Problem 11 - Chemistry and Chemical Reactivity

Question

Ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) has a vapor pressure of \(59 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C} .\) What quantity of heat energy 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

Approximately 90.26 kJ of heat energy is required to evaporate 125 mL of ethanol.

Step 1: Calculate the Mass of Ethanol

First, calculate the mass of ethanol. Using the volume of ethanol given as \(125 \text{ mL}\) and the density as \(0.7849 \text{ g/mL}\), multiply them together to find the mass.\[\text{Mass} = 125 \text{ mL} \times 0.7849 \text{ g/mL} = 98.1125 \text{ g}\]

Step 2: Convert Mass of Ethanol to Moles

Next, convert the mass of ethanol to moles. The molar mass of ethanol \(\mathrm{C_2H_5OH}\) can be calculated as follows: \(2 \times 12.01\text{ (C)} + 6 \times 1.008\text{ (H)} + 16.00\text{ (O)} = 46.08 \text{ g/mol}\). Now, use this to convert grams to moles:\[\text{Moles of ethanol} = \frac{98.1125 \text{ g}}{46.08 \text{ g/mol}} \approx 2.13 \text{ mol}\]

Step 3: Compute Heat Energy Required for Vaporization

Use the enthalpy of vaporization to find the total heat energy required. Multiply the moles of ethanol by the given enthalpy of vaporization \(42.32 \text{ kJ/mol}\):\[\text{Heat energy} = 2.13 \text{ mol} \times 42.32 \text{ kJ/mol} \approx 90.26 \text{ kJ}\]

Key Concepts

Ethanol Vaporization

When ethanol (\(\mathrm{C_2H_5OH}\)) vaporizes, it changes from liquid to gas. This physical change requires energy, known as the enthalpy of vaporization. For ethanol at \(25^{\circ} \mathrm{C}\), this is \(42.32\, \mathrm{kJ/mol}\).
Understanding ethanol vaporization involves knowing why energy is necessary. In the liquid state, ethanol molecules attract each other tightly. They must overcome these forces to become gas molecules that are far apart. This requires inputting energy.

• Vaporization is endothermic. Energy gets absorbed during the process.
• The \(\mathrm{kJ/mol}\) value tells us how much energy is needed per mole of ethanol.

Given \(125\, \mathrm{mL}\) of ethanol, the goal is to calculate how much heat is necessary for the complete vaporization of this volume.

Density Calculation

To find how much heat is needed for \(125\, \mathrm{mL}\) of ethanol, first calculate its mass. Density is the key here, usually expressed as grams per milliliter, \(\mathrm{0.7849\, g/mL}\)eral approach for determining mass:

• Use the formula: Mass = Volume × Density.
• For our ethanol, it means: Mass = \(125\, \mathrm{mL}\) × \(0.7849\, \mathrm{g/mL}\),which results in \(98.1125\, \mathrm{g}\).

This mass calculation is crucial as it forms the basis for converting to moles later on. Always remember, density relates mass and volume, making it a vital step in processes involving liquid substances.

Molar Mass Conversion

After determining the mass of ethanol, convert this mass into moles. This is because vaporization energy is calculated on a molar basis. Knowing the molar mass of ethanol, \(\mathrm{46.08\, g/mol}\),is essential. Here's a simple way to calculate moles:

• The formula is: Moles = Mass / Molar Mass.
• Therefore, \(\mathrm{98.1125\, g} ÷ \mathrm{46.08\, g/mol} = \approx 2.13\, \mathrm{mol}\)

This conversion sets up the next and final step: finding out exactly how much heat energy is required for the vaporization of this amount of ethanol.

Heat Energy Calculation

Finally, determine the heat energy required for the vaporization process using the enthalpy of vaporization. You already know there are about \(2.13\, \mathrm{mol}\)of ethanol needing to be vaporized.
The calculation involves:

• Multiplying the moles of ethanol by the enthalpy of vaporization, \(42.32\, \mathrm{kJ/mol}\).
• This gives: Heat energy = \(2.13\ \mathrm{mol} \times 42.32\, \mathrm{kJ/mol} \approx 90.26\, \mathrm{kJ}\).

So, \(90.26\, \mathrm{kJ}\)is the total amount of heat energy necessary to vaporize \(125\, \mathrm{mL}\)of ethanol at \(25^{\circ} \mathrm{C}\).This step combines all previous calculations to provide the final energy requirement.