Problem 42 - Chemistry and Chemical Reactivity

Question

If you place 1.0 L of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in a small laboratory that is \(3.0 \mathrm{m}\) long, \(2.5 \mathrm{m}\) wide, and \(2.5 \mathrm{m}\) high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethanol at \(25^{\circ} \mathrm{C}\) is \(59 \mathrm{mm} \mathrm{Hg}\) and the density of the liquid at this temperature is \(0.785 \mathrm{g} / \mathrm{cm}^{3}\).

Step-by-Step Solution

Answer

All the ethanol will evaporate.

Step 1: Calculate Laboratory Volume

First, calculate the volume of the laboratory room in cubic meters. The formula for the volume of a rectangular box is given by: \[ V = ext{Length} \times ext{Width} \times ext{Height} \] Substitute the values: \[ V = 3.0 \, \text{m} \times 2.5 \, \text{m} \times 2.5 \, \text{m} = 18.75 \, \text{m}^3 \] Convert the volume to liters (1 m³ = 1000 L): \[ V = 18.75 \, \text{m}^3 \times 1000 = 18750 \, \text{L} \]

Step 2: Calculate Ethanol Mass in Grams

To determine the mass of 1.0 L of ethanol, use the given density: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]Rearrange to find mass: \[ \text{Mass} = \text{Density} \times \text{Volume} \]Substitute the values: \[ \text{Mass} = 0.785 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 785 \, \text{g} \]

Step 3: Calculate Moles of Ethanol

Convert the mass of ethanol to moles using the molar mass of ethanol (C_2 H_5 OHe) which is approximately 46.07g/mol: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]\[ \text{Moles} = \frac{785 \, \text{g}}{46.07 \, \text{g/mol}} \approx 17.03 \, \text{mol} \]

Step 4: Calculate Maximum Moles Possible in Vapor Phase

Use the ideal gas law to determine the maximum moles of ethanol vapor that the air can hold: \[ PV = nRT \]Where: - \( P \) is the vapor pressure (converted to atm: 59 mmHg = 59/760 atm), - \( V \) is the volume of the room in liters (18750 L), - \( R \) is the gas constant (0.0821 L·atm/mol·K), - \( T \) is the temperature in Kelvin (298 K at 25°C). Rearrange to solve for \( n \) (moles): \[ n = \frac{PV}{RT} \]Substitute the values: \[ n = \frac{(59/760) \times 18750}{0.0821 \times 298} \approx 57.2 \, \text{mol} \]

Step 5: Determine if All Ethanol Evaporates

Compare the moles of ethanol that can exist as vapor (57.2 mol) with the moles available initially (17.03 mol). Since 57.2 mol is greater than 17.03 mol, all of the ethanol will evaporate.

Key Concepts

Vapor Pressure

Vapor pressure is an important concept when studying the behavior of liquids, especially in relation to evaporation. It refers to the pressure exerted by a vapor in thermodynamic equilibrium with its liquid phase at a given temperature. This concept helps us understand how a liquid might behave under different conditions.
In our exercise, the vapor pressure of ethanol at 25°C is given as 59 mm Hg. This means that at this temperature, ethanol will continue to evaporate until the pressure created by its vapor reaches 59 mm Hg. If the surrounding pressure is lower than this, ethanol molecules will keep evaporating, attempting to reach this equilibrium state.
Understanding vapor pressure helps in predicting whether a given amount of liquid can completely evaporate in a specific environment. In this scenario, since the vapor pressure of ethanol was compared to the capacity of the room to hold ethanol vapor, it was concluded that all the ethanol would indeed evaporate.

Ideal Gas Law

The Ideal Gas Law is a crucial equation in chemistry that relates pressure, volume, temperature, and number of moles of a gas. It is represented as: \( PV = nRT \)where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume of the gas,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (0.0821 L·atm/mol·K),
• \( T \) is the temperature in Kelvin.

In our problem, this law was used to compute how many moles of ethanol vapor the laboratory room can hold at 25°C. By applying the values provided, we determined that the room could hold approximately 57.2 moles of ethanol. This comparison was essential to decide whether all of the ethanol would evaporate. Since the calculated capacity is higher than the initial moles of ethanol (17.03 moles), the room can accommodate all of the ethanol as a vapor, confirming complete evaporation.

Density Calculation

Density is a measure of how much mass a substance has per unit volume. It is often expressed in units like grams per cubic centimeter (g/cm³). The formula to find density is:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
For ethanol in our exercise, the density was given as 0.785 g/cm³. This value was crucial for converting the volume of ethanol (1.0 L) into its mass, which is a necessary step for further calculations. By rearranging the density formula to solve for mass:\[ \text{Mass} = \text{Density} \times \text{Volume} \]we found that 1.0 L of ethanol corresponds to a mass of 785 grams.
This calculation was pivotal to estimating how many moles of ethanol were initially present, enabling us to further analyze its potential evaporation using the Ideal Gas Law.

Molar Mass Calculation

Molar mass is an essential quantity in chemistry, as it allows you to convert between the mass of a substance and the number of moles. The molar mass is the mass of one mole of a substance, and its unit is grams per mole (g/mol).
In the exercise, the molar mass of ethanol \((\text{C}_2 \text{H}_5 \text{OH})\) was given as approximately 46.07 g/mol. To find the number of moles in the 1.0 L of ethanol, we used the mass (785 grams) obtained from the density calculation and the following formula:\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]By substituting the values, the calculation showed that there are about 17.03 moles of ethanol present.
This number of moles provided the baseline for determining whether or not all the ethanol could evaporate in the given laboratory conditions.