Problem 46 - Chemistry The Central Science In Si Units

Question

Due to the environmental concern of fluorocarbons as refrigerants, a refrigerant based on a mixture of hydrocarbons was used as a replacement. It is a patented blend of ethane, propane, butane, and isobutane. Isobutane has a normal boiling point of \(-12^{\circ} \mathrm{C}\). The molar specific heat of liquid phase and gas phase isobutane are \(129.7 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) and \(95.2 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) respectively. The heat of vaporization for this compound is \(21.3 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(25.0 \mathrm{~g}\) of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at \(40^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The heat required is 13.399 kJ.

Step 1: Calculate Molar Mass of Isobutane

Isobutane's chemical formula is \( C_4H_{10} \). Its molar mass can be calculated by adding the atomic masses of its constituent atoms: - Carbon (C) has an atomic mass of 12.01 g/mol. - Hydrogen (H) has an atomic mass of 1.008 g/mol. The molar mass is thus \( (4 \times 12.01) + (10 \times 1.008) = 58.14 \text{ g/mol} \).

Step 2: Convert Mass to Moles

To find the number of moles of isobutane, use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] Substitute the values: \[ \text{moles} = \frac{25.0 \text{ g}}{58.14 \text{ g/mol}} \approx 0.430 \text{ moles} \]

Step 3: Calculate Heat for Temperature Increase in Liquid Phase

Use specific heat capacity to calculate the heat needed to raise the temperature from \(-50^{\circ}C\) (liquid) to \(-12^{\circ}C\) (boiling point). The formula is: \[ q_1 = m \cdot C_{\text{liquid}} \cdot \Delta T \] Where: - \( m = 0.430 \text{ moles} \), - \( C_{\text{liquid}} = 129.7 \text{ J/mol-K} \), - \( \Delta T = (-12 - (-50)) \text{C} = 38 \text{K} \). Calculate \( q_1 \): \[ q_1 = 0.430 \cdot 129.7 \cdot 38 = 2115.03 \text{ J} \approx 2.115 \text{ kJ} \]

Step 4: Calculate Heat for Vaporization

Calculate the heat required to vaporize the liquid at the boiling point (phase change). Use the formula: \[ q_2 = n \cdot \Delta H_{\text{vap}} \] Where: - \( n = 0.430 \text{ moles} \), - \( \Delta H_{\text{vap}} = 21.3 \text{ kJ/mol} \). Calculate \( q_2 \): \[ q_2 = 0.430 \cdot 21.3 = 9.159 \text{ kJ} \]

Step 5: Calculate Heat for Temperature Increase in Gas Phase

Use specific heat capacity to calculate the heat needed to raise the temperature from \(-12^{\circ}C\) to \(40^{\circ}C\) in the gas phase. The formula is: \[ q_3 = m \cdot C_{\text{gas}} \cdot \Delta T \] Where: - \( m = 0.430 \text{ moles} \), - \( C_{\text{gas}} = 95.2 \text{ J/mol-K} \), - \( \Delta T = (40 - (-12)) \text{C} = 52 \text{K} \). Calculate \( q_3 \): \[ q_3 = 0.430 \cdot 95.2 \cdot 52 = 2125.456 \text{ J} \approx 2.125 \text{ kJ} \]

Step 6: Calculate Total Heat Required

Add up all the individual heats to find the total heat required: \[ q_{\text{total}} = q_1 + q_2 + q_3 \] Substitute the values: \[ q_{\text{total}} = 2.115 + 9.159 + 2.125 = 13.399 \text{ kJ} \]

Key Concepts

Specific Heat Capacity

The specific heat capacity is a fundamental property of matter. It tells us the amount of heat required to raise the temperature of a given quantity of substance by one degree Celsius (or one Kelvin). Every material has its unique specific heat capacity, so this property plays a critical role in understanding how different substances absorb and lose heat.
For isobutane, the problem specifies the specific heat capacities for both its liquid and gaseous states:

• Liquid phase: 129.7 J/mol-K
• Gas phase: 95.2 J/mol-K

To calculate the heat required for temperature changes, we use the formulas:

• For the liquid phase: \( q = m \cdot C_{\text{liquid}} \cdot \Delta T \)
• For the gas phase: \( q = m \cdot C_{\text{gas}} \cdot \Delta T \)

Where \( q \) represents the heat added, \( m \) is the number of moles, \( C \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. By knowing these values, we can precisely calculate the energy required to change the temperature of isobutane as it transitions between phases.

Phase Change

Phase changes are transformations between different states of matter, such as from liquid to gas or solid to liquid. These transformations occur without any change in temperature, despite the exchange of heat.
For isobutane in the original problem, the critical phase change is vaporization—the transition from liquid to gas. This occurs at its boiling point, -12°C, and requires a specific amount of heat known as the heat of vaporization.
The heat of vaporization for isobutane is specified as 21.3 kJ/mol. This means that each mole of isobutane requires 21.3 kJ of energy to change from liquid to gas at the boiling point. The formula used for the heat required for vaporization is:

• \( q = n \cdot \Delta H_{\text{vap}} \)

Where \( q \) is the heat added, \( n \) is the number of moles, and \( \Delta H_{\text{vap}} \) is the heat of vaporization. This calculation is essential for accurately determining the total energy needs for phase changes, playing a critical role in thermodynamic calculations.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's a crucial property used to convert between the mass of a substance and the number of moles, which is essential for stoichiometric calculations in chemistry.
For isobutane, the chemical formula is \( C_4H_{10} \). To find its molar mass, sum the atomic masses of all the atoms contained in the molecule:

• Carbon (C): 12.01 g/mol, with four atoms contributing to a total of \( 4 \times 12.01 \) g/mol.
• Hydrogen (H): 1.008 g/mol, with ten atoms contributing to a total of \( 10 \times 1.008 \) g/mol.

This results in a molar mass of 58.14 g/mol for isobutane. Having the molar mass allows us to convert a given mass of isobutane into moles using the formula:

• \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \)

This is a critical step in the calculation, ensuring accurate conversion from mass to moles, which is necessary for calculating the required heat for phase changes and temperature changes.

Boiling Point

The boiling point is the temperature at which a liquid changes to a gas, under a given pressure (usually 1 atmosphere for standard boiling points). At this temperature, the vapor pressure of the liquid equals the external pressure surrounding the liquid.
For isobutane, the boiling point is given as -12°C. This is the temperature at which isobutane will begin to vaporize. Understanding the boiling point is crucial for thermodynamic calculations, as it indicates when a phase change from liquid to gas will occur, and lets us apply the heat of vaporization at the correct moment in these calculations.
When a substance reaches its boiling point, any heat input will be used solely to transform the liquid into a gas, rather than increasing the temperature. This temperature plateau during a phase change is a significant characteristic of thermodynamics. The boiling point helps define the conditions under which a substance will change its phase, which is vital for any process involving heating or cooling of materials.