4.2 Enthalpy of reactions and Hess's Law

Enthalpy of Reactions and Hess's Law

Enthalpy and Hess's Law are central tools in thermochemistry for quantifying heat flow in chemical reactions. They connect the energetics of bond-breaking and bond-forming at the molecular level to the macroscopic heat effects you can measure with a calorimeter.

Because enthalpy is a state function, you can calculate the energy change for a reaction you can't easily run in the lab by combining simpler reactions whose enthalpy changes are known. This principle underlies everything from industrial process design to predicting reaction feasibility.

Enthalpy in Thermochemistry

Definition and Significance of Enthalpy

Enthalpy ($H$) is a thermodynamic state function defined as:

$H = U + PV$

where $U$ is the internal energy, $P$ is pressure, and $V$ is volume. The reason enthalpy is so useful: at constant pressure, the enthalpy change $\Delta H$ equals the heat exchanged between the system and surroundings ($\Delta H = q_P$). That's what makes it directly measurable in most laboratory and real-world conditions.

Because $H$ is a state function, $\Delta H$ depends only on the initial and final states of the system, not on the path connecting them. This property is what makes Hess's Law work.

Types of Enthalpy Changes

• Exothermic reactions have $\Delta H < 0$. The system releases heat to the surroundings. Example: combustion of methane releases 890.4 kJ/mol.
• Endothermic reactions have $\Delta H > 0$. The system absorbs heat from the surroundings. Example: the thermal decomposition of calcium carbonate requires energy input.
• Standard enthalpy of formation ($\Delta H_f^\circ$) is the enthalpy change when one mole of a compound forms from its constituent elements, all in their standard states, at a specified temperature (usually 298 K) and 1 bar pressure.

By convention, $\Delta H_f^\circ = 0$ for any element in its most stable allotrope under standard conditions. For carbon, that's graphite (not diamond). For hydrogen and oxygen, that's $H_2(g)$ and $O_2(g)$, respectively.

Calculating Enthalpy Changes

Using Standard Enthalpies of Formation

The standard enthalpy of reaction can be calculated from tabulated formation enthalpies:

$\Delta H_{rxn}^\circ = \sum n \cdot \Delta H_f^\circ(\text{products}) - \sum n \cdot \Delta H_f^\circ(\text{reactants})$

where $n$ is the stoichiometric coefficient of each species.

Step-by-step process:

1. Write the balanced equation.
2. Look up $\Delta H_f^\circ$ for every product and reactant. Remember: elements in their standard states have $\Delta H_f^\circ = 0$.
3. Multiply each $\Delta H_f^\circ$ by its stoichiometric coefficient.
4. Sum the product terms, sum the reactant terms, and subtract reactants from products.

Example: For $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$:

$\Delta H_{rxn}^\circ = [\Delta H_f^\circ(CO_2) + 2\,\Delta H_f^\circ(H_2O)] - [\Delta H_f^\circ(CH_4) + 2\,\Delta H_f^\circ(O_2)]$

$= [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)] = -890.3 \text{ kJ/mol}$

Notice that $\Delta H_f^\circ(O_2) = 0$ because $O_2(g)$ is an element in its standard state.

Using Hess's Law

Hess's Law states that the total enthalpy change for a reaction is independent of the pathway, as long as the initial and final states are the same. This is a direct consequence of enthalpy being a state function.

In practice, you combine known reactions to construct a target reaction. The rules for manipulating reactions:

• Reverse a reaction → flip the sign of $\Delta H$
• Multiply a reaction by a factor $n$ → multiply $\Delta H$ by $n$
• Add reactions together → add their $\Delta H$ values

Hess's Law for Enthalpy

Applying Hess's Law

1. Write the target reaction clearly.
2. Identify known reactions that contain the same species.
3. Manipulate each known reaction (reverse, scale) so that when added, intermediates cancel and you recover the target reaction.
4. Apply the same manipulations to the $\Delta H$ values and sum them.

A useful strategy: focus on species that appear in only one known reaction. Fix those first, then adjust the remaining reactions to cancel intermediates.

