Question

For the vaporization of one mole of bromine at \(59^{\circ} \mathrm{C}\), determine (a) \(\Delta H\) (Table 8.2) (b) \(\Delta P V\) (c) \(\Delta E\)

Short answer

Question: Calculate the changes in enthalpy, pressure times volume, and internal energy for the vaporization of one mole of bromine at 59°C. Answer: (a) Change in enthalpy (∆H) = 29.54 kJ/mol (b) Change in Pressure times Volume (∆PV) = 2764.988 J/mol (c) Change in internal energy (∆E) = 26775.012 J/mol

Step-by-step solution

Find the Change in Enthalpy (∆H)

To solve this part, we need to refer to Table 8.2, which gives the change in enthalpy values for various processes. According to this table, the change in enthalpy for the vaporization of one mole of bromine is 29.54 kJ/mol. So, ∆H = 29.54 kJ/mol

Convert temperature to Kelvin

The exercise gives the temperature in Celsius (59°C). We need to convert it to Kelvin for further calculations. To convert Celsius to Kelvin, add 273.15 to the Celsius value. Temperature in Kelvin (T) = 59 + 273.15 = 332.15 K

Calculate the Change in Pressure times Volume (∆PV)

Use the Clapeyron equation to find the change in PV. The Clapeyron Equation is: \(\frac{dP}{dT} = \frac{\Delta H}{T \Delta V}\) We also need the ideal gas equation (for one mole of gas): \(P V = R T\), where R is the ideal gas constant, 8.314 J/(mol*K). So, \(\Delta P V = P_{vapor} V_{molar} - P_{liquid} V_{molar}\) Since liquid bromine is much more dense than its vapor, we can assume that the molar volume of liquid bromine is negligible compared to the molar volume of the vapor, i.e., \(V_{molar(liquid)} \approx 0\). Thus, we have \(\Delta P V = P_{vapor} V_{molar}\) Now, we can use the Clapeyron equation and the ideal gas equation to determine the value for ∆PV: \(\Delta P V = \Delta H (\frac{R T}{\Delta H}) = R T\) So, ∆PV = (8.314 J/(mol*K)) * (332.15 K) = 2764.988 J/mol

Calculate the Change in Internal Energy (∆E)

Now, we can find the change in internal energy using the formula ∆E = ∆H - ∆(PV). ∆E = ∆H - ∆PV = (29.54 kJ/mol - 2764.988 J/mol) Since, 29.54 kJ/mol = 29540 J/mol (1 kJ = 1000 J) ∆E = 29540 J/mol - 2764.988 J/mol = 26775.012 J/mol Now, we have all the answers from the exercise. (a) Change in enthalpy (∆H) = 29.54 kJ/mol (b) Change in Pressure times Volume (∆PV) = 2764.988 J/mol (c) Change in internal energy (∆E) = 26775.012 J/mol

Key concepts

Change in Enthalpy

The change in enthalpy, denoted as \(\Delta H\), is a measure of the total energy change in a system during a process, such as vaporization, at constant pressure. When considering the vaporization of bromine, \(\Delta H\) represents the amount of heat required to convert one mole of liquid bromine into bromine vapor without changing the temperature. In our specific example, the \(\Delta H\) for the vaporization of one mole of bromine at 59°C is given as 29.54 kJ/mol.

Understanding \(\Delta H\) is crucial for thermodynamic calculations as it indicates how much energy is involved in phase changes. This information is pertinent to fields like chemistry, physics, and engineering, where heat transfer and phase transitions play a pivotal role in the behavior of substances and the design of processes.

Clapeyron Equation

The Clapeyron Equation is a powerful tool for studying phase transitions, such as melting, boiling, and sublimation. It relates the pressure \(P\), temperature \(T\), and change in enthalpy (\(\Delta H\)) to the change in volume that occurs during a phase transition (\(\Delta V\)).

The equation is typically expressed as:\[\frac{dP}{dT} = \frac{\Delta H}{T \Delta V}\]
It assumes that the transition occurs reversibly and at equilibrium. In the context of vaporization, the equation can help determine the change in pressure times volume (\(\Delta PV\)) by employing the ideal gas law for the vapor phase and considering the volume of the liquid phase negligible in comparison. An understanding of the Clapeyron Equation is fundamental for students because it offers a framework for predicting how changes in temperature affect phase equilibrium and vice versa.

Internal Energy

Internal energy, denoted as \(\Delta E\), encapsulates the total energy contained within a system. It includes kinetic energy, due to the motion of molecules, and potential energy, arising from intermolecular forces. In a phase change like vaporization, internal energy changes as molecules absorb heat and overcome intermolecular forces to transition from liquid to vapor phase.

The relationship between enthalpy and internal energy is given by the equation \(\Delta E = \Delta H - \Delta(PV)\). This indicates that internal energy change considers both the heat absorbed (\(\Delta H\)) and the work done on or by the system (\(\Delta(PV)\)). For vaporization, work is done to expand against atmospheric pressure, which must be accounted for to find the change in internal energy. This concept is vital for comprehending how energy transfers and transformations facilitate physical processes, informing students about the underlying mechanics of thermodynamics.