Problem 93 - Chemistry: Principles and Reactions

Question

The normal boiling point for ethyl alcohol is \(78.4^{\circ} \mathrm{C} .5^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) is \(282.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\). At what temperature is the vapor pressure of ethyl alcohol \(357 \mathrm{~mm} \mathrm{Hg} ?\)

Step-by-Step Solution

Answer

Answer: The temperature at which the vapor pressure of ethyl alcohol is 357 mm Hg is approximately 66.4°C.

Step 1: Find the enthalpy of vaporization at T = 78.4°C using molar heat capacity

Since we are given the molar heat capacity at constant volume for C2H5OH(g), we can find the enthalpy of vaporization at T1 = 78.4°C (the given normal boiling point) using the following equation: \(\Delta H_{vap} = C_{v}(T_{2} - T_{1}) \text{(Assumption: constant molar heat capacity)}\) We have the molar heat capacity and T1, but we need to find T2. At the boiling point, the vapor pressure of the liquid is equal to the atmospheric pressure. Assuming standard conditions, the atmospheric pressure is 1 atm, which can be converted to mm Hg: \(1 \ \text{atm} = 760 \ \text{mm Hg}\) Since we know the vapor pressure is 357 mm Hg, we can find T2 by solving for it using the Clausius-Clapeyron equation. However, we first need the enthalpy of vaporization.

Step 2: Convert temperatures to Kelvin

To use the Clausius-Clapeyron equation and find the enthalpy of vaporization, we need to convert the given Celsius temperatures to Kelvin: \(T_{1(K)} = T_{1(°C)} + 273.15\) \(T_{1(K)} = 78.4 + 273.15 = 351.55 \ \text{K}\)

Step 3: Calculate the enthalpy of vaporization

Now we can substitute the conversion factor from Step 2 into the first equation for enthalpy of vaporization: \(\Delta H_{vap} = C_{v}(T_{2} - T_{1})\) \(\Delta H_{vap} = 282.7 \ \text{J/mol K} \cdot (760 - 351.55) \ \text{K}\) \(\Delta H_{vap} \approx 115189.485 \ \text{J/mol}\)

Step 4: Apply the Clausius-Clapeyron equation to find T2

With all necessary values to use the Clausius-Clapeyron equation, we can solve for the temperature at which the vapor pressure of ethyl alcohol is 357 mm Hg: \(ln\frac{P_2}{P_1}=-\frac{\Delta H_{vap}}{R}\left(\frac{1}{T2}-\frac{1}{T1}\right)\) \(ln\frac{357}{760}=-\frac{115189.485}{8.314}\left(\frac{1}{T2}-\frac{1}{351.55}\right)\) Now, solve for the temperature T2: \(T2 \approx 339.55 \ \text{K}\)

Step 5: Convert the final temperature to Celsius

Finally, we'll convert the temperature back to Celsius for the final answer: \(T_{2(°C)} = T_{2(K)} - 273.15\) \(T_{2(°C)} = 339.55 - 273.15\) \(T_{2(°C)} \approx 66.4 ^{\circ}\text{C}\) The temperature at which the vapor pressure of ethyl alcohol is 357 mm Hg is approximately 66.4°C.

Key Concepts

Ethyl Alcohol Boiling Point

The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. For ethyl alcohol, also known as ethanol, this occurs at 78.4°C under standard atmospheric pressure, which is equivalent to 760 mm Hg.
This means that at 78.4°C, ethyl alcohol's vapor pressure is precisely matching atmospheric pressure, allowing it to transition from a liquid to a gas.
Understanding boiling points helps us explore how different substances respond to heat, which is essential in fields like chemistry and thermodynamics.

• **Standard Atmospheric Pressure:** 760 mm Hg = 1 atm
• **Normal Boiling Point:** The temperature at which a liquid boils under 1 atm
• **Ethanol:** Boils at 78.4°C under normal pressure

Recognizing boiling points assists in predicting how a substance behaves under various temperatures and pressures, which is crucial for applications involving distillation and refrigeration.

Vapor Pressure

Vapor pressure is the pressure created by the vapor when in equilibrium with its liquid or solid form in a closed system. It measures the tendency of particles to escape from the liquid or solid state into the gaseous state.
For ethyl alcohol, the vapor pressure increases with temperature. At its boiling point of 78.4°C, the vapor pressure equals 760 mm Hg. However, when the temperature decreases, so does the vapor pressure.
The Clausius-Clapeyron equation allows us to relate vapor pressure to temperature mathematically. This equation is used to estimate the vapor pressure at different temperatures by finding the enthalpy of vaporization.

• **Equilibrium:** Vapor pressure reaches equilibrium at a given temperature when the rate of evaporation equals the rate of condensation
• **Clausius-Clapeyron Equation:** Assesses how vapor pressure changes with temperature
• **Observation:** Lower temperatures lead to lower vapor pressures

Understanding these concepts allows us to manipulate conditions in industrial processes and better comprehend weather patterns.

Enthalpy of Vaporization

Enthalpy of vaporization is the heat required to convert a given amount of a liquid into gas at a constant temperature and pressure. It essentially measures the energy needed to break intermolecular forces and transition a liquid into its gaseous form.
For ethyl alcohol, this property is vital to determining its volatility and how readily it evaporates. In the context of the exercise, the enthalpy of vaporization (\(\Delta H_{vap}=115189.485 \ ext{J/mol}\)) was calculated based on known values.
This value is crucial for applying the Clausius-Clapeyron equation to find the vapor pressure at different temperatures.

• **Energy Requirement:** The measure of energy needed to vaporize the liquid
• **Application:** Important for calculating other thermodynamic properties
• **Relevance:** Used to predict behavior in dynamic temperature environments

Recognizing the enthalpy of vaporization helps us gauge how substances behave when heated, playing a role in things like distillation and designing efficient energy systems.