Problem 17 - Chemistry The Molecular Science

Question

Some camping stoves contain liquid butane, \(\mathrm{C}_{4} \mathrm{H}_{10} .\) They work only when the outside temperature is warm enough to allow the butane to have a reasonable vapor pressure (so they are not very good for camping in temperatures below about \(\left.0{ }^{\circ} \mathrm{C}\right)\). Assume the vaporization enthalpy of butane is \(22.44 \mathrm{~kJ} / \mathrm{mol}\) and the camp stove fuel tank contains \(190 . \mathrm{g}\) liquid \(\mathrm{C}_{4} \mathrm{H}_{10} .\) Calculate the heat energy transfer required to vaporize all of the butane.

Step-by-Step Solution

Answer

The heat energy transfer required is 73.36 kJ.

Step 1: Identify Given Information

We are given: - Enthalpy of vaporization of butane, \( \Delta H_{vap} = 22.44 \, \text{kJ/mol} \).- Mass of liquid butane, \( 190 \, \text{g} \). - Molecular weight of butane, \( \mathrm{C}_{4}\mathrm{H}_{10} \), which is \( 4\times12.01 + 10\times1.01 = 58.12 \, \text{g/mol} \).

Step 2: Convert Mass to Moles

To find the amount of energy required for vaporization, we first need to convert the mass of butane to moles using the formula:\[\text{moles of } \mathrm{C}_4\mathrm{H}_{10} = \frac{\text{mass}}{\text{molecular weight}} = \frac{190 \, \text{g}}{58.12 \, \text{g/mol}} \approx 3.27 \, \text{mol}\]

Step 3: Calculate Heat Energy Transfer

Using the moles of butane and the enthalpy of vaporization, we calculate the total heat energy required to vaporize all the butane:\[Q = n \times \Delta H_{vap} = 3.27 \, \text{mol} \times 22.44 \, \text{kJ/mol} = 73.36 \, \text{kJ}\]

Step 4: Conclusion

Thus, the heat energy transfer required to vaporize all 190 g of butane is approximately 73.36 kJ.

Key Concepts

Heat energy transfer

Understanding heat energy transfer is crucial for analyzing the amount of energy needed to change the state of a substance, such as turning liquid butane into gas. This process involves the transfer of heat energy from the surroundings to the butane to provide the necessary energy for phase change.

When a substance absorbs heat, its molecules begin to move faster, leading to a change in state—such as from liquid to gas—once the right amount of energy is absorbed. This energy transfer is essential in processes like cooking, heating, and even using a camping stove with butane.

For butane to fully vaporize, the appropriate amount of energy must be supplied. This means understanding not only the heat involved but also the other factors impacting the process, such as temperature, pressure, and the substance's properties.

Enthalpy of vaporization

The enthalpy of vaporization is a specific type of enthalpy change and a critical factor in phase changes from liquid to gas. It quantifies the amount of energy required to vaporize one mole of a substance at its boiling point under standard pressure conditions.

For butane, the enthalpy of vaporization is given as "22.44 kJ/mol." This means that to convert one mole of liquid butane completely into a gas at its boiling point, 22.44 kJ of energy is needed. This value is crucial for engineers and scientists when evaluating the energy needs for processes involving butane, particularly in industrial settings.

In practical terms, knowing the enthalpy of vaporization helps predict how a chemical substance will behave under different temperature conditions—like when camping at lower temperatures where butane's vapor pressure may be insufficient for vaporization without extra energy.

Molecular weight calculation

Molecular weight calculation is a foundational concept in chemistry that helps to determine how much of a substance is needed or used in a reaction or process. It is particularly useful for converting between mass and moles—a common task in stoichiometry and thermochemistry.

For butane, the molecular formula is \(\mathrm{C}_{4}\mathrm{H}_{10}\), and its molecular weight is calculated by adding up the atomic weights of carbon and hydrogen in butane's chemical structure:

• Carbon (C): 12.01 g/mol
• Hydrogen (H): 1.01 g/mol

Thus, the molecular weight becomes \(4 \times 12.01 + 10 \times 1.01 = 58.12\,\text{g/mol}.\)

Knowing the molecular weight allows conversion from grams to moles using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molecular weight}}.\) This step is essential for computing quantities like the number of moles of butane needed for energy transfer calculations.