Calculating The Heat Of Reaction For 2OF₂(g) + 2S(s) → SO₂(g) + SF₄(g) Using Hess's Law

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Introduction

In the realm of chemical thermodynamics, understanding the heat of reaction, denoted as ΔH°, is crucial for predicting the feasibility and energy changes associated with chemical reactions. The heat of reaction, also known as the enthalpy change, quantifies the amount of heat absorbed or released during a chemical reaction at constant pressure. A negative ΔH° indicates an exothermic reaction, where heat is released, while a positive ΔH° signifies an endothermic reaction, where heat is absorbed. This article delves into the process of determining the heat of reaction for a specific chemical reaction using Hess's Law, a fundamental principle in thermochemistry. We will focus on the reaction: 2OF₂(g) + 2S(s) → SO₂(g) + SF₄(g), and calculate its ΔH° at 298 K using a set of given reactions and their respective ΔH° values.

Hess's Law: A Cornerstone of Thermochemistry

Hess's Law, named after the Swiss chemist Germain Hess, states that the enthalpy change for a chemical reaction is independent of the path taken between the initial and final states. In simpler terms, if a reaction can be expressed as the sum of a series of steps, the enthalpy change for the overall reaction is the sum of the enthalpy changes for each individual step. This law is a direct consequence of the fact that enthalpy is a state function, meaning its value depends only on the initial and final states of the system, not on the pathway taken to reach those states. This principle allows us to calculate the heat of reaction for complex reactions by manipulating and combining the enthalpy changes of simpler, known reactions.

The power of Hess's Law lies in its ability to bypass the need for direct calorimetric measurements for every reaction. By strategically combining known thermochemical equations, we can indirectly determine the enthalpy change for reactions that are difficult or impossible to measure directly. This approach is particularly useful in scenarios involving unstable intermediates, very slow reactions, or reactions that produce multiple products, making direct measurement challenging. To effectively utilize Hess's Law, we must carefully manipulate the given thermochemical equations, ensuring that the target reaction is obtained by summing the modified equations. Remember that when an equation is multiplied by a factor, its ΔH° value must also be multiplied by the same factor. Similarly, when an equation is reversed, the sign of its ΔH° value must be changed.

Problem Statement: Calculating the Heat of Reaction

Our objective is to determine the heat of reaction (ΔH°) at 298 K for the following reaction:

2OF₂(g) + 2S(s) → SO₂(g) + SF₄(g)

We are provided with the following thermochemical equations and their corresponding ΔH° values:

  1. OF₂(g) + H₂O(l) → O₂(g) + 2HF(g) ΔH° = -276.6 kJ/mol
  2. SF₄(g) + 2H₂O(l) → SO₂(g) + 4HF(g) ΔH° = -827.5 kJ/mol
  3. H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -285.8 kJ/mol
  4. S(s) + O₂(g) → SO₂(g) ΔH° = -296.8 kJ/mol
  5. ½H₂(g) + ½F₂(g) → HF(g) ΔH° = -271.1 kJ/mol

By strategically manipulating these equations and applying Hess's Law, we can calculate the ΔH° for the target reaction. The following steps outline the process:

Step-by-Step Solution Using Hess's Law

To determine the heat of reaction for the target reaction, we need to manipulate the given equations in such a way that they add up to the target reaction. This involves multiplying equations by coefficients and/or reversing equations, while correspondingly adjusting their ΔH° values.

Step 1: Analyze the Target Reaction

The target reaction is:

2OF₂(g) + 2S(s) → SO₂(g) + SF₄(g)

We need to identify which of the given equations contain the reactants (OF₂ and S) and products (SO₂ and SF₄) of the target reaction.

Step 2: Manipulate the Given Equations

  1. Equation 1: OF₂(g) + H₂O(l) → O₂(g) + 2HF(g) ΔH° = -276.6 kJ/mol

    • We need 2 OF₂(g) on the reactant side, so we multiply this equation by 2:

    2OF₂(g) + 2H₂O(l) → 2O₂(g) + 4HF(g) ΔH° = 2 * (-276.6 kJ/mol) = -553.2 kJ/mol

  2. Equation 2: SF₄(g) + 2H₂O(l) → SO₂(g) + 4HF(g) ΔH° = -827.5 kJ/mol

    • We need 1 SF₄(g) on the product side, so we keep this equation as is.

  3. Equation 3: H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -285.8 kJ/mol

    • This equation doesn't directly contribute to the target reaction, but it contains H₂O(l), which needs to be canceled out. We'll address this later.

  4. Equation 4: S(s) + O₂(g) → SO₂(g) ΔH° = -296.8 kJ/mol

    • We need 2 S(s) on the reactant side and 1 SO₂(g) on the product side, so we multiply this equation by 2:

    2S(s) + 2O₂(g) → 2SO₂(g) ΔH° = 2 * (-296.8 kJ/mol) = -593.6 kJ/mol

    • However, we only need 1 SO₂(g) on the product side in the target reaction. Since Equation 2 already provides SO₂(g), we will eventually need to reverse this equation to cancel out the extra SO₂(g).

