Problem 21 - General Chemistry: Principles and Modern Applications

Question

The enthalpy of vaporization of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l}),\) is \(33.9 \mathrm{kJmol}^{-1}\) at \(298 \mathrm{K}\). How many liters of \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})\) measured at \(298 \mathrm{K}\) and \(95.1 \mathrm{mmHg}\), are formed when \(1.54 \mathrm{kJ}\) of heat is absorbed by \(\mathrm{C}_{6} \mathrm{H}_{6}(1)\) at a constant f 298 K?

Step-by-Step Solution

Answer

Therefore, 10.9 liters of C6H6(g) are formed when 1.54 kJ of heat is absorbed by C6H6(l) under the given conditions.

Step 1: Calculate the number of moles of benzene that can be vaporized with given heat

As we know that enthalpy of vaporization is the energy required to convert 1 mole of a substance from the liquid to the gas phase without any change in temperature. Therefore, we can calculate the number of moles of benzene which can be vaporized by the given heat as \[ moles = \frac{Heat}{ Enthalpy \: of \: vaporization} \]. Inputting the given values, we get \[moles = \frac{1.54 kJ}{33.9 kJ/mol} = 0.045 mol\]

Step 2: Applying Ideal Gas Law

Ideal Gas Law is given by \[ P * V = n * R * T \], where P represents the pressure, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.08206 L atm / K mol), and T is the temperature in Kelvin. We are required to find V, therefore we need to rearrange the equation to express V: \[V = \frac{n * R * T}{P}\] Now, we can substitute the given variables into this equation. But before that, we need to convert pressure from mmHg to atm because the gas constant we are using requires the pressure to be in atm. We know, 1 atm = 760 mmHg. Hence, \[P = \frac{95.1 mmHg}{760 mmHg/atm} = 0.125 atm\]

Step 3: Substituting values into the Ideal Gas Law equation

Now we substitute the calculated and given values into the Ideal Gas Law equation: \[ V = \frac{0.045 mol * 0.08206 L atm / K mol * 298 K }{0.125 atm} = 10.9 L\]

Key Concepts

Enthalpy of Vaporization

The enthalpy of vaporization is an important concept in understanding how substances change from liquid to gas. It measures the energy required to vaporize one mole of a substance at constant temperature and pressure.
For substances like benzene, which is \( \text{C}_6 \text{H}_6 \), this energy requirement is crucial when calculating how much heat is needed to transform a specific amount of liquid into its gaseous state.
In the given problem, the enthalpy of vaporization for benzene is \( 33.9 \text{kJ/mol} \) at a temperature of \( 298 \text{K} \). This means that to vaporize \( 1 \text{mol} \) of benzene, \( 33.9 \text{kJ} \) of energy is required.
By knowing the enthalpy of vaporization, you can calculate how much benzene can be vaporized when a certain amount of energy is absorbed.

Benzene Vaporization

Benzene is a common organic compound with the formula \( \text{C}_6 \text{H}_6 \). Its vaporization process is the conversion from the liquid to the gaseous phase. This transformation is highly dependent on energy, which is provided as heat at a constant temperature.
In this exercise, benzene vaporizes in response to an absorbed heat of \( 1.54 \text{kJ} \). We use the enthalpy of vaporization to determine the amount of benzene that can undergo this change.
By dividing the given heat by the enthalpy of vaporization, \( 0.045 \text{mol} \) of benzene can be vaporized. This insight is essential in practical applications such as distillation processes, where controlling the vaporization of benzene is crucial for product separation.

Pressure Conversion

Pressure conversion is a critical step when dealing with gas equations, especially when applying the Ideal Gas Law. The Ideal Gas Law requires pressure in \( \text{atm} \) because the constant used (\( R = 0.08206 \text{L atm / K mol} \)) assumes this unit.
In this problem, the initial pressure is given as \( 95.1 \text{mmHg} \). To convert this to \( \text{atm} \), divide by \( 760 \text{mmHg/atm} \), since \( 1 \text{atm} \) is equivalent to \( 760 \text{mmHg} \).
After performing this calculation, the pressure is found to be \( 0.125 \text{atm} \). Making sure pressure is in the correct units is crucial when solving problems involving gases, as it ensures that the calculations use consistent units throughout.

Moles Calculation

Calculating moles is a fundamental step in solving gas-related problems. It involves determining how many moles of a substance are present, based on a given amount of heat energy or mass.
In the context of the Ideal Gas Law, knowing the number of moles is necessary because it directly affects the volume of gas produced. For the vaporization of benzene, \( 0.045 \text{mol} \) was determined by dividing the absorbed heat (\( 1.54 \text{kJ} \)) by the enthalpy of vaporization (\( 33.9 \text{kJ/mol} \)).
This conversion is vital in chemical equations and reactions, as the mole serves as the bridge between energy and physical quantity. Understanding how to calculate moles enables you to manipulate and predict the behavior of gases under various conditions.