Bond enthalpy

### The energy it takes to break a bond

In the section on enthalpy, we saw that we can calculate the overall enthalpy change of a reaction by summing up the known enthalpy changes of a mix of simpler reactions. We called that thermochemistry. We further saw that we could also calculate enthalpy changes using standard enthalpies of formation (ΔHfo). Recall that the enthalpy change for any reaction is just the sum of the standard enthalpies of formation of the products minus the sum of those of the reactants.

$$\Delta H^o = \sum \Delta H_{f \; products}^o - \sum \Delta H_{f \; reactants}^o$$

Well, it turns out that we can go even further down that road — right to the end, actually.

Now we define the enthalpy of a chemical bond, the bond enthalpy, which is the amount of energy it takes to break any particular bond, such as a C—C bond or an O=O bond. Generally we work with molar bond enthalpies, the energy required to break one mole of a bond. The negative of that bond enthalpy is the energy released upon forming the corresponding bond.

As an example, consider the synthesis of phosgene gas (COCl2) from carbon monoxide (CO) and chlorine gas (Cl2):

#### CO(g) + Cl2 (g) → COCl2 (g)

It takes energy to form a chemical bond, and energy (enthalpy) is released when a bond is formed. A chemical reaction is exothermic if less energy goes into the breaking of the bonds of the reactants than is released in forming the bonds of the products. Conversely, if less energy is released in forming the bonds of the product(s) than it takes to break the bonds of the reactant(s), thermal energy will need to be absorbed from the surroundings in order to make up the deficit — an endothermic reaction.

The figure below illustrates the case for the exothermic synthesis of phosgene. We imagine that in a first step, all of the CO and Cl2 bonds are broken, and that in a second, the bonds of the product is formed. The overall enthalpy of the reaction is just the difference between those. The picture above is, of course, a purely theoretical one. It's not often true that all bonds of reactants are completely broken in order to form products. In fact, the transition state structure — a structure somewhere between reactants and products — often involves a mixture of stretched, partial bonds and bonds that never break during the reaction process.

But enthalpy is a state function, meaning that the precise pathway from one state to another (like from reactant(s) to product(s)) doesn't matter, only the difference in total enthalpy of both states.

This table gives a number of bond enthalpies for single, double and triple bonds. Notice that, on average, single bonds are weaker than double bonds (it takes less energy to break them) and double bonds are weaker than triple bonds. #### Bond enthalpy

The bond enthalpy of a particular bond is the amount of energy (SI unit Joules, J) required to break one mole of that bond.

The enthalpy change of a system is a state function. It doesn't depend on how that change came about. It is simply a difference between the final and initial enthalpies of the system.

### Example 1

Use bond enthalpies to estimate the overall enthalpy change for the water-forming reaction:
2 H2 + O2 → 2 H2O.

Solution: On the reactant side, we have two moles of H-H and one mole of O=O double bonds to break. Using our table, that's

\begin{align} (2 \; mol)\left(\frac{436 \; KJ}{mol}\right) + (1 \; mol)&\left(\frac{498 \; KJ}{mol} \right) \\[5pt] &= 1370 \;KJ/mol \end{align}

On the product side, we break two moles of O-H bonds, with a bond enthalpy of 463 KJ/mol:

$$(4 \; mol)\left(\frac{463 \; KJ}{mol}\right) = 1852 \; KJ.$$

The overall enthalpy change is the sum of the bond enthalpies of the products minus the sum of the bond enthalpies of the reactants:

\begin{align} \Delta H &= (1852 - 1370) \; KJ \\[5pt] &= {\bf 482 \; KJ} \end{align}

Pretty simple. It might be nicer to express this enthalpy change in terms of enthalpy per 1 mole of water formed, or 241 KJ/mol of water formed.

### Example 2

Use bond enthalpies to estimate the overall enthalpy change for the combustion of one mole of n-octane:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O.

Solution: First we'll add up the bond enthalpies on the reactant side. There are 7 C-C bonds and 18 H-C bonds, but we have to multiply by the coefficient 2 from the balanced equation. That's

$$2[7(346) + 18(413)] \; KJ = 19,712 \; KJ$$

Add to that the 25 O-O bonds that must be broken:

$$25(498) \; KJ = 12,450 \; KJ$$

The total energy of bond-breaking on the left is

$$19,712 \; KJ + 12,450 \; KJ = 32,162 \; KJ$$

Now the bonds on the product side include 32 C=O bonds and 36 O-H bonds:

$$32(799) + 36(463) \; KJ = 42,236 \; KJ$$

Now the enthalpy of the reaction is

$$32,854 \; KJ - 42,236 \; KJ = \bf -9,382 \; KJ$$

That's 4,691 KJ of energy released for every octane burned. Octane, by the way, is a component of most gasoline mixtures, and is part of the reason we measure the quality of gas by its octane rating.

X

### SI units

SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units

 length meter (m) mass Kilogram (Kg) time second (s) force Newton N energy Joule J

### Practice problems

Use bond enthalpies from the table above to estimate the enthalpy changes of the following reactions:

 1 C4H10 (g) + O2 (g) → CO2 (g) + H2O (g) 2 H2O2 (l) → H2 (g) + O2 (g) 3 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016,2022 Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.