Problem 108
Question
The molar heat of vaporization of acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O},\) is \(30.3 \mathrm{~kJ} / \mathrm{mol}\) at its boiling point. How many kilojoules of heat would be liberated by the condensation of \(5.00 \mathrm{~g}\) of acetone?
Step-by-Step Solution
Verified Answer
10.4 kJ of heat would be liberated by the condensation of 5.00 g of acetone.
1Step 1: Determine the moles of acetone
To find out the number of moles of acetone, use its molecular weight (\text{C}_{3} \text{H}_{6}\text{O}) and the given mass. The molecular weight of acetone is 58.08 g/mol. Calculate the moles using the formula: \[ \text{moles of acetone} = \frac{\text{mass of acetone}}{\text{molecular weight of acetone}} \]
2Step 2: Calculate the heat liberated during condensation
Use the molar heat of vaporization to find the total heat liberated during the condensation. The formula to use is: \[ \text{heat} = \text{moles of acetone} \times \text{molar heat of vaporization} \]
3Step 3: Calculate the final answer
Multiply the moles of acetone by the molar heat of vaporization to get the heat in kilojoules.
Key Concepts
Condensation of AcetoneCalculating MolesEnthalpy Change
Condensation of Acetone
When we talk about the condensation of acetone, we're referring to the process where acetone changes from its gaseous form back into a liquid. This is the opposite of vaporization, where a liquid turns into a gas. Condensation is an exothermic process, meaning it releases heat into the surrounding environment.
In our exercise, the key figure is the 'molar heat of vaporization' of acetone, which is given as 30.3 kJ/mol. Though it may sound counterintuitive, the molar heat of vaporization is also used when considering condensation. This is because the amount of heat required to vaporize a mole of a substance is the same amount released when a mole of the substance condenses. Thus, when 5.00 grams of acetone condense, they release heat, which can be calculated using the substance's molar heat of vaporization.
In our exercise, the key figure is the 'molar heat of vaporization' of acetone, which is given as 30.3 kJ/mol. Though it may sound counterintuitive, the molar heat of vaporization is also used when considering condensation. This is because the amount of heat required to vaporize a mole of a substance is the same amount released when a mole of the substance condenses. Thus, when 5.00 grams of acetone condense, they release heat, which can be calculated using the substance's molar heat of vaporization.
Calculating Moles
Calculating moles is a foundational concept in chemistry that allows us to work with the actual number of molecules or atoms we have in a given sample. Moles relate mass to the molecular weight of a substance, enabling us to convert between the macroscopic world that we can measure, and the microscopic world of individual molecules.
The molecular weight of acetone, which is 58.08 g/mol, tells us how much one mole of acetone weighs. To find the number of moles in our 5.00 gram sample, we divide the mass of acetone by its molecular weight: \[ \text{moles of acetone} = \frac{\text{mass of acetone}}{\text{molecular weight of acetone}} \] By doing so, we correlate the given mass of a compound to the number of moles, which then can be used to calculate various thermodynamic properties, including the heat released or absorbed under different conditions.
The molecular weight of acetone, which is 58.08 g/mol, tells us how much one mole of acetone weighs. To find the number of moles in our 5.00 gram sample, we divide the mass of acetone by its molecular weight: \[ \text{moles of acetone} = \frac{\text{mass of acetone}}{\text{molecular weight of acetone}} \] By doing so, we correlate the given mass of a compound to the number of moles, which then can be used to calculate various thermodynamic properties, including the heat released or absorbed under different conditions.
Enthalpy Change
Enthalpy change, often denoted as \( \Delta H \), represents the heat absorbed or released by a system at constant pressure. It's a measure of the energy change within a system, excluding work done by the system on its surroundings. In the context of condensation, the enthalpy change is negative because heat is being released, hence it is often referenced as the 'heat of condensation'.
To find the enthalpy change of the condensation process for acetone, we use the equation: \[ \text{heat} = \text{moles of acetone} \times \text{molar heat of vaporization} \] Once we've calculated the moles of acetone in our sample, we multiply it by the molar heat of vaporization to find the heat in kilojoules. This value corresponds to the heat that would be liberated, or the negative enthalpy change, for the process of condensation.
To find the enthalpy change of the condensation process for acetone, we use the equation: \[ \text{heat} = \text{moles of acetone} \times \text{molar heat of vaporization} \] Once we've calculated the moles of acetone in our sample, we multiply it by the molar heat of vaporization to find the heat in kilojoules. This value corresponds to the heat that would be liberated, or the negative enthalpy change, for the process of condensation.
Other exercises in this chapter
Problem 106
What intermolecular attractions will be formed when the following substances condense from the gas phase? (a) hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\righ
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We can determine the heat of vaporization by using the Clausius-Clapeyron equation if we know the vapor pressures of a substance at two different temperatures.
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