Question

The standard enthalpy of formation of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(298 \mathrm{~K}\) is \(-285.8\) \(\mathrm{kJ} / \mathrm{mol} .\) Calculate the change in internal energy for the following process at \(298 \mathrm{~K}\) and 1 atm: $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \Delta E^{\circ}=? $$ (Hint: Using the ideal gas equation, derive an expression for work in terms of \(n, R\), and \(T\).)

Short answer

The change in internal energy for the given process at 298K and 1 atm is approximately -284.6 kJ/mol.

Step-by-step solution

From thermodynamics, we know that the change in enthalpy, \(\Delta H\), is related to the change in internal energy, \(\Delta E\), and the work done in a process, \(W\), as follows: $$ \Delta H = \Delta E + W $$ We want to find \(\Delta E\). So we need to calculate \(W\) and rearrange the equation: $$ \Delta E = \Delta H - W $$ #Step 2: Derive the work done using the ideal gas equation#

The ideal gas equation is given by: $$ PV=nRT $$ Where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. Now, let's consider the work done, \(W\), during the expansion of the gas. It can be expressed as: $$ W = -P\Delta V $$ Using the ideal gas equation, we can derive an expression for \(\Delta V\) in terms of \(n\), \(R\), and \(T\): $$ \Delta V = \frac{\Delta (nRT)}{P} $$ Substitute this expression back into the work equation: $$ W = -P \left(\frac{\Delta (nRT)}{P}\right) $$ Cancel out the pressure terms: $$ W = -\Delta(nRT) $$ #Step 3: Calculate the work done in the given process#

In the given process, we have the following stoichiometric changes in number of moles: $$ \Delta n = 1 + \frac{1}{2} - 1 = \frac{1}{2} $$ Now, we can substitute the values of \(\Delta n\), \(R\), and \(T\) into the work equation: $$ W = - \left(\frac{1}{2}\cdot 8.314 \ \mathrm{J/(mol\cdot K)} \cdot 298 \ \mathrm{K} \right) $$ Calculate the work done: $$ W \approx -1243 \ \mathrm{J/mol} $$ #Step 4: Calculate the change in internal energy#

Now that we have the work done, we can calculate the change in internal energy using the equation from Step 1: $$ \Delta E = \Delta H - W $$ Substitute the given values for \(\Delta H\) and the calculated value for \(W\): $$ \Delta E = -285.8 \ \mathrm{kJ/mol} - (-1243 \ \mathrm{J/mol}) $$ Convert the second term to kJ/mol: $$ \Delta E = -285.8 \ \mathrm{kJ/mol} + 1.243 \ \mathrm{kJ/mol} $$ Calculate the change in internal energy: $$ \Delta E \approx -284.6 \ \mathrm{kJ/mol} $$ Thus, the change in internal energy for the given process at 298K and 1 atm is approximately -284.6 kJ/mol.

Key concepts

Enthalpy of Formation

Enthalpy of formation refers to the heat change when one mole of a compound forms from its constituent elements in their standard states. For water, the standard enthalpy of formation at 298 K is \(-285.8 \, \text{kJ/mol}\).
This value provides a basis for understanding energy changes in chemical reactions. When reactions occur, substances often release or absorb heat. This heat change is crucial in thermodynamics to determine whether a reaction is spontaneous or requires external energy.
Understanding enthalpy helps in studying energy conservation and conversion in chemical processes. It provides insight into how energy is stored and transformed in molecular structures.

Internal Energy

Internal energy, denoted as \(\Delta E\), represents the total energy contained within a system. It's a key concept in thermodynamics, where it is analyzed along with heat and work.
When we look at internal energy change, we focus on how energy is transferred between systems, which can occur through heat transfer or work done on or by the system.
The relationship between enthalpy change \((\Delta H)\), internal energy change \((\Delta E)\), and work \((W)\) is given by the equation:

• \(\Delta H = \Delta E + W\)

This equation helps us find the internal energy change when we know the enthalpy change and work involved.
Internal energy change is essential for understanding how energy is managed in chemical reactions and processes.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry, expressed as \(PV = nRT\), describing the behavior of ideal gases. Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature.
This law helps predict how gases will respond under varying conditions of temperature, volume, and pressure. In the context of the exercise, it assists in determining the change in volume, \(\Delta V\), during the process.
For calculating work done by a gas, we express \(\Delta V\) in terms of \(n\), \(R\), and \(T\):

• \(\Delta V = \frac{\Delta(nRT)}{P}\)

By substituting this into the work formula, we help determine the work done during the expansion or compression of the gas.

Work Calculation

Work calculation in thermodynamics involves understanding how energy is transferred during gas expansion or compression. In this exercise, work is calculated using:

• \(W = -P\Delta V\)

By substituting the ideal gas law expression for \(\Delta V\), work done in a chemical process is given by:

• \(W = -\Delta(nRT)\)

In our case, through mole balance, we calculate:

• \(\Delta n = \frac{1}{2}\)

Subsequently, substituting values for \(\Delta n\), \(R\), and \(T\):

• \(W \approx -1243 \, \text{J/mol}\)

Understanding work helps in assessing how systems perform energy transfer and transformation, essential for engineering and scientific applications.