Problem 11 - Chemistry and Chemical Reactivity

Question

Ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) has a vapor pressure of \(59 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\). What quantity of energy as heat is required to evaporate \(125 \mathrm{mL}\) of the alcohol at \(25^{\circ} \mathrm{C} ?\) The enthalpy of vaporization of the alcohol at \(25^{\circ} \mathrm{C}\) is 42.32 \(\mathrm{kJ} / \mathrm{mol} .\) The density of the liquid is \(0.7849 \mathrm{g} / \mathrm{mL}.\)

Step-by-Step Solution

Answer

90.15 kJ of energy is required to evaporate 125 mL of ethanol at 25°C.

Step 1: Calculate the Mass of Ethanol

First, calculate the mass of 125 mL of ethanol using its density. Since the density is 0.7849 g/mL: \[ \text{Mass} = \text{Volume} \times \text{Density} = 125 \text{ mL} \times 0.7849 \text{ g/mL} = 98.1125 \text{ g} \]

Step 2: Convert Mass to Moles

Next, convert the mass of ethanol to moles. The molar mass of ethanol is calculated as: \[ \text{Molar mass of ethanol (C}_2\text{H}_5\text{OH)} = 2 \times 12.01 + 6 \times 1.01 + 16.00 = 46.08 \text{ g/mol} \]Now, use this molar mass to convert grams to moles: \[ \text{Moles of ethanol} = \frac{98.1125 \text{ g}}{46.08 \text{ g/mol}} = 2.13 \text{ moles} \]

Step 3: Calculate the Heat Required for Vaporization

Use the enthalpy of vaporization to determine the energy required to vaporize the ethanol. The enthalpy of vaporization is 42.32 kJ/mol, so: \[ \text{Energy required (q)} = \text{Moles} \times \text{Enthalpy of vaporization} = 2.13 \text{ moles} \times 42.32 \text{ kJ/mol} = 90.15 \text{ kJ} \]

Step 4: Finalize the Calculation

Thus, 90.15 kJ of energy is required to evaporate 125 mL of ethanol at 25°C.

Key Concepts

Vapor Pressure

Vapor pressure is an important concept when understanding how liquids evaporate. It refers to the pressure exerted by the vapor when it is in equilibrium with its liquid form at a given temperature. This pressure increases with temperature because more molecules have enough energy to escape the liquid's surface and enter the gaseous phase.
In this exercise, we deal with ethanol which has a vapor pressure of 59 mm Hg at 25°C. This shows how readily ethanol evaporates under these conditions. Higher vapor pressure indicates that the liquid can evaporate more easily, which is an important detail when calculating the energy required for vaporization or processes involving enthalpy changes.
Understanding vapor pressure helps us to appreciate the balance between a substance's gaseous and liquid states and provides insight into the liquid's evaporation rate.

Mole Calculation

Calculating moles is essential in chemistry as it allows us to count particles by weighing them. The mole is a unit that measures the amount of a substance. One mole corresponds to Avogadro's number, or approximately \(6.022 \times 10^{23}\) particles of that substance.
For ethanol, the exercise involves converting its mass to moles using its molar mass, which is determined by the formula \(2 \times 12.01 + 6 \times 1.01 + 16.00 = 46.08\) g/mol. Once the molar mass is known, the number of moles can be calculated by the formula: \[ \text{Moles of a substance} = \frac{\text{Mass of the substance (in grams)}}{\text{Molar mass of the substance (g/mol)}} \]
By knowing the number of moles, you can calculate other properties such as the energy required for a chemical process, making mole calculation a pivotal step in solving many chemistry problems.

Density of Liquids

Density is a measure of how much mass is contained in a given volume. It is an intrinsic property that can help identify substances and predict how they will behave in different situations. The density of a liquid can affect its buoyancy and its flow properties.
In this exercise, the density of ethanol is used to determine how much it weighs, which is the first step in calculating the energy required to evaporate a given quantity. The calculation involves using the formula: \( \text{Mass} = \text{Density} \times \text{Volume} \).
With ethanol's density being 0.7849 g/mL, we multiply by its volume to find the mass. Understanding how density interacts with volume and mass allows for a clearer comprehension of the substance's characteristics and aids in further quantitative analysis required in chemistry.