Example of Hess's Law Application

Target reaction: $C(s) + 2H_2(g) \rightarrow CH_4(g)$, find $\Delta H$

Known reactions:

1. $C(s) + O_2(g) \rightarrow CO_2(g)$, $\Delta H_1 = -393.5 \text{ kJ/mol}$
2. $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$, $\Delta H_2 = -890.4 \text{ kJ/mol}$
3. $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$, $\Delta H_3 = -285.8 \text{ kJ/mol}$

Manipulations:

• Keep reaction 1 as written (we need $C(s)$ on the left): $\Delta H_1 = -393.5 \text{ kJ/mol}$
• Reverse reaction 2 (we need $CH_4$ as a product): $CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)$, $\Delta H_2' = +890.4 \text{ kJ/mol}$
• Multiply reaction 3 by 2 (we need $2H_2$): $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$, $\Delta H_3' = 2(-285.8) = -571.6 \text{ kJ/mol}$

Add them up and verify cancellation:

$C(s) + O_2(g) + CO_2(g) + 2H_2O(l) + 2H_2(g) + O_2(g) \rightarrow CO_2(g) + CH_4(g) + 2O_2(g) + 2H_2O(l)$

Cancel $CO_2$, $2H_2O$, and $2O_2$ from both sides:

$C(s) + 2H_2(g) \rightarrow CH_4(g)$

$\Delta H = (-393.5) + (+890.4) + (-571.6) = -74.7 \text{ kJ/mol}$

This matches the known $\Delta H_f^\circ$ of methane, confirming the calculation.

Enthalpy Diagrams

Constructing Enthalpy Diagrams

An enthalpy diagram plots enthalpy on the vertical axis against reaction progress on the horizontal axis. These diagrams give you a visual picture of the energy landscape of a reaction.

• Place reactants and products at their respective enthalpy levels. For exothermic reactions, products sit lower than reactants; for endothermic reactions, products sit higher.
• The vertical distance between reactant and product levels represents $\Delta H$ for that step.
• Intermediate states can be included at their own enthalpy levels, showing each step of a multi-step pathway.

Using Enthalpy Diagrams to Visualize Hess's Law

Enthalpy diagrams make Hess's Law intuitive. You can draw two different pathways from the same reactants to the same products, and the total vertical displacement is identical regardless of which path you follow.

For the methane formation example above, one path goes directly from $C(s) + 2H_2(g)$ to $CH_4(g)$ ($\Delta H = -74.7$ kJ/mol). Another path goes through the combustion products $CO_2(g) + 2H_2O(l)$ as intermediates. Both paths yield the same overall $\Delta H$, which is exactly what Hess's Law predicts.

Bond Energies and Enthalpy Changes

Relationship Between Bond Energies and Enthalpy Changes

Bond dissociation energy is the energy required to homolytically break one mole of a specific bond in the gas phase. The reverse process (bond formation) releases the same amount of energy.

You can estimate $\Delta H$ for a gas-phase reaction using average bond energies:

$\Delta H \approx \sum(\text{bond energies of bonds broken}) - \sum(\text{bond energies of bonds formed})$

Pay attention to the sign convention here: breaking bonds costs energy (positive), forming bonds releases energy (negative). The formula already accounts for this by subtracting bonds formed.

Limitations of the Bond Energy Approach

This method gives only an estimate of $\Delta H$, and there are a few reasons why:

• Average vs. actual bond energies. Tabulated values are averages across many different molecules. The actual O-H bond energy in water differs slightly from the O-H bond energy in methanol, for instance.
• Gas-phase assumption. The method strictly applies to gas-phase reactions. If reactants or products are liquids or solids, you'd need to account for intermolecular interactions (enthalpies of vaporization, etc.) separately.
• No accounting for molecular environment. Resonance stabilization, ring strain, and other structural effects can make real bond energies deviate from tabulated averages.

For the combustion of methane, the bond energy estimate typically comes out within 5-10% of the experimental value. When you need precise $\Delta H$ values, standard enthalpies of formation or direct calorimetric measurements are more reliable.