Problem 44 - General Chemistry

Question

Liquid butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is stored in cylinders to be used as a fuel. Suppose \(39.3 \mathrm{~g}\) of butane gas is removed from a cylinder. How much heat must be provided to vaporize this much gas? The heat of vaporization of butane is \(21.3 \mathrm{~kJ} / \mathrm{mol}\)

Step-by-Step Solution

Answer

To vaporize 39.3 g of butane, approximately 14.4 kJ of heat is needed.

Step 1: Determine moles of butane

First, we have to determine how many moles of butane are in 39.3 g. To do this, we need the molar mass of butane. The formula for butane is \(C_4H_{10}\), which has a molar mass of \((4 \times 12.01 + 10 \times 1.01) \ \text{g/mol} = 58.12 \ \text{g/mol}\). Divide the mass of butane by its molar mass to find the moles: \(\text{moles of butane} = \frac{39.3 \ \text{g}}{58.12 \ \text{g/mol}} \approx 0.676 \ \text{mol}.\)

Step 2: Calculate heat needed

We use the heat of vaporization to find the heat required to vaporize the butane. Multiply the moles of butane by the heat of vaporization: \(\text{Heat} = 0.676 \ \text{mol} \times 21.3 \ \text{kJ/mol} \approx 14.4 \ \text{kJ}.\) Thus, the amount of heat needed to vaporize 39.3 g of butane is approximately 14.4 kJ.

Key Concepts

Understanding the Molar Mass of Butane

The molar mass of a substance is an important concept in chemistry that helps us convert between grams and moles. For butane, a hydrocarbon with the chemical formula \(C_4H_{10}\), calculating its molar mass involves summing the atomic weights of all the atoms in a molecule of butane. The element carbon (C) has an atomic mass of approximately 12.01 grams per mole, and hydrogen (H) about 1.01 grams per mole.

To determine the molar mass of butane, you multiply the number of carbon atoms (4) by the atomic mass of carbon (12.01 g/mol) and the number of hydrogen atoms (10) by the atomic mass of hydrogen (1.01 g/mol). Thus:

• Mass of carbon in butane = 4 × 12.01 = 48.04 g/mol
• Mass of hydrogen in butane = 10 × 1.01 = 10.10 g/mol

Add these values together to find the molar mass of butane:

• Molar mass of butane = 48.04 + 10.10 = 58.14 g/mol

This calculation is essential for converting a given mass of butane into moles, which is a step needed for further calculations.

Calculating the Heat for Vaporization

Vaporization is the process where a liquid changes into vapor. For any given substance, the heat of vaporization is the amount of heat required to turn one mole of liquid into vapor without changing the temperature. For butane, the heat of vaporization is given as 21.3 kJ/mol.

To calculate the total heat required to vaporize a specific amount of butane, such as the 39.3 grams referenced here, you will first need to convert the mass to moles using its molar mass. We found out that 39.3 g of butane is equivalent to approximately 0.676 moles.

Once we know the number of moles, the calculation involves multiplying these moles by the heat of vaporization:

• Heat required = Moles of butane × Heat of vaporization
• Heat required = 0.676 mol × 21.3 kJ/mol = 14.41 kJ

Therefore, you need approximately 14.4 kJ of energy to vaporize 39.3 g of butane. This gives us an understanding of how much energy is needed to convert the liquid butane into gas.

Explaining Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the relationships between heat and other forms of energy associated with chemical reactions and physical changes of state. It's about understanding how energy is absorbed or released by substances and how it affects the behavior of matter during these transformations.

In the context of our butane vaporization problem, chemical thermodynamics helps determine the energy change involved when butane transitions from a liquid to a gaseous state. This specific energy change is quantified by the heat of vaporization, which is a form of latent heat.

Thermodynamics involves principles such as:

• Conservation of Energy: Energy cannot be created or destroyed; it can only be transformed from one form to another.
• Equilibrium: Reactions achieve a balance where the forward and reverse reactions occur at the same rate.
• Spontaneity: Determines if a process will occur naturally under certain conditions.

Understanding thermodynamics is crucial when calculating energy requirements for processes like vaporization, which in turn helps us manage fuels and many industrial applications.