculation straightforward:

• Mass of water = 1.00 g/mL × 0.500 mL = 0.500 g

These calculations provide the foundation for finding how much energy is required for phase changes.

Exploring the Concept of Boiling Point

The boiling point is the temperature at which a liquid's vapor pressure equals atmospheric pressure, causing it to transition to vapor. Mercury has a notable boiling point of 357°C at standard atmospheric conditions, implying it needs substantial energy to vaporize.

This temperature factor plays a crucial role in determining the enthalpy change, as seen in the exercise. Understanding boiling points helps in predicting energy needs and behaviors of substances during heating.

In comparison, water boils at 100°C, which explains the differences in their vaporization energy requirements despite its higher molar enthalpy.

Heat Energy Transfer Explained

Heat energy transfer is the flow of thermal energy from one object to another, often caused by temperature differences. In vaporization, calculating heat energy transfer reveals the energy needed for a liquid to become gas.

For mercury, using the formula \( q = \text{mass} \times \text{enthalpy change} \), we found the energy transfer required:

• \( q = 6.8 \, \text{g} \times 294 \, \text{J/g} = 1999.2 \, \text{J} \)

This means nearly 2000 Joules are needed to vaporize 0.500 mL of mercury.

For water, the calculation differs slightly due to its molar enthalpy:

• \( q = 0.0278 \, \text{mol} \times 40.7 \, \text{kJ/mol} = 1.13 \, \text{kJ} \)
• Converting to Joules, \( q = 1130 \, \text{J} \)

These calculations highlight the importance of understanding both mass and mole-based enthalpy changes in energy transfer.