Standard enthalpy changes

The enthalpy of reaction

The enthalpy of reaction, $\Delta_r H$, is the enthalpy of the products minus that of the reactants. It depends on the conditions. So chemists agree on a standard state for comparing values.

• The standard state of a substance is its pure form at 1 bar, at the stated temperature.
• Data are usually given at $298\ \text{K}$.
• A standard value carries the sign $\ominus$, as in $\Delta_r H^\ominus$.

For example, the standard state of water at $298\ \text{K}$ is pure liquid water at 1 bar. For carbon it is graphite.

Thermochemical equations

A thermochemical equation is a balanced equation with the physical state of every substance and the value of $\Delta_r H$:

Read the coefficients as moles. Three rules follow.

1. The states matter. Water as gas gives a different value from water as liquid.
2. Multiply the equation by a number, and $\Delta_r H$ is multiplied by the same number.
3. Reverse the equation, and $\Delta_r H$ changes sign.

Enthalpies of phase change

Melting, boiling and subliming change no bonds inside molecules. They still take in heat, to pull the molecules apart.

• Fusion, solid to liquid, $\Delta_{fus}H^\ominus$. For ice it is $6.00\ \text{kJ/mol}$ at $273\ \text{K}$.
• Vaporisation, liquid to gas, $\Delta_{vap}H^\ominus$. For water it is $40.79\ \text{kJ/mol}$ at $373\ \text{K}$.
• Sublimation, solid straight to gas, $\Delta_{sub}H^\ominus$. It is the sum of the other two: $\Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H$.

Vaporisation needs far more heat than fusion. Melting only loosens the molecules. Boiling must separate them completely. Solid carbon dioxide, dry ice, sublimes straight to gas.

Standard enthalpy of formation

The standard enthalpy of formation, $\Delta_f H^\ominus$, is the enthalpy change when one mole of a compound forms from its elements in their standard states.

By agreement, $\Delta_f H^\ominus$ of an element in its standard state is zero. So $\text{O}_2\text{(g)}$, $\text{H}_2\text{(g)}$ and graphite all have zero.

A table like this gives the enthalpy of any reaction among these substances:

Here $a_i$ and $b_i$ are the coefficients in the equation.

Worked example. Burning methane. Products: $-393.5 + 2(-285.8) = -965.1\ \text{kJ}$. Reactants: $-74.8 + 2(0) = -74.8\ \text{kJ}$. So $\Delta_r H^\ominus = -965.1 - (-74.8) = -890.3\ \text{kJ mol}^{-1}$.

Practise with new numbers

Other standard enthalpies

• Enthalpy of combustion, $\Delta_c H^\ominus$: one mole of a substance burns completely in oxygen. It is always negative. For methane it is $-890.3\ \text{kJ mol}^{-1}$. Fuels are compared by it.
• Enthalpy of atomisation, $\Delta_a H^\ominus$: one mole of a substance breaks completely into gaseous atoms. It is always positive. For $\text{CH}_4\text{(g)}$ it is $1665\ \text{kJ mol}^{-1}$.
• Enthalpy of solution, $\Delta_{sol}H^\ominus$: one mole of a substance dissolves in a large amount of solvent. For an ionic solid, the lattice must break up and the ions must be surrounded by water. So $\Delta_{sol}H = \Delta_{lattice}H + \Delta_{hyd}H$. Course b1 of Chemical Bonding built the lattice enthalpy of NaCl, $788\ \text{kJ mol}^{-1}$, with a Born-Haber cycle.
• Enthalpy of dilution: the heat when a solution is diluted further. It depends on how much solvent is added, and becomes zero once the solution is very dilute.

Try it in Python

The program stores enthalpies of formation in a dictionary. A function sums products minus reactants for any reaction.

Common mistakes

• Forgetting the coefficients. Multiply each $\Delta_f H^\ominus$ by its number in the equation.
• Giving elements a value. An element in its standard state has $\Delta_f H^\ominus = 0$. Diamond and ozone are not standard states, so theirs are not zero.
• Taking reactants minus products. It is always products minus reactants.
• Ignoring states. $\text{H}_2\text{O(l)}$ and $\text{H}_2\text{O(g)}$ have different enthalpies of formation.
• Making more than one mole. A formation reaction makes exactly one mole of the compound, even if that needs fractions such as $\frac{1}{2}\text{O}_2$.

Remember these

• Standard state: pure substance at 1 bar, usually at $298\ \text{K}$.
• Reverse an equation: change the sign. Multiply it: multiply $\Delta H$.
• $\Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H$. Vaporisation needs more heat than fusion.
• $\Delta_f H^\ominus$ of an element in its standard state is zero.
• $\Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) - \sum b_i \Delta_f H^\ominus(\text{reactants})$.

Check yourself

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

1. Write the equation for the standard enthalpy of formation of $\text{NH}_3\text{(g)}$.

2. Why is $\Delta_{vap}H$ of water much larger than $\Delta_{fus}H$?

3. What is $\Delta_f H^\ominus$ of $\text{O}_2\text{(g)}$?

4. Use the table to find $\Delta_r H^\ominus$ for $\text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$.

5. Is an enthalpy of atomisation positive or negative? Why?