Problem 103 - Chemistry The Central Science In Si Units

Question

The corrosion (rusting) of iron in oxygen-free water includes the formation of iron(II) hydroxide from iron by the follow. ing reaction: $$ \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g) $$ If 1 mol of iron reacts at \(298 \mathrm{~K}\) under \(101.3 \mathrm{kPa}\) pressure, the reaction performs \(2.48 \mathrm{~J}\) of \(P-V\) work, pushing back the atmosphere as the gaseous \(\mathrm{H}_{2}\) forms. At the same time, \(11.73 \mathrm{~kJ}\) of heat is released to the environment. What are the values of \(\Delta H\) and of \(\Delta E\) for this reaction?

Step-by-Step Solution

Answer

\(\Delta H = -11.73 \; \text{kJ}\), \(\Delta E = -11.73 \; \text{kJ}\)

Step 1: Understanding the Known Values

We are given that the reaction performs \(2.48 \; \text{J}\) of \(P-V\) work, and \(11.73 \; \text{kJ}\) of heat is released. We need to calculate \(\Delta H\) and \(\Delta E\).

Step 2: Relating Work, Heat, and Energy Change

We use the relationship \(\Delta E = q + w\), where \(q\) is the heat exchanged and \(w\) is the work done. Here, \(q = -11.73 \; \text{kJ}\) (since heat is released) and \(w = -2.48 \; \text{J}\) (work done by the system, so energy is lost).

Step 3: Converting Work Units

The work done \(w = -2.48 \; \text{J}\) needs to be converted to \(\text{kJ}\). Since \(1 \; \text{kJ} = 1000 \; \text{J}\), \(w = -0.00248 \; \text{kJ}\).

Step 4: Calculating Change in Internal Energy (\(\Delta E\))

Substitute the values into the formula: \(\Delta E = q + w = -11.73 \; \text{kJ} - 0.00248 \; \text{kJ} = -11.73248 \; \text{kJ}\).

Step 5: Calculating Enthalpy Change (\(\Delta H\))

Since \(\Delta H = q\) under constant pressure, and heat exchange is \(q = -11.73 \; \text{kJ}\), we have \(\Delta H = -11.73 \; \text{kJ}\).

Key Concepts

Enthalpy

Enthalpy is a key concept in thermochemistry. It represents the heat content of a system at constant pressure. When reactions occur, they often release or absorb heat, and the change in enthalpy (94H) helps us understand this energy flow.

In a chemical reaction, if the enthalpy change is negative, it signifies that the system is releasing heat to the surroundings, as with many exothermic processes. Conversely, a positive enthalpy change indicates that the system is absorbing heat, typical of endothermic reactions.

• Standard Conditions: 94H is often measured under standard conditions (298 K and 101.3 kPa).
• Relation to Heat: Under constant pressure, 94H is equal to the heat exchanged (94H = q).

In the exercise concerning iron reacting in water, we observe that 94H is -11.73 kJ, indicating an exothermic reaction where energy is released as heat.

This heat release can be a critical factor in understanding how spontaneously corrosion processes occur in certain environments.

Internal Energy

Internal energy (94E) is the total energy contained within a system. In thermochemistry, it is crucial as it accounts for both kinetic and potential energies of molecules.

The change in internal energy, 94E, can be calculated using the formula: \[94E = q + w \]where \(q\) is the heat exchanged and \(w\) is the work done by or on the system. This equation shows that 94E considers both heat transfer and mechanical work changes.

• Heat Exchange: It can be gained or lost by the system (94E decreases if the system loses heat).
• Work Done: The system does work if it expands against external pressure (energy is lost).

In our example, the corrosion reaction performed 2.48 J of work and released 11.73 kJ of heat. The conversion of work into kJ and adding it to the heat value gives 94E = -11.73248 kJ, indicating an overall energy decrease in the system.

This negative internal energy shows that the system has less energy post-reaction, aligning with the concept that energy is transferred to the surroundings.

P-V Work

Pressure-Volume work, commonly referred to as P-V work, occurs when a system changes its volume against an external pressure. It's a vital concept in understanding energy changes in reactions involving gases.

In the given exercise, the formation of hydrogen gas (92(g)) from the reaction results in the system expanding against the atmospheric pressure, performing P-V work:

• Formula: P-V work is calculated as \(w = -P\Delta V\).
• Work Done by the System: If the system does work, energy is lost, as shown by a negative value of \(w\).

The exercise tells us that 2.48 J of P-V work was done as the hydrogen gas expanded. Converting this to kJ helps in combining it with heat to find 94E.

Understanding P-V work allows us to see how volume changes influence energy distribution in chemical processes. It exemplifies how systems interact with their surroundings, significantly affecting their internal energy.