Determine The Standard Enthalpy Change (Δ_r H°) And Standard Internal Energy Change (Δ_r U°) For Reaction (3), Given Reactions (1) And (2) With Their Respective Δ_r H° Values.

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In the realm of thermodynamics, understanding energy changes associated with chemical reactions is crucial. Thermochemistry, a branch of chemical thermodynamics, delves into the heat absorbed or released during chemical and physical transformations. Two key thermodynamic properties used to quantify these energy changes are enthalpy change (Δ_r H°) and internal energy change (Δ_r U°). This article aims to elucidate the calculation of Δ_r H° and Δ_r U° for a specific reaction (R3) given the enthalpy changes of two other related reactions (R1 and R2). We will explore Hess's Law, a fundamental principle that allows us to determine enthalpy changes for reactions that can be expressed as a sum of other reactions. Additionally, we will discuss the relationship between enthalpy and internal energy changes, considering the work done by the system due to volume changes.

Background: Enthalpy and Internal Energy

Before diving into the calculations, it's essential to define enthalpy and internal energy. Internal energy (U) is the total energy of a system, including kinetic and potential energies of its constituent particles. Enthalpy (H), on the other hand, is a thermodynamic property that combines internal energy with the product of pressure (P) and volume (V): H = U + PV. For reactions occurring at constant pressure, enthalpy change (ΔH) represents the heat absorbed or released by the system. Exothermic reactions release heat (ΔH < 0), while endothermic reactions absorb heat (ΔH > 0).

The standard enthalpy change of reaction (Δ_r H°) refers to the enthalpy change when a reaction is carried out under standard conditions (298 K and 1 atm) with all reactants and products in their standard states. The standard internal energy change of reaction (Δ_r U°) is the change in internal energy when a reaction is carried out under standard conditions. The relationship between Δ_r H° and Δ_r U° is given by the equation:

Δ_r H° = Δ_r U° + Δ(PV) = Δ_r U° + (Δn_g)RT

where Δn_g is the change in the number of moles of gaseous species in the reaction, R is the ideal gas constant (8.314 J/mol·K), and T is the temperature in Kelvin. This equation highlights that the difference between enthalpy and internal energy changes is primarily due to the work done by the system against the constant pressure of the atmosphere, which is significant only when there is a change in the number of moles of gas.

Problem Statement

Consider the following chemical reactions with their respective standard enthalpy changes:

R1: A(g) + B(s) → 2C(g) Δ_r H° = -54.96 kJ/mol R2: 2A(g) + D(g) → 2E(g) Δ_r H° = +482.64 kJ/mol

Our objective is to determine the standard enthalpy change (Δ_r H°) and the standard internal energy change (Δ_r U°) for the following reaction:

R3: 4C(g) + D(g) → 2B(s) + 2E(g)

Applying Hess's Law to Determine Δ_r H° for Reaction R3

Hess's Law states that the enthalpy change for a reaction is independent of the pathway taken and depends only on the initial and final states. This means that if a reaction can be expressed as the sum of two or more other reactions, the enthalpy change for the overall reaction is the sum of the enthalpy changes for the individual reactions. To determine Δ_r H° for reaction R3, we need to manipulate reactions R1 and R2 in such a way that their sum yields reaction R3.

First, observe that reaction R3 involves 4 moles of C(g) on the reactant side, while reaction R1 produces 2 moles of C(g). To match the stoichiometry, we need to reverse reaction R1 and multiply it by 2. Reversing a reaction changes the sign of Δ_r H°, and multiplying a reaction by a factor multiplies Δ_r H° by the same factor. Therefore, the modified reaction R1 (R1') and its enthalpy change are:

R1': 2C(g) → A(g) + B(s) Δ_r H°' = -(-54.96 kJ/mol) * 2 = +109.92 kJ/mol

Multiplying the reverse reaction by 2:

2 * [2C(g) → A(g) + B(s)] which gives 4C(g) → 2A(g) + 2B(s) Δ_r H° = 2 * (+54.96 kJ/mol) = +109.92 kJ/mol

Next, we need to consider reaction R2, which involves reactants and products present in reaction R3. Reaction R2 is:

