Heat of Reaction

Core Concepts

In this article you will learn about heat of reaction, its meaning in thermodynamics, and how to calculate it.

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Enthalpy of Reaction ( \Delta H_{rxn}\degree) is another name for Heat of Reaction. The enthalpy of reaction is the change in enthalpies brought about by a chemical reaction under conditions of constant pressure. Importantly, there are many other forms of enthalpy. We may also describe enthalpy as a thermodynamical unit that gauges the energy per mole that a chemical process produces or releases. Since this function is derived by internal energy, pressure, temperature, and volume, it is viewed as a state function.

enthalpy-heat of reaction

Since it tends to be measured at constant pressure, the change of enthalpy is often expressed as:

     \begin{gather*} \Delta H = q \end{gather*}

The standard enthalpy of reaction is commonly shown as  \Delta H_{rxn}\degree or  \Delta H\degree. Its units of measure are kilojoules per mole. The standard conditions for this function are  1\text{atm} and 25\degree \text{C}.

How to calculate the Heat of Reaction

Calculating Experimentally

To quantify the change in enthalpy experimentally, you need a calorimeter, which is an isolated system with constant pressure. You can even make your own calorimeter.

Calculating Numerically

To calculate the heat of reaction numerically, you’ll need the standard enthalpies of formation for all products and reactants. You can usually find this information in thermodynamic charts found in your textbook. Once you have those numbers, you just need to follow this equation:

     \begin{gather*} \Delta H_{rxn}\degree = \sum \Delta v_{p}\Delta H_{f} \text{(products)} - \sum \Delta v_{r}\Delta H_{f} \text{(reactants)} \end{gather*}

where  v_{p} is the stoichiometric coefficient of the balanced reaction’s product,  v_{r} is the reactant’s stoichiometric coefficient, and  \Delta H_{f} is the standard enthalpy of formation for each product and reactant.

Additionally, since the Heat of Reaction is a state function, it only depends on final and initial steps, not on the path. As a result, using Hess’ Law, you can calculate the overall enthalpy of a multistep reaction as the sum of the  \Delta H‘s of each individual step.

The value of  \Delta H might be either positive or negative because it represents the amount of energy exchanged throughout the reaction. If the value is negative, then it means that heat energy is going out of the system, going into the surroundings. It also means that it is an exothermic reaction. If the value is positive, then it means that heat energy from the surroundings is entering the system, indicating an endothermic reaction.

Example of the Heat Reaction calculation

Calculate the heat reaction of the acetylene combustion:

  1. Firstly, find and balance the chemical equation.

     \begin{gather*} 2C_{2}H_{2}\text{(g)} + 5O_{2} \text{(g)} \rightarrow 4CO_{2} \text{(g)} + 2H_{2}O \text{(g)} \end{gather*}

2. Secondly, find each heat of formation. You can use a resource like this table.

3. Thirdly, multiply each enthalpy by it’s stoichiometric coefficient.

3.1. Starting with the products

     \begin{align*} {\Delta H_{f, CO_{2}} &= -396.5 \text{kJ/mol}} \\ {\Delta H_{f, H_{2}O} &= -241.8 \text{kJ/mol}} \\ {v_{p} \Delta H_{f, CO_{2}} &= 4\text{mol}(-396.5 \text{kJ/mol}) = -1574\text{kJ}} \\ {v_{p}\Delta H_{f, H_{2}O} &= 2\text{mol}(-241.8 \text{kJ/mol}) = -483.6 \text{kJ}} \end{align*}

3.2. Then the reactants

     \begin{align*} {\Delta H_{f, C_{2}H_{2}} &= 227 \text{kJ/mol}} \\ {\Delta H_{f, O_{2}} &= 0 \text{kJ/mol}} \\ {v_{p} \Delta H_{f, C_{2}H_{2}} &= 2\text{mol}(227 \text{kJ/mol}) = 454\text{kJ}} \\ {v_{p}\Delta H_{f, O_{2}} &= 5\text{mol}(0 \text{kJ/mol}) = 0 \text{kJ}} \end{align*}

4. Fourthly, add both values to reactants and products to obtain the sum of each.

     \begin{align*} {\sum v_{p} \Delta H_{f} \text{(products)} &= -1574 \text{kJ} + (-483.6 \text{kJ}) = -2057.6 \text{kJ}} \\ {\sum v_{p} \Delta H_{f} \text{(reactants)} &= 454 \text{kJ} + 0 = 454 \text{kJ}} \end{align*}

5. Finally, it is now possible to use Hess’ Law to find the heat of reaction.

     \begin{align*} {\Delta H_{rxn}\degree &= \sum \Delta v_{p}\Delta H_{f} \text{(products)} - \sum \Delta v_{r}\Delta H_{f} \text{(reactants)}} \\ {\Delta H_{rxn} &= (-2057.6 \text{kJ}) - (454 \text{kJ}) = -2511.6 \text{kJ}} \end{align*}

Important points of the example

As you may notice O2‘s standard enthalpy formation is 0. This is because pure molecules don’t have enthalpy of formation. Elemental oxygen is naturally found as a diatomic structure, so O2 would be the pure molecule. You may also notice that the result was negative. This result tells us that energy is coming out of the system, making it an exothermic reaction. In fact, every combustion reaction is exothermic.

Understanding Heat of Reaction Graphs

Graphs are very useful to understand the heat of reaction mechanism. In the graph below, we can see how the heat of reaction is exactly the change in energy. On the same graph, it is possible to see the relative energy of the transition state at the top of the graph. During the transition state, bonds in the compounds break to form new ones, generating the products. The reaction requires a certain amount of energy, called the activation energy, to initiate this change in chemical bonding. Finally, the energy released will depend on the specifics of the reaction; the graph below represents an exothermic reaction. An endothermic reaction would likely have a bigger activation energy and heat would be absorbed, rather than released.

Heat of Reaction Practice Problems

Problem 1

What is the heat of reaction for aluminum oxide formation?

     \begin{gather*} {Al + O_{2} \rightarrow Al_{2}O_{3}} \\ {\DeltaH_{f, Al_{2}O_{3}} = -1675 \text{kJ/mol}} \end{gather*}

Problem 2

Consider the combustion of benzene:

     \begin{gather*} 2C_{6}H_{6}\text{(g)} + 15O_{2} \text{(g)} \rightarrow 12CO_{2} \text{(g)} + 6H_{2}O \text{(g)} \end{gather*}

If the heat of reaction is  -6370 \text{kJ} what is the heat of formation of benzene?

Heat of Reaction Practice Problem Solutions

1:  -3350 \text{kJ/mol}

2:  80.6 \text{kJ/mol}