Hess's law and bond enthalpies

Hess's law

Some enthalpy changes cannot be measured. Carbon burnt in a little oxygen gives a mixture of carbon monoxide and carbon dioxide, never pure CO. So how can we find $\Delta_f H^\ominus$ of CO?

Germain Hess answered this in 1840. Hess's law of constant heat summation says:

If a reaction takes place in several steps, its standard enthalpy change is the sum of the standard enthalpy changes of the steps.

It follows because enthalpy is a state function. The change depends only on where you start and where you end.

A worked cycle

Both of these can be measured, because each burns completely:

• $\text{C(graphite)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \qquad \Delta H_2 = -393.5\ \text{kJ mol}^{-1}$
• $\text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \qquad \Delta H_3 = -283.0\ \text{kJ mol}^{-1}$

Carbon can reach carbon dioxide directly, or first become CO and then burn on. Both routes start and end at the same place:

So $\Delta H_1 = -393.5 - (-283.0) = -110.5\ \text{kJ mol}^{-1}$. That is $\Delta_f H^\ominus$ of CO.

Adding equations

You can get the same answer by treating the equations like algebra.

1. Write the target equation.
2. Take each known equation as it is, reverse it, or multiply it, so that the substances you need land on the right side.
3. Add the equations. Cancel what appears on both sides.
4. Add the enthalpy changes in the same way, changing the sign of any reversed one.

For CO: keep the burning of carbon, and reverse the burning of CO. Adding them, $\text{CO}_2$ cancels and half a mole of $\text{O}_2$ cancels. What is left is $\text{C} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}$, with $-393.5 + 283.0 = -110.5\ \text{kJ mol}^{-1}$.

Practise with new numbers

Bond enthalpies

A reaction breaks some bonds and makes new ones. Breaking a bond always takes energy in. Making a bond always gives energy out.

The bond dissociation enthalpy is the enthalpy change when one mole of a bond breaks, with everything in the gas phase. For hydrogen, $\text{H}_2\text{(g)} \rightarrow 2\text{H(g)}$ takes $435\ \text{kJ mol}^{-1}$.

In a molecule with several bonds of one kind, each takes a different energy to break. Methane needs $1665\ \text{kJ mol}^{-1}$ to break all four C-H bonds, though the steps differ. So chemists use the mean bond enthalpy: $\dfrac{1665}{4} \approx 416\ \text{kJ mol}^{-1}$ for C-H in methane. Averaged over many compounds, tables give values like these:

For a reaction where every substance is a gas:

Here $\sum H_{\text{broken}}$ adds the bond enthalpies of all the bonds broken, and $\sum H_{\text{formed}}$ those of all the bonds formed.

Worked example. $\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}$. Broken: $435 + 242 = 677\ \text{kJ}$. Formed: $2 \times 431 = 862\ \text{kJ}$. So $\Delta_r H \approx 677 - 862 = -185\ \text{kJ mol}^{-1}$. More energy comes out than goes in, so the reaction is exothermic.

The answer is an estimate, because mean values are averages. It works only when every substance is a gas. A liquid needs its enthalpy of vaporisation too.

Practise with new numbers

Try it in Python

The program keeps mean bond enthalpies in a dictionary and estimates two reaction enthalpies from the bonds broken and formed.

Common mistakes

• Formed minus broken. For bond enthalpies it is broken minus formed, the opposite order to formation enthalpies.
• Forgetting to reverse the sign. A reversed equation has the opposite sign of $\Delta H$.
• Scaling the equation but not $\Delta H$. If you double an equation, double its enthalpy change too.
• Using bond enthalpies for liquids. They hold only for gases.
• Expecting an exact answer. Mean bond enthalpies give an estimate, not the measured value.

Remember these

• Hess's law: the enthalpy change is the same by any route, because $H$ is a state function.
• Equations add like algebra. Reverse: change the sign. Multiply: multiply $\Delta H$.
• Bond breaking takes energy in, bond making gives it out.
• For gases, $\Delta_r H \approx \sum(\text{broken}) - \sum(\text{formed})$.
• Mean bond enthalpies are averages, so the result is an estimate.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why does Hess's law hold?

2. If $\text{A} \rightarrow \text{B}$ has $\Delta H = -40\ \text{kJ}$, what is $\Delta H$ for $2\text{B} \rightarrow 2\text{A}$?

3. Why is a reaction enthalpy from bond enthalpies only an estimate?

4. Use the table to estimate $\Delta_r H$ for $\text{C}_2\text{H}_4\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow \text{C}_2\text{H}_4\text{Cl}_2\text{(g)}$. Take C-Cl as $326\ \text{kJ/mol}$.

5. Why can carbon monoxide's enthalpy of formation not be measured directly?