Question

The corrosion (rusting) of iron in oxygen-free water includes the formation of iron(II) hydroxide from iron by the following 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?

Short answer

The change in enthalpy (ΔH) for this reaction is approximately \(-11,480.548 \: J\) and the change in internal energy (ΔE) is \(-11,727.52 \: J\).

Step-by-step solution

Calculate the work done

Given that the reaction performs 2.48 J of P-V work, we have w = 2.48 J.

Calculate the heat released

Given that 11.73 kJ of heat is released to the environment, we have q = -11.73 kJ (the negative sign indicates heat is released). To keep the units consistent, convert q to J: q = -11.73 kJ * 1000 J/kJ = -11730 J.

Calculate the change in internal energy (ΔE)

Using the relation ΔE = q + w, ΔE = (-11730 J) + (2.48 J) = -11727.52 J

Calculate the change in volume (ΔV)

Given that 1 mol of iron reacts, we have 1 mol of H₂ gas produced. Using the ideal gas law, PV = nRT, with pressure P = 101.3 kPa, n = 1 mol, R = 8.314 J/(mol*K), and temperature T = 298 K, ΔV = nRT/P = (1 mol) * (8.314 J/(mol*K)) * (298 K) / (101.3 kPa * (1 kJ/(kJ * J))) = 0.0244 m³/mol

Calculate ΔH

Using the relation ΔH = ΔE + PΔV, ΔH = (-11727.52 J) + (101.3 kPa * 0.0244 m³/mol) * (1 J/(1 kPa * m³)) ΔH = -11727.52 J + 246.972 J = -11480.548 J (approximately) Thus, the change in enthalpy (ΔH) for this reaction is approximately -11480.548 J and the change in internal energy (ΔE) is -11727.52 J.

Key concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a measure of heat energy released or absorbed during a chemical reaction at constant pressure. It helps us understand whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). In our exercise, the reaction is exothermic since heat is given off to the surroundings. This is indicated by the negative value assigned to \( q \), the heat, at \(-11.73 \text{ kJ}\) or \(-11730 \text{ J}\).

When we talk about enthalpy in terms of chemical reactions, it's essential to consider both the internal energy and the work done by the system as it expands or contracts. For reactions producing gases like hydrogen, work is done when the gas pushes against atmospheric pressure. In this scenario, the \( P-V \) work caused by gas expansion is \(2.48 \text{ J}\).

• The formula for enthalpy change is \( \Delta H = \Delta E + P \Delta V \).
• Here, \( \Delta E \) is the change in internal energy.
• \( P \Delta V \) represents the work done by the system due to volume change.

Therefore, the enthalpy change \( \Delta H \) calculated was approximately \(-11480.548 \text{ J}\), confirming that the reaction is strongly exothermic.

Internal Energy

Internal energy, symbolized by \( \Delta E \), encompasses all the energy contained within a system. This includes kinetic and potential energies of all atoms and molecules, and it is influenced directly by heat transfer and work done by the system.

In the corrosion problem, the internal energy change is determined using the formula \( \Delta E = q + w \), where \( q \) is the heat exchanged, and \( w \) is the work done. Our previous values help us calculate:

• \( q = -11730 \text{ J} \), indicating heat release.
• \( w = 2.48 \text{ J} \), denoting work done.

Thus, the internal energy change \( \Delta E \) was calculated to be \(-11727.52 \text{ J}\). This negative value indicates an overall loss of energy within the system, aligning with the reaction being exothermic.

Internal energy's role is crucial in thermochemistry, as it provides insight into both heat transfer and the balance of energy forms during reactions.

First Law of Thermodynamics

The First Law of Thermodynamics, a fundamental concept in thermochemistry, is essentially a law of energy conservation. It postulates that energy cannot be created or destroyed, only transformed from one form to another. Applied to chemical reactions, it means the energy change of a system is the sum of energy transferred as heat and work.

In the case of the corrosion reaction, the First Law is applied as "\( \Delta E = q + w \)," where:

• \( \Delta E \) represents the change in internal energy.
• \( q \) is the heat absorbed or released \(-11730 \text{ J}\).
• \( w \) is the work done \(2.48 \text{ J}\).

This principle ensures that the different energy forms during the reaction, whether absorbed as heat or used in doing work, are comprehensively accounted for. It gives us the tools to compute internal energy changes and further delve into subsequent changes in enthalpy.

By using the First Law, we're provided a clear understanding of how energy behaves in chemical processes, reinforcing the vital concept that energy is always conserved, even as it changes form during a reaction.