  5. Equation 5: ½H₂(g) + ½F₂(g) → HF(g) ΔH° = -271.1 kJ/mol

    • This equation provides HF(g), which is a byproduct in other equations. We need to eliminate HF(g). We'll address this later.

Step 3: Sum the Manipulated Equations

At this point, let's sum the modified Equations 1, 2, and 4:

  • 2OF₂(g) + 2H₂O(l) → 2O₂(g) + 4HF(g) ΔH° = -553.2 kJ/mol
  • SF₄(g) + 2H₂O(l) → SO₂(g) + 4HF(g) ΔH° = -827.5 kJ/mol
  • 2S(s) + 2O₂(g) → 2SO₂(g) ΔH° = -593.6 kJ/mol

Adding these equations gives:

2OF₂(g) + 2S(s) + 4H₂O(l) → SF₄(g) + 2SO₂(g) + 4HF(g) + 2O₂(g) + 4HF(g)

ΔH° = -553.2 kJ/mol + (-827.5 kJ/mol) + (-593.6 kJ/mol) = -1974.3 kJ/mol

This isn't quite the target reaction yet. We have extra H₂O(l), SO₂(g), and HF(g) that need to be eliminated.

Step 4: Eliminate Byproducts

  1. Eliminate SO₂(g): We have 2 SO₂(g) on the product side, but we only need 1. To reduce it to 1, let's reverse Equation 4:

    • Reverse of Equation 4: SO₂(g) → S(s) + O₂(g) ΔH° = +296.8 kJ/mol

    Now, adding this reversed equation to the sum from Step 3 reduces the SO₂(g) by one and also removes 1 S(s) and 1 O₂(g).

  2. Eliminate H₂O(l): We have 4 H₂O(l) on the reactant side. To eliminate this, we can use Equation 3 (H₂(g) + ½O₂(g) → H₂O(l)) and multiply it by 4 and reverse it:

    • 4 * Reverse of Equation 3: 4H₂O(l) → 4H₂(g) + 2O₂(g) ΔH° = 4 * (+285.8 kJ/mol) = +1143.2 kJ/mol

    This will cancel out the H₂O(l) but introduces 4 H₂(g).

  3. Eliminate HF(g): We have 8 HF(g) on the product side. To eliminate this, we use Equation 5 (½H₂(g) + ½F₂(g) → HF(g)), multiply it by 8, and reverse it:

    • 8 * Reverse of Equation 5: 8HF(g) → 4H₂(g) + 4F₂(g) ΔH° = 8 * (+271.1 kJ/mol) = +2168.8 kJ/mol

    This will eliminate the HF(g) and also cancel out the 4 H₂(g) introduced in the previous step.

Step 5: Final Summation

Now, let's sum all the manipulated equations:

  • 2OF₂(g) + 2H₂O(l) → 2O₂(g) + 4HF(g) ΔH° = -553.2 kJ/mol
  • SF₄(g) + 2H₂O(l) → SO₂(g) + 4HF(g) ΔH° = -827.5 kJ/mol
  • 2S(s) + 2O₂(g) → 2SO₂(g) ΔH° = -593.6 kJ/mol
  • SO₂(g) → S(s) + O₂(g) ΔH° = +296.8 kJ/mol
  • 4H₂O(l) → 4H₂(g) + 2O₂(g) ΔH° = +1143.2 kJ/mol
  • 8HF(g) → 4H₂(g) + 4F₂(g) ΔH° = +2168.8 kJ/mol

Summing these equations and canceling out the common terms, we get:

2OF₂(g) + 2S(s) → SO₂(g) + SF₄(g)

ΔH° = -553.2 kJ/mol + (-827.5 kJ/mol) + (-593.6 kJ/mol) + 296.8 kJ/mol + 1143.2 kJ/mol + 2168.8 kJ/mol

ΔH° = 1634.5 kJ/mol

Final Answer

The heat of reaction (ΔH°) at 298 K for the reaction 2OF₂(g) + 2S(s) → SO₂(g) + SF₄(g) is 1634.5 kJ/mol. This positive value indicates that the reaction is endothermic, meaning it requires heat to proceed.

Conclusion

This comprehensive walkthrough demonstrates the application of Hess's Law in determining the heat of reaction for a complex chemical process. By strategically manipulating and combining known thermochemical equations, we successfully calculated the ΔH° for the target reaction. Understanding and applying Hess's Law is essential for thermochemical calculations and provides valuable insights into the energy changes associated with chemical reactions. In this specific case, we found that the reaction 2OF₂(g) + 2S(s) → SO₂(g) + SF₄(g) is endothermic, requiring a significant amount of energy input to occur. The accurate determination of enthalpy changes is vital in various fields, including chemical engineering, materials science, and environmental chemistry, where energy considerations play a crucial role in process design and optimization. Mastering Hess's Law allows chemists and engineers to predict the thermal behavior of chemical reactions, leading to safer and more efficient chemical processes.