Problem 29 - Chemical Principles: The Quest for Insight

Question

(a) At its boiling point, the vaporization of \(0.235 \mathrm{~mol} \mathrm{CH}_{4}(1)\) requires \(1.93 \mathrm{~kJ}\) of heat. What is the enthalpy of vaporization of methane? (b) An electric heater was immersed in a flask of boiling crhanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), and \(22.45 \mathrm{~g}\) of ethanol was vaporized when \(21.2 \mathrm{~kJ}\) of energy was supplied. What is the enthalpy of vaporization of ethanol?

Step-by-Step Solution

Answer

The enthalpy of vaporization of methane is \(8.21 \mathrm{~kJ/mol}\), and for ethanol, it is \(43.5 \mathrm{kJ/mol}\).

Step 1: Determine the Enthalpy of Vaporization for Methane

Use the formula for enthalpy change, which is the heat absorbed or released per mole of substance. The formula is: \[\text{Enthalpy of vaporization} (\Delta H_{vap}) = \frac{Q}{n}\] where \(Q\) is the amount of heat energy in kilojoules and \(n\) is the number of moles of the substance. For methane (CH4), \(Q = 1.93\) kJ and \(n = 0.235\) mol. Plug these values into the formula to find the enthalpy of vaporization.

Step 2: Calculate the Enthalpy of Vaporization of Methane

Insert the given values into the enthalpy formula: \[\Delta H_{vap} = \frac{1.93\, \text{kJ}}{0.235\, \text{mol}}\] and calculate \(\Delta H_{vap}\) for methane.

Step 3: Determine the Enthalpy of Vaporization for Ethanol

First, convert the mass of ethanol to moles using its molar mass (\text{46.07 g/mol}): \[n = \frac{m}{M}\] where \(m\) is the mass of ethanol and \(M\) is the molar mass of ethanol. Then use the formula from Step 1 to find the enthalpy of vaporization for ethanol.

Step 4: Calculate the Moles of Ethanol

Insert the given values to find the number of moles of ethanol: \[n = \frac{22.45\, \text{g}}{46.07\, \text{g/mol}}\] and solve for \(n\).

Step 5: Calculate the Enthalpy of Vaporization of Ethanol

With the number of moles and the amount of heat supplied, use the enthalpy formula to find \(\Delta H_{vap}\) for ethanol: \[\Delta H_{vap} = \frac{Q}{n}\] where \(Q = 21.2\) kJ.

Key Concepts

Enthalpy Change Formula

Understanding the enthalpy change formula is crucial when studying thermodynamics in chemistry. This formula allows us to calculate the enthalpy change, \[\begin{equation}ΔH = \frac{Q}{n} \end{equation}\], which represents the amount of heat energy absorbed or released during a phase change, per mole of substance. Our primary focus is on the phase change known as vaporization—the transition from liquid to gas. When a substance vaporizes, it absorbs heat from the surroundings; this heat absorbed per mole is the enthalpy of vaporization (\[\begin{equation}ΔH_{vap}\end{equation}\]). For example, if 1.93 kJ of heat (\[\begin{equation}Q\end{equation}\]) is required to vaporize 0.235 moles (\[\begin{equation}n\end{equation}\]) of methane, the enthalpy of vaporization would be calculated using the given values in the enthalpy change formula.

Mol to Grams Conversion

When working with chemical reactions or processes like vaporization, it's often necessary to convert between moles and grams—an essential skill in both chemistry and stoichiometry. The mole is a fundamental unit in chemistry for quantifying the amount of a substance, and converting it to grams requires the molar mass of the substance, which is the mass of one mole of that substance, expressed in grams per mole (\[\begin{equation}\text{g/mol}\end{equation}\]).To perform a mol to grams conversion, use the formula: \[\begin{equation}n = \frac{m}{M}\end{equation}\], where \[\begin{equation}n\end{equation}\] is the number of moles, \[\begin{equation}m\end{equation}\] is the mass of the substance in grams, and \[\begin{equation}M\end{equation}\] is the molar mass. For instance, if you have 22.45 g of ethanol with a molar mass of 46.07 g/mol, dividing the mass by the molar mass gives you the amount in moles, necessary for calculating the enthalpy of vaporization.

Heat Energy in Chemistry

Heat energy, often symbolized as \[\begin{equation}Q\end{equation}\], plays a pivotal role in various chemical processes, especially those involving phase changes such as vaporization. In chemistry, heat is a form of energy transfer due to a temperature difference and is measured in joules (J) or kilojoules (kJ).When a substance like methane or ethanol vaporizes, it absorbs a quantity of heat energy. The amount of energy absorbed reflects the energetic requirement for the molecules to overcome intermolecular forces and transition to the gaseous state. Calculating the heat required to vaporize a substance at its boiling point, as seen in our examples, helps determine the enthalpy of vaporization, a key thermodynamic quantity illustrating the energy changes during this phase transition.

Physical Chemistry Principles

Physical chemistry integrates the principles of physics and chemistry to understand the physical properties and behavior of molecules. In the context of enthalpy of vaporization, physical chemistry principles dictate that the heat absorbed during vaporization relates directly to the strength of intermolecular forces within the liquid; stronger forces require more energy to break. Other relevant principles include the concept of molar mass and its role in determining stoichiometric relationships in mol to grams conversions, as well as the laws of thermodynamics, which govern the flow and transformation of energy in chemical processes.In the exercise presented, these principles are applied to find the enthalpy of vaporization for methane and ethanol, demonstrating the practical application of theoretical concepts in quantifying the energy dynamics of phase changes. Understanding these core principles is foundational in mastering material in physical chemistry and beyond.