Thermochemical Equations and Hess’s Law: Rearrangement and Calculation of Enthalpy Changes

Rearrangement of Thermochemical Equations

Introduction to Thermochemical Equations

A thermochemical equation is a chemical equation that includes an enthalpy of reaction (ΔH). The enthalpy change is directly proportional to the stoichiometry of the reaction as written. Any change to the original reaction will cause the same change in the ΔH value.

• Enthalpy of Reaction (ΔH): The heat change associated with a chemical reaction at constant pressure.

• Thermochemical Equation Example: 2 Mg (s) + O2 (g) → 2 MgO (s)    ΔH = –1204 kJ

Rearrangement Methods

Thermochemical equations can be manipulated mathematically, and the enthalpy change must be adjusted accordingly:

Operation	How to Adjust ΔH	Example
Multiplication	Multiply ΔH by the same factor as the equation	$\text{If: } \quad \mathrm{A} \rightarrow \mathrm{B} \quad \Delta H = x$ $2\mathrm{A} \rightarrow 2\mathrm{B} \quad \Delta H = 2x$
Division	Divide ΔH by the same factor as the equation	$\text{If: } \quad 2\mathrm{A} \rightarrow 2\mathrm{B} \quad \Delta H = y$ $\mathrm{A} \rightarrow \mathrm{B} \quad \Delta H = \frac{y}{2}$
Reversing	Change the sign of ΔH	$\text{If: } \quad \mathrm{A} \rightarrow \mathrm{B} \quad \Delta H = z$ $\mathrm{B} \rightarrow \mathrm{A} \quad \Delta H = -z$

Example: Rearranging a Formation Equation

Given the formation equation for boron trioxide:

• 4 B (s) + 3 O2 (g) → 2 B2O3 (s)    ΔHrxn = –2547 kJ

To find the enthalpy when the reaction is reversed:

• 2 B2O3 (s) → 4 B (s) + 3 O2 (g)    ΔHrxn = +2547 kJ

Additional info: Reversing the reaction changes the sign of ΔH.

Practice Example

Calculate ΔHrxn for:

• S (s) + (3/2) O2 (g) → SO3 (g)

Given:

• (1/2) S (s) + (1/2) O2 (g) → (1/2) SO2 (g)    ΔH = –296.8 kJ

• 2 SO3 (g) → 2 SO2 (g) + O2 (g)    ΔH = 198.4 kJ

Additional info: Multiply the first equation by 3 and the second by –1, then sum to get the target equation and ΔH.

Applying Hess’s Law

Hess’s Law Overview

Hess’s Law states that the enthalpy change of an overall reaction is the sum of the enthalpy changes of the individual steps (partial reactions):

• ΔHoverall = Σ ΔHsteps

This allows calculation of ΔH for reactions that cannot be measured directly, by combining known reactions.

Example: Using Hess’s Law

Given:

• XeF2 (s) + F2 (g) → XeF4 (s)    ΔH = +123 kJ

• Xe (g) + 2 F2 (g) → XeF4 (s)    ΔH = –262 kJ

To find ΔH for:

• XeF2 (s) + F2 (g) → Xe (g) + 2 F2 (g)

Reverse the second reaction and add to the first:

• ΔHrxn = +123 kJ + 262 kJ = +385 kJ

Additional info: Always adjust the sign of ΔH when reversing a reaction.

Stepwise Approach to Hess’s Law Problems

1. Identify and locate each compound in the overall equation within the set of partial reactions.

2. Reverse or multiply partial reactions as needed to match the overall equation, adjusting ΔH accordingly.

3. Sum the equations and cancel intermediates (compounds that appear on both sides).

Reaction Intermediates: Compounds that are produced in one step and consumed in another, and thus do not appear in the overall reaction.

Practice Problems

• Given a set of reactions and their ΔH values, manipulate and sum them to find the ΔH for a target reaction.

• Always check that the final equation matches the target reaction before summing ΔH values.

Key Terms and Concepts

• Thermochemical Equation: A balanced chemical equation that includes the enthalpy change.

• Enthalpy (ΔH): The heat content of a system at constant pressure.

• Hess’s Law: The total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in.

• Reaction Intermediate: A substance produced in one step and consumed in another during a multi-step reaction.

Summary Table: Manipulating Thermochemical Equations

Action	Effect on ΔH	Example
Multiply equation by n	Multiply ΔH by n	$\mathrm{A} \rightarrow \mathrm{B}, \Delta H = x$ $2\mathrm{A} \rightarrow 2\mathrm{B}, \Delta H = 2x$
Divide equation by n	Divide ΔH by n	$2\mathrm{A} \rightarrow 2\mathrm{B}, \Delta H = y$ $\mathrm{A} \rightarrow \mathrm{B}, \Delta H = \frac{y}{2}$
Reverse equation	Change sign of ΔH	$\mathrm{A} \rightarrow \mathrm{B}, \Delta H = z$ $\mathrm{B} \rightarrow \mathrm{A}, \Delta H = -z$

Additional info: These principles are essential for solving enthalpy problems in GOB Chemistry, especially when direct measurement is not possible.