Standard Enthalpy Change of Reaction (ΔHᵣ°)
What is Energy in a Chemical Reaction?
Every chemical reaction involves a change in energy. Think about lighting a candle. The wax burns, produces a flame, and releases heat and light. This is a clear example of energy being released. Conversely, when you bake a cake, you have to put the batter in a hot oven; you are adding heat energy to make the reaction happen. In chemistry, we call this energy enthalpy, symbolized by the letter $ H $. We are usually more interested in the change in enthalpy during a reaction, which we write as $ \Delta H $ (the Greek letter Delta, Δ, means "change in").
The Standard Enthalpy Change of Reaction, $ \Delta H_r^\circ $, is the specific enthalpy change when the number of moles of reactants shown in the balanced chemical equation react completely to form products under standard conditions. Standard conditions are defined as a pressure of $ 100 \text{kPa} $ (approximately 1 atmosphere) and a specified temperature, usually $ 25^\circ\text{C} $ ($ 298 \text{K} $). The little circle (°) next to the $ \Delta H $ symbol reminds us that the measurement was taken under these standard conditions.
Exothermic vs. Endothermic Reactions
All chemical reactions can be placed into one of two categories based on whether they release heat to the surroundings or absorb heat from them.
Exothermic Reactions ($ \Delta H_r^\circ < 0 $): These reactions release energy, usually as heat or light, into their surroundings. The products have less energy than the reactants. This means the enthalpy change, $ \Delta H $, is negative. The reaction feels hot. Combustion, like burning fuel or a candle, is a classic exothermic process.
Endothermic Reactions ($ \Delta H_r^\circ > 0 $): These reactions absorb energy from their surroundings. The products have more energy than the reactants. This means the enthalpy change, $ \Delta H $, is positive. The reaction feels cold. Photosynthesis, where plants use energy from sunlight to make glucose, is a vital endothermic reaction.
| Feature | Exothermic Reaction | Endothermic Reaction |
|---|---|---|
| Energy Flow | Releases energy to surroundings | Absorbs energy from surroundings |
| Sign of $ \Delta H_r^\circ $ | Negative ($ - $) | Positive ($ + $) |
| Temperature Change | Surroundings get warmer | Surroundings get colder |
| Energy of Products | Lower than reactants | Higher than reactants |
| Common Examples | Combustion, neutralization, respiration | Photosynthesis, melting ice, cooking an egg |
How to Calculate ΔHᵣ°: Hess's Law and Standard Enthalpy of Formation
It is often difficult or dangerous to measure the enthalpy change of a reaction directly. Fortunately, chemists have powerful tools to calculate it. The two most important methods involve Hess's Law and standard enthalpy of formation data.
Hess's Law states that the total enthalpy change for a reaction is the same, no matter which route is taken from reactants to products. This means if we can break down a complex reaction into a series of simpler steps whose enthalpy changes are known, we can add up those changes to find the overall $ \Delta H_r^\circ $.
For example, imagine you want to find the enthalpy change for the reaction: $ A + B \rightarrow D $. You know two other reactions:
1. $ A + B \rightarrow C $ with $ \Delta H = -100 \text{kJ} $
2. $ C \rightarrow D $ with $ \Delta H = +50 \text{kJ} $
Adding these steps gives $ A + B \rightarrow D $, and the overall $ \Delta H = (-100) + (+50) = -50 \text{kJ} $.
The most common and straightforward method uses the Standard Enthalpy of Formation ($ \Delta H_f^\circ $)[1]. This is the enthalpy change when one mole of a compound is formed from its elements in their standard states. For example, the $ \Delta H_f^\circ $ for water is the energy change for the reaction: $ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) $.
The formula to calculate the standard enthalpy change of reaction using these values is:
This means you add up the standard enthalpies of formation for all the products and subtract the sum of the standard enthalpies of formation for all the reactants.
A Practical Example: Calculating the Enthalpy of Combustion of Methane
Let's apply the formula to a real-world reaction: the combustion of natural gas (methane, $ CH_4 $). This is the reaction that heats many homes.
The balanced chemical equation is:
$ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) $
We look up the standard enthalpy of formation ($ \Delta H_f^\circ $) values in a reference table:
- $ \Delta H_f^\circ[CH_4(g)] = -74.8 \text{kJ/mol} $
- $ \Delta H_f^\circ[O_2(g)] = 0 \text{kJ/mol} $ (by definition, for any element in its standard state)
- $ \Delta H_f^\circ[CO_2(g)] = -393.5 \text{kJ/mol} $
- $ \Delta H_f^\circ[H_2O(l)] = -285.8 \text{kJ/mol} $
Now, we plug these values into our formula:
$ \Delta H_r^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) $
$ \Delta H_r^\circ = \left[1(-393.5) + 2(-285.8)\right] - \left[1(-74.8) + 2(0)\right] $
$ \Delta H_r^\circ = \left[ -393.5 - 571.6 \right] - \left[ -74.8 \right] $
$ \Delta H_r^\circ = \left[ -965.1 \right] - \left[ -74.8 \right] $
$ \Delta H_r^\circ = -965.1 + 74.8 $
$ \Delta H_r^\circ = -890.3 \text{kJ/mol} $
The negative sign confirms that the reaction is exothermic. When one mole of methane gas burns completely, it releases $ 890.3 \text{kJ} $ of energy under standard conditions.
Important Questions
An element in its standard state (like $ O_2(g) $, $ C(s) $ as graphite, or $ Na(s) $) is the most stable form of that element at standard conditions. By definition, forming an element from itself involves no chemical change and therefore no enthalpy change. So, $ \Delta H_f^\circ $ for any element in its standard state is zero.
The state of matter (solid, liquid, or gas) is crucial because energy is required to change states. For example, the $ \Delta H_f^\circ $ for water vapor ($ H_2O(g) $) is $ -241.8 \text{kJ/mol} $, which is less negative than for liquid water ($ -285.8 \text{kJ/mol} $). This difference of $ 44.0 \text{kJ/mol} $ is the energy needed to vaporize the liquid. Using the wrong state in a calculation will give an incorrect $ \Delta H_r^\circ $.
Not on its own. A negative $ \Delta H_r^\circ $ (exothermic) often suggests a reaction might be spontaneous, but it is not a guarantee. Spontaneity is determined by another quantity called Gibbs Free Energy ($ \Delta G $), which considers both enthalpy ($ \Delta H $) and entropy ($ \Delta S $), a measure of disorder. Some endothermic reactions (positive $ \Delta H $) are spontaneous because they cause a large increase in disorder.
Conclusion
Footnote
[1] Standard Enthalpy of Formation (ΔHf°): The enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (100 kPa and usually 298 K).