R2: 2A(g) + D(g) → 2E(g) Δ_r H° = +482.64 kJ/mol

Now, we can add the modified reaction R1' (multiplied by 2) and reaction R2 to see if their sum yields reaction R3. Adding these reactions:

4C(g) → 2A(g) + 2B(s) Δ_r H° = +109.92 kJ/mol 2A(g) + D(g) → 2E(g) Δ_r H° = +482.64 kJ/mol

Adding these two reactions, we get:

4C(g) + 2A(g) + D(g) → 2A(g) + 2B(s) + 2E(g)

Simplifying by canceling out the 2A(g) on both sides:

4C(g) + D(g) → 2B(s) + 2E(g)

This is exactly reaction R3. Therefore, the enthalpy change for reaction R3 (Δ_r H°_R3) can be calculated by adding the enthalpy changes for the modified reactions:

Δ_r H°_R3 = 2 * Δ_r H°' + Δ_r H°_R2 = +109.92 kJ/mol + 482.64 kJ/mol = +592.56 kJ/mol

So, the standard enthalpy change for reaction R3 is +592.56 kJ/mol. This positive value indicates that reaction R3 is endothermic, meaning it absorbs heat from the surroundings.

Calculating Δ_r U° for Reaction R3

Now that we have determined Δ_r H° for reaction R3, we can calculate Δ_r U° using the relationship between enthalpy and internal energy changes:

Δ_r H° = Δ_r U° + (Δn_g)RT

where:

  • Δ_r H° = +592.56 kJ/mol = 592560 J/mol
  • Δn_g = (moles of gaseous products) - (moles of gaseous reactants) in reaction R3
  • R = 8.314 J/mol·K
  • T = 298 K (standard conditions)

In reaction R3:

4C(g) + D(g) → 2B(s) + 2E(g)

The number of moles of gaseous reactants is 4 (from 4C(g)) + 1 (from D(g)) = 5 moles. The number of moles of gaseous products is 2 (from 2E(g)). Note that B(s) is a solid and does not contribute to Δn_g.

Therefore, Δn_g = 2 - 5 = -3 moles.

Now we can rearrange the equation to solve for Δ_r U°:

Δ_r U° = Δ_r H° - (Δn_g)RT

Plugging in the values:

Δ_r U° = 592560 J/mol - (-3 mol)(8.314 J/mol·K)(298 K)

Δ_r U° = 592560 J/mol + (3 mol)(8.314 J/mol·K)(298 K)

Δ_r U° = 592560 J/mol + 7432.716 J/mol

Δ_r U° = 600000 J/mol (approximately)

Δ_r U° = 599992.716 J/mol

Converting to kJ/mol:

Δ_r U° ≈ 599.99 kJ/mol

Thus, the standard internal energy change for reaction R3 is approximately 599.99 kJ/mol.

Summary of Results

For the reaction R3: 4C(g) + D(g) → 2B(s) + 2E(g), we have determined the following:

  • Standard enthalpy change (Δ_r H°) = +592.56 kJ/mol
  • Standard internal energy change (Δ_r U°) ≈ 599.99 kJ/mol

Conclusion

In conclusion, by applying Hess's Law and the relationship between enthalpy and internal energy changes, we have successfully calculated Δ_r H° and Δ_r U° for reaction R3. The positive value of Δ_r H° indicates that the reaction is endothermic, while the calculated Δ_r U° provides insight into the energy change at a constant volume. Understanding these thermodynamic properties is crucial for predicting the feasibility and energy requirements of chemical reactions in various chemical processes and applications. Mastering these concepts allows chemists and engineers to design and optimize chemical reactions for efficiency and sustainability. These calculations are foundational in fields ranging from industrial chemistry to environmental science, underscoring the broad applicability of thermochemical principles.

  • Thermochemistry
  • Enthalpy Change (Δ_r H°)
  • Internal Energy Change (Δ_r U°)
  • Hess's Law
  • Endothermic Reactions
  • Exothermic Reactions
  • Standard Conditions
  • Ideal Gas Constant (R)
  • Stoichiometry
  • Thermodynamics