Problem 46 - Chemistry: The Central Science in SI Units, Global Edition

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 to convert \(25.0 \mathrm{~g}\) of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at \(40^{\circ} \mathrm{C}\) is \(13,523 \mathrm{~J}\).

Step 1: Calculate the heat required to heat up the liquid to its boiling point

First, we need to find the moles of isobutane given the mass: \(n = \frac{m}{M}\) where \(n\) represents the number of moles, \(m\) is the mass of isobutane, and \(M\) is the molar mass of isobutane. For isobutane, the molecular formula is \(C_4H_{10}\), and its molar mass is: \(M = 4 \times 12.01 + 10 \times 1.01 = 58.12 \mathrm{~g/mol}\) Now we can calculate the moles of isobutane: \(n = \frac{25.0 ~\mathrm{g}}{58.12 ~\mathrm{g/mol}} = 0.430 ~\mathrm{mol}\) Next, we need to calculate the heat required to heat the liquid from its initial temperature (-50°C) to the boiling point (-12°C). To do this, we will use the molar specific heat of the liquid phase and the formula: \(q_1 = nC_{p_{liquid}} \Delta T\) where \(q_1\) represents the heat required for this step, \(C_{p_{liquid}}\) is the molar specific heat of liquid isobutane, and \(\Delta T\) is the temperature change. \(\Delta T = T_{bp} - T_{initial} = -12 - (-50) = 38^{\circ} \mathrm{C}\) Now we can find the heat required: \(q_1 = (0.430 ~\mathrm{mol})(129.7 ~\mathrm{J/mol~K})(38~\mathrm{K}) = 2127 ~\mathrm{J}\)

Step 2: Calculate the heat required to vaporize the liquid

Now we need to calculate the heat required for the phase change, vaporization. We will use the heat of vaporization and the formula: \(q_2 = n\Delta H_{vap}\) where \(q_2\) represents the heat required for this step, and \(\Delta H_{vap}\) is the heat of vaporization for isobutane. \( q_2 = (0.430 ~\mathrm{mol})(21.3 ~\mathrm{kJ/mol}) = 9.16 ~\mathrm{kJ}\) Convert to Joules: \(q_2 = 9160 ~\mathrm{J}\)

Step 3: Calculate the heat required to heat up the gas to the final temperature

Finally, we need to calculate the heat required to heat the gas from the boiling point (-12°C) to the final temperature (40°C). To do this, we will use the molar specific heat of the gas phase and the formula: \(q_3 = nC_{p_{gas}}\Delta T\) where \(q_3\) represents the heat required for this step, \(C_{p_{gas}}\) is the molar specific heat of isobutane in the gas phase, and \(\Delta T\) is the temperature change. \(\Delta T = T_{final} - T_{bp} = 40 - (-12) = 52^{\circ} \mathrm{C}\) Now we can find the heat required: \(q_3 = (0.430 ~\mathrm{mol})(95.2 ~\mathrm{J/mol~K})(52 ~\mathrm{K}) = 2236 ~\mathrm{J}\)

Step 4: Calculate the total heat required

To find the total heat required, we need to sum the heats calculated for the three steps: \(q_{total} = q_1 + q_2 + q_3\) \(q_{total} = 2127 ~\mathrm{J} + 9160 ~\mathrm{J} + 2236 ~\mathrm{J} = 13523 ~\mathrm{J}\) The heat required to convert 25.0 g of isobutane from a liquid at -50°C to a gas at 40°C is 13,523 J.

Key Concepts

Heat of Vaporization

When a substance transforms from a liquid to a vapor, it absorbs a specific amount of heat without changing its temperature; this is known as the heat of vaporization. This energy breaks intermolecular forces in the liquid, allowing the molecules to transition to a vapor state. For isobutane, the heat of vaporization is noted as 21.3 kJ/mol. This indicates the amount of energy needed per mole for its phase change from liquid to gas at its boiling point. Understanding this helps in determining the energy requirements for entire vaporization processes, crucial for applications like refrigeration, where control of temperature and phase shifts are important for functionality.

Specific Heat Capacity

The specific heat capacity of a substance is the amount of heat needed to change the temperature of one mole of substance by one degree Celsius. This concept is important to understand because it influences how much energy substances require to increase their temperature. For isobutane, the molar specific heat is different in its liquid ( 129.7 J/mol·K) and gas ( 95.2 J/mol·K) phases. Hence, different energy amounts are required to increase the temperature of liquid or gaseous isobutane. This difference is vital in industrial processes where temperature control is necessary during both phases.

Phase Change

A phase change occurs when a substance transitions between solid, liquid, and gas phases. For isobutane, the main phase changes involved in the problem include the liquid-to-vapor phase at the boiling point. During this phase change, the temperature remains constant despite the input of energy, as the energy is used to change the phase rather than increase the temperature. The boiling point of isobutane is -12°C, which means this is the temperature at which it starts converting from liquid to gas. Properly calculating the energy required for this transition is essential in the context of energy efficiency and cost-effective processes.

Molar Mass Calculation

Molar mass calculation is fundamental in determining the number of moles of a substance from its given mass. Understanding the molar mass allows one to convert masses of elements or compounds into moles, which is a crucial step in most chemistry calculations. For isobutane, with its chemical formula C_4H_{10} , the molar mass is calculated using the atomic masses of carbon and hydrogen, resulting in 58.12 g/mol. Knowing this lets us compute how many moles are present in a given mass of isobutane, which in turn helps in calculating other thermodynamic properties like heat capacity or heat of vaporization based on the number of moles being heated or vaporized.