8.5: Bond Energies and Chemical Reactions

Introduction

What happens when you take a basketball, place it halfway up a playground slide, and then let it go? The basketball rolls down the slide. What happens if you do it again? Does the basketball roll down the slide? It should. If you were to perform this experiment over and over again, do you think the basketball would ever roll up the slide? Probably not. Why not? Well, for starters, in all of our experience, the basketball has always moved to a lower position when given the opportunity. The gravitational attraction of Earth exerts a force on the basketball, and given the chance, the basketball will move down. We say that the basketball is going to a lower gravitational potential energy. The basketball can move up the slide, but only if someone exerts some effort (that is, work) on the basketball. A general statement can be made based on countless observations over centuries of study:

A similar statement can be made about atoms in compounds:

In making compounds, atoms act like a basketball on a playground slide; they move in the direction of decreasing energy. Thus, when atoms combine to make a compounds, a quantity of energy, equal to the difference between the energies of the bonded atoms and the energies of the separated atoms, is released, usually as heat. 

We can reverse this process, just as with the basketball. If we put energy into a molecule, we can cause its bonds to break, separating a molecule into individual atoms. The energy needed to break the bond is the same amount of energy that is liberated when the atoms made the chemical bond in the first place. 

Enthalpy Change or Heat of Reaction, ΔH

When a chemical reaction occurs, the atoms in the reactants rearrange their chemical bonds to make products. The new arrangement of bonds does not have the same total energy as the bonds in the reactants. Therefore, when chemical reactions occur, there will always be an accompanying energy change. 

Energy changes in chemical reactions are usually measured as changes in enthalpy. In this process, one adds energy to the reaction to break bonds, and extracts energy for the bonds that are formed.

\[\text{enthalpy change, ΔH} ≈ \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \label{Eqn1} \]

The enthalpy change, ΔH, for a given reaction can be calculated using the bond energy values from Table \(\PageIndex{1}\) by using Equation \ref{Eqn1} . The ≈ sign is used because we are adding together average bond energies (i.e., over many different molecules). Hence, this approach does not give exact values for the enthalpy change, ΔH.

Example \(\PageIndex{1}\): Formation of H₂O

Let’s consider the reaction in Equation \ref{Eqn2} of 2 mols of hydrogen gas (H₂) with 1 mol of oxygen gas (O₂) to give 2 mol water:

\[\ce{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \label{Eqn2} \]

or shown graphically in Figure \(\PageIndex{1}\):

Average bond energies involved with this reaction are:

• \(\ce{H–H}\) = 105 kcal/mol
• \(\ce{O=O}\) = 119 kcal/mol
• \(\ce{O–H}\) = 110 kcal/mol

In this reaction, 2 H–H bonds and 1 O=O bonds are broken, while 4 O–H bonds (two for each H₂O) are formed. The energy changes can be tabulated and calculated as follows Table \(\PageIndex{2}\)  

Table \(\PageIndex{2}\): Bond Energy Changes

Bonds Broken (kcal/mol)		Bonds Formed (kcal/mol)
2 H–H	2 mol x 105 kcal/mol = 210 kcal	4 O–H	4 mols x 110 kcal/mol = 440 kcal
1 O=O	1 mol x 119 kcal/mol = 119 kcal
	Total = 329 kcal		Total = 440 kcal

\[\begin{align*} \Delta H &≈ \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \\[4pt] &≈ 329 \: \text{kcal} - 440 \: \text{kcal} \\[4pt] &≈ −111 \: \text{kcal} \end{align*} \nonumber \]The enthalpy change (ΔH) of the reaction is approximately −111 kcal/mol.

We can also re-write the reaction equation with the ΔH information (see below). 

\[\ce{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \ \: \: \: \: \: \Delta H = -111 \: \text{kcal} \nonumber \] 

Note that an exothermic reaction has a negative ΔH value.This means that bonds in the products (440 kcal) are stronger than the bonds in the reactants (329 kcal) by about 111 kcal/mol. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes. This excess energy is released as heat, so the reaction is exothermic. Hence, we can re-write the reaction with the heat released (111 kcal) on the product side of the equation (Figure \(\PageIndex{2}\)):

An endothermic reaction on the other hand will have a positive ΔH value, because energy in the form of heat is absorbed from the surroundings, and the absorbed heat will be written on the reactant side Figure \(\PageIndex{5}\).

