Problem 24 - Chemistry and Chemical Reactivity

Question

A 25.0 -mL sample of benzene at \(19.9^{\circ} \mathrm{C}\) was cooled to its melting point, \(5.5^{\circ} \mathrm{C},\) and then frozen. How much energy was given off as heat in this process? (The density of benzene is \(0.80 \mathrm{g} / \mathrm{mL},\) its specific heat capacity is \(1.74 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}, \text { and its heat of fusion is } 127 \mathrm{J} / \mathrm{g} .)\)

Step-by-Step Solution

Answer

The energy given off as heat is approximately 3040 J.

Step 1: Calculate the Mass of Benzene

First, calculate the mass of the benzene sample. Since we have the volume and the density, use the formula: \( \text{mass} = \text{volume} \times \text{density} \). \[ \text{mass} = 25.0 \, \text{mL} \times 0.80 \, \text{g/mL} = 20.0 \, \text{g} \]

Step 2: Calculate Heat Released During Cooling

Next, calculate the energy released when the benzene cools from \(19.9^{\circ}\mathrm{C}\) to its melting point, \(5.5^{\circ}\mathrm{C}\). Use the formula: \( q = m \cdot c \cdot \Delta T \)where \( m = 20.0 \, \text{g} \), \( c = 1.74 \, \text{J/g} \cdot \mathrm{K} \), and \( \Delta T = (5.5 - 19.9)^{\circ}\mathrm{C} = -14.4 \)K.\[q = 20.0 \, \text{g} \times 1.74 \, \text{J/g} \cdot \mathrm{K} \times (-14.4) \, \mathrm{K} = -500.16 \, \mathrm{J}\]The negative sign indicates heat is released.

Step 3: Calculate Heat Released During Freezing

Now, calculate the energy released when the benzene changes phase from liquid to solid at its melting point. Use the heat of fusion: \( q = m \cdot \Delta H_f \)where \( m = 20.0 \, \text{g} \) and \( \Delta H_f = 127 \, \text{J/g} \).\[q = 20.0 \, \text{g} \times 127 \, \text{J/g} = 2540 \, \text{J}\]

Step 4: Calculate Total Energy Released

Finally, add the absolute values of the energies calculated from cooling and freezing to find the total energy released.\[q_{\text{total}} = |q_{\text{cooling}}| + |q_{\text{freezing}}| = 500.16 \, \text{J} + 2540 \, \text{J} = 3040.16 \, \text{J}\]Thus, the energy given off as heat in this process was approximately \(3040 \, \text{J}\).

Key Concepts

Specific Heat Capacity

The specific heat capacity is a critical concept in thermodynamics that describes how much energy is required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). It tells us how a substance reacts to the absorption or release of heat. In our benzene example, the specific heat capacity is given as \(1.74 \, \text{J/g} \cdot \text{K}\). This means that for every gram of benzene, \(1.74\) joules of energy will change its temperature by \(1\) Kelvin.

When benzene was cooled from \(19.9^{\circ}\text{C}\) to \(5.5^{\circ}\text{C}\), we used the formula: \( q = m \cdot c \cdot \Delta T \), where \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. This equation helps calculate the heat transfer during the cooling process. For benzene, this provides a way to determine the energy lost as it cools to its melting point, which is crucial when large temperature changes are involved.

Heat of Fusion

The heat of fusion is the amount of energy needed to change a substance from solid to liquid at its melting point, without changing its temperature. It's an essential factor when considering phase changes, like melting and freezing.

In our case, benzene changes from liquid to solid at its melting point of \(5.5^{\circ}\text{C}\). The heat of fusion is given as \(127 \, \text{J/g}\), which signifies that each gram of benzene needs \(127\) joules to be released when it solidifies.

• Energy required per gram: \(127 \, \text{J/g}\)
• Mass of benzene: \(20.0 \, \text{g}\)

Using \( q = m \cdot \Delta H_f \), we calculate the total heat released during the freezing process. Hence, when the benzne freezes, it releases this heat as it moves from a more energetic liquid state to a less energetic solid state, without changing its temperature.

Energy Calculation

Energy calculation often combines the concepts of temperature change energy and phase change energy. When calculating energy released or absorbed in a process, it's essential to account for both the cooling or heating and the phase change.

In the benzene problem, we combined the energy from cooling (using the specific heat capacity) and the energy from freezing (using the heat of fusion). The total energy released was calculated by adding the absolute values of both the heat released during cooling and freezing:

• Heat due to cooling: approximately \(500\,\text{J}\)
• Heat due to freezing: \(2540\,\text{J}\)

By adding these together, \(q_\text{total} = 500\,\text{J} + 2540\,\text{J} = 3040\,\text{J}\) was found to be the total amount of energy released during the cooling and freezing process. This comprehensive approach ensures all variables affecting energy transfer are considered, leading to accurate results.