Enthalpy and calorimetry

Why a new quantity

In course e1 you met the internal energy $U$ and the first law, $\Delta U = q + w$. In a sealed rigid vessel no expansion work is done. Then the heat taken in equals $\Delta U$.

Most reactions, though, run in open flasks and beakers. The pressure stays at that of the air, and the volume can change. A reaction that makes gas pushes the air back and does work. Some of the heat is used up in that work, so $q$ no longer equals $\Delta U$.

Chemists therefore use a new quantity, the enthalpy $H$:

Like $U$, $p$ and $V$, enthalpy is a state function. Its change depends only on the start and the end, not on the path between them.

Enthalpy is heat at constant pressure

At constant pressure, the expansion work is $w = -p\Delta V$. Put this in the first law: $\Delta U = q_p - p\Delta V$. So the heat at constant pressure is

That is why enthalpy matters. The heat taken in at constant pressure is the enthalpy change. If heat is given out, $\Delta H$ is negative and the reaction is exothermic. If heat is taken in, $\Delta H$ is positive and the reaction is endothermic.

Delta H and Delta U for gases

For solids and liquids, the volume hardly changes, so $\Delta H$ and $\Delta U$ are almost equal. Gases are different. For an ideal gas, $pV = nRT$. If a reaction changes the moles of gas by $\Delta n_g$ at a fixed temperature, then

Here $\Delta n_g$ is the moles of gaseous products minus the moles of gaseous reactants. Solids and liquids do not count. $R$ is $8.314\ \text{J K}^{-1}\text{mol}^{-1}$, so turn joules into kilojoules before you add.

Worked example. For boiling water, $\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(g)}$, $\Delta_{vap}H$ is $40.79\ \text{kJ/mol}$ at $373\ \text{K}$. One mole of gas forms, so $\Delta n_g = 1$.

• $\Delta n_g R T = 1 \times 8.314 \times 373 = 3101\ \text{J} = 3.10\ \text{kJ}$.
• $\Delta U = 40.79 - 3.10 = 37.69\ \text{kJ/mol}$.

The rest of the heat, $3.10\ \text{kJ}$, is the work the steam does as it pushes back the air.

Extensive and intensive properties

An extensive property depends on the amount of matter. Mass, volume, internal energy, enthalpy and heat capacity are extensive. Double the sample, and each doubles.

An intensive property does not depend on the amount. Temperature, pressure, density and molar heat capacity are intensive. A cup of tea and a pot of the same tea have the same temperature.

Dividing one extensive property by another gives an intensive one. Mass divided by volume gives density.

Heat capacity

When a body takes in heat $q$, its temperature rises by $\Delta T$. The two are proportional:

The constant $C$ is the heat capacity. A large heat capacity means a small rise for a given heat.

• The molar heat capacity $C_m$ is the heat capacity of one mole, in $\text{J K}^{-1}\text{mol}^{-1}$. For water it is $75.3\ \text{J K}^{-1}\text{mol}^{-1}$.
• The specific heat capacity $c$ is the heat capacity of one gram, in $\text{J g}^{-1}\text{K}^{-1}$. For water it is $4.18\ \text{J g}^{-1}\text{K}^{-1}$. Then $q = m c \Delta T$.

A gas warmed at constant pressure also expands and does work. So it needs more heat than at constant volume. For one mole of an ideal gas, the difference is exactly $R$:

Practise with new numbers

Measuring heat: calorimetry

A calorimeter measures the heat of a reaction from a temperature change. The reaction gives its heat to a known amount of water. You measure the rise and use the heat capacity.

• The bomb calorimeter. The sample burns in oxygen inside a sealed steel vessel, the bomb, which sits in water. The volume cannot change, so no expansion work is done. The heat measured is $\Delta U$.
• The coffee-cup calorimeter. The reaction runs in a solution in an insulated cup open to the air. The pressure stays constant, so the heat measured is $\Delta H$.

In both, the heat the reaction gives out is the heat the calorimeter takes up. So the sign changes: $\Delta H = -q_{\text{water}}$, divided by the moles that reacted.

Try it in Python

The program works through a coffee-cup measurement: the temperature rise, the heat taken up by the water, and the enthalpy change per mole.

Common mistakes

• Counting liquids in $\Delta n_g$. Only gases count. In the burning of methane the liquid water does not count.
• Mixing joules and kilojoules. $\Delta n_g R T$ comes out in joules. Divide by 1000 before you add it to kilojoules.
• Losing the sign in calorimetry. The water takes up the heat the reaction gives out. A warming means a negative $\Delta H$.
• Calling heat capacity intensive. Heat capacity is extensive. Molar and specific heat capacity are intensive.
• Thinking a bomb calorimeter gives $\Delta H$. It works at constant volume, so it gives $\Delta U$.

Remember these

• $H = U + pV$ is a state function. At constant pressure, $q_p = \Delta H$.
• $\Delta H = \Delta U + \Delta n_g R T$, with $\Delta n_g$ counting gases only.
• Exothermic: $\Delta H < 0$. Endothermic: $\Delta H > 0$.
• $q = C\Delta T = m c \Delta T$. For an ideal gas, $C_p - C_v = R$.
• Bomb calorimeter: constant volume, $\Delta U$. Coffee-cup calorimeter: constant pressure, $\Delta H$.

Check yourself

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

1. Is the heat taken in at constant pressure equal to $\Delta U$ or to $\Delta H$?

2. For $\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}$, how do $\Delta H$ and $\Delta U$ compare?

3. Which of these are intensive: volume, temperature, enthalpy, density?

4. Why does a gas need more heat to warm by 1 K at constant pressure than at constant volume?

5. Which calorimeter would you use to measure $\Delta U$ for burning sugar?