Example \(\PageIndex{2}\): Determining the Enthalpy Change of a Reaction

What is the enthalpy change for the reaction shown in Equation \ref{Eqn3}? Is the reaction exothermic or endothermic?

\[\ce{H_2(g) + Br_2(g) \rightarrow 2HBr(g)} \label{Eqn3} \]

Solution

Step 1- First look at the equation and identify which bonds exist on in the reactants (bonds broken).

• one H-H bond and
• one Br-Br bond

Step 2- Do the same for the products (bonds formed)

• two H-Br bonds

Step 3- Identify the bond energies of these bonds from Table \(\PageIndex{1}\):

• H-H bonds: 105 kcal/mol
• Br-Br bonds: 46 kcal/mol

Step 4- Set up the table (Table \(\PageIndex{3}\)) and apply the formula for enthalpy change.

Table \(\PageIndex{3}\): Solutions to Example 7.4.1

Bonds Broken (kcal/mol)		Bonds Formed (kcal/mol)
1 H–H	1 mol x 105 kcal/mol = 105 kcal	2 H–Br	2 mols x 87 kcal/mol = 174 kcal
1 Br–Br	1 mol x 46 kcal/mol = 46 kcal
	Total = 151 kcal		Total = 174 kcal

\[\begin{align*} \Delta H &≈ 151 \: \text{kcal} - 174 \: \text{kcal} \\[4pt] &≈ −23 \: \text{kcal} \end{align*} \nonumber \]

Step 5- Since \(ΔH\) is negative (−23 kcal), the reaction is exothermic.

Endothermic and Exothermic Reactions

As mentioned above, endothermic and exothermic reactions can be thought of as having energy as either a reactant or a product.

Endothermic reactions require energy, so energy is a reactant. Heat flows from the surroundings to the system (reaction mixture) and the enthalpy of the system increases (ΔH is positive).

Exothermic reactions on the other hand release energy, so energy is a product. The enthalpy of the system decreases (ΔH is negative).

• In the course of an endothermic process, the system gains heat from the surroundings and so the temperature of the surroundings decreases (gets cold).
• A chemical reaction is exothermic if heat is released by the system into the surroundings. Because the surroundings is gaining heat from the system, the temperature of the surroundings increases. See Figure \(\PageIndex{3}\). 

We will look at two examples:

• Endothermic Reaction: When \(1 \: \text{mol}\) of calcium carbonate decomposes into \(1 \: \text{mol}\) of calcium oxide and \(1 \: \text{mol}\) of carbon dioxide, \(177.8 \: \text{kJ}\) of heat is absorbed. Because the heat is absorbed by the system, the \(177.8 \: \text{kJ}\) is written as a reactant Figure \(\PageIndex{4}\).

\[\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = +177.8 \: \text{kJ} \nonumber \]

• Exothermic Reaction: When methane gas is combusted, heat is released, making the reaction exothermic. Specifically, the combustion of \(1 \: \text{mol}\) of methane releases 890.4 kilojoules of heat energy. This information can be shown as part of the balanced equation in two ways. First, the amount of heat released can be written in the product side of the reaction Figure \(\PageIndex{5}\). Another way is to write the heat of reaction (ΔH) information with a negative sign, \(-890.4 \: \text{kJ}\).

\[\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -890.4 \: \text{kJ} \nonumber \]

Energy Diagrams

Endothermic and exothermic reactions can be visually represented by energy-level diagrams like the ones in Figure \(\PageIndex{6A and 6B}\).

In endothermic reactions, the reactants have higher bond energy (stronger bonds) than the products. Strong bonds have lower potential energy than weak bonds. Hence, the energy of the reactants is lower than that of the products. This type of reaction is represented by an "uphill" energy-level diagram shown in Figure \(\PageIndex{6A}\). For an endothermic chemical reaction to proceed, the reactants must absorb energy from their environment to be converted to products.

In an exothermic reaction, the bonds in the product have higher bond energy (stronger bonds) than the reactants. In other words, the energy of the products is lower than the energy of the reactants, hence is energetically downhill, shown in Figure \(\PageIndex{6B}\). Energy is given off as reactants are converted to products. The energy given off is usually in the form of heat (although a few reactions give off energy as light). In the course of an exothermic reaction, heat flows from the system to its surroundings, and thus, gets warm.

Key Takeaways

• Atoms are held together by a certain amount of energy called bond energy.
• Energy is required to break bonds. Energy is released when chemical bonds are formed because atoms become more stable.
• Chemical processes are labeled as exothermic or endothermic based on whether they give off or absorb energy, respectively.