7 Thermochemistry

Energy transfers and internal energy

Begin by choosing the system, the matter being studied, and the surroundings, everything else. Energy is conserved: energy transferred out of one part of the universe enters another. Thermochemistry tracks these transfers during chemical and physical changes.

Heat is energy transferred because of a temperature difference; work is another way energy can move between system and surroundings. Temperature describes a thermal property of matter, while heat describes energy in transit. Heat is represented by $q$, and energy is commonly measured in joules or kilojoules.

Use the chemistry sign convention:

• If heat enters the system, $q>0$; if heat leaves it, $q<0$.

• If work is done on the system, $w>0$; if work is done by the system, $w<0$.

For expansion or compression against a constant external pressure, pressure–volume work is

where $\Delta V=V_{\text{final}}-V_{\text{initial}}$. Expansion gives $\Delta V>0$, so the system does work on its surroundings and $w<0$. Compression gives $\Delta V<0$, so work is done on the system and $w>0$.

The internal energy $U$ is the total microscopic energy of the system. The first law of thermodynamics accounts for changes in it:

For example, if a gas expands by $2.0\ \text{L}$ against an external pressure of $1.0\ \text{atm}$, then

If it also absorbs $500\ \text{J}$ of heat, then $\Delta U=500\ \text{J}-203\ \text{J}=297\ \text{J}$. The negative work indicates energy transferred from the gas to its surroundings.

Internal energy is a state function: its change depends only on the initial and final states. Heat and work are path-dependent because their amounts depend on how the process occurs.

Takeaway: Define the system first, then apply the signs for heat and work consistently in $\Delta U=q+w$.

Enthalpy and reaction heat

Enthalpy is a state function defined by

At constant pressure, if pressure–volume work is the only work, the heat transferred to the system equals its enthalpy change:

The sign of the enthalpy change describes the direction of reaction heat transfer:

• An exothermic reaction releases heat from the system, so $\Delta H<0$.

• An endothermic reaction absorbs heat into the system, so $\Delta H>0$.

A thermochemical equation pairs a balanced reaction with its enthalpy change. That value applies to the reaction as written. Reversing the reaction changes the sign of $\Delta H$; multiplying every coefficient by a factor multiplies $\Delta H$ by the same factor. For instance, if the reaction as written has $\Delta H=-100\ \text{kJ}$, carrying out half that amount gives $\Delta H=-50\ \text{kJ}$.

Reaction enthalpy can be calculated from standard enthalpies of formation:

Here, $n$ is the coefficient of each substance in the balanced equation. Alternatively, Hess’s law allows enthalpy changes for individual steps to be added to obtain the overall enthalpy change. This works because enthalpy is a state function.

Takeaway: Keep the balanced equation tied to its enthalpy value, and use the reaction direction and coefficients to adjust that value.

Measuring heat with calorimetry

Calorimetry determines heat transfer by measuring temperature changes. For a substance that changes temperature without changing phase, use

where $m$ is mass, $c$ is specific heat capacity, and $\Delta T=T_{\text{final}}-T_{\text{initial}}$. For water, a commonly used value is $c=4.184\ \text{J g}^{-1}\,^{\circ}\text{C}^{-1}$. A temperature difference of $1$ Celsius degree has the same size as a difference of $1$ kelvin.

In an insulated experiment, energy conservation gives

If a reaction warms the water around it, the water gains heat while the reaction loses the same amount, assuming negligible heat exchange with the calorimeter and outside environment.

For example, suppose a reaction warms $50.0\ \text{g}$ of water by $3.00\ ^\circ\text{C}$. Treating the solution like water gives

With negligible heat exchange elsewhere, $q_{\text{rxn}}=-627.6\ \text{J}$. If $0.0100\ \text{mol}$ of limiting reactant reacts, the heat released per mole of that reactant is $-62.8\ \text{kJ mol}^{-1}$. At constant pressure, the reaction heat corresponds to $\Delta H$ for the amount that reacted.

The calorimeter itself may also absorb heat. If its heat capacity is $C_{\text{cal}}$, then $q_{\text{cal}}=C_{\text{cal}}\Delta T$, and the energy balance becomes

A coffee-cup calorimeter operates approximately at constant pressure, so it can measure reaction enthalpy. A bomb calorimeter is rigid and operates at constant volume; the reaction heat measured there corresponds to $\Delta U$, not directly to $\Delta H$.

Takeaway: Calculate heat gained by the measured surroundings, then use the energy balance to find heat lost or gained by the reaction. Include the calorimeter term when it is not negligible.

A dependable problem-solving approach

A reliable calculation keeps the physical setup, signs, and units connected:

1. Identify the system and surroundings, and state assumptions such as negligible heat loss.

2. Choose the relevant relationship: $\Delta U=q+w$, $w=-P_{\text{ext}}\Delta V$, $q=mc\Delta T$, or $q=C\Delta T$.

3. Define temperature change as final minus initial, $\Delta T=T_{\text{final}}-T_{\text{initial}}$, and track the signs of heat and work.

4. Use energy conservation to relate the measured surroundings to the system or reaction.

5. Convert units and, when needed, use the balanced equation to report heat per mole or per reaction as written.

The key connections are that heat and work change internal energy, constant-pressure heat can equal an enthalpy change under the stated conditions, and calorimetry uses temperature changes to quantify those transfers.