Enthalpy and Calorimetry

What is it?

Chemical reactions almost always release or absorb heat. The heat change of a reaction at constant pressure is its enthalpy change, $\Delta H$ , usually given in kJ/mol.

Exothermic reaction: heat is released to the surroundings, which warm up. $\Delta H$ is negative. Examples: combustion, neutralization, respiration.
Endothermic reaction: heat is absorbed from the surroundings, which cool down. $\Delta H$ is positive. Examples: dissolving ammonium nitrate (instant cold packs), photosynthesis, thermal decomposition.

Key idea

$\Delta H = H_\text{products} - H_\text{reactants}$ . Negative: exothermic (heat out). Positive: endothermic (heat in). The sign is always from the point of view of the reaction (the system).

Why does it matter?

Energy. Fuels are compared by the heat their combustion releases per gram or per mole.
Food. The energy content of food (in kJ or kcal) is measured by burning it in a calorimeter.
Safety and design. Chemists must know how much heat a reaction gives out to cool it safely, or how much to supply to keep it going.

How does it work?

1. Thermochemical equations

A thermochemical equation gives $\Delta H$ for the amounts in the balanced equation:

\begin{gathered} \small \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \\[4pt] \Delta H = -890.5\ \text{kJ} \end{gathered}

Doubling the equation doubles $\Delta H$ .
Reversing the equation changes the sign of $\Delta H$ .
States matter: forming $\ce{H2O(g)}$ instead of $\ce{H2O(l)}$ releases less heat.

2. Specific heat capacity and q = mcΔT

The specific heat capacity, $c$ , is the heat needed to raise the temperature of 1 g of a substance by 1 °C (or 1 K). For water, $c = 4.184$ J/(g·°C). The heat $q$ gained or lost is:

q = m \times c \times \Delta T

where $m$ is the mass in g and $\Delta T = T_\text{final} - T_\text{initial}$ . A change of 1 °C equals a change of 1 K, so $\Delta T$ is the same on both scales.

3. Calorimetry

A calorimeter measures heat changes. In a simple coffee-cup calorimeter, a reaction happens in water inside insulated cups:

Measure the temperature change of the solution.
Calculate $q_\text{solution} = m c \Delta T$ (assume the solution behaves like water: density 1.00 g/mL, $c$ = 4.184 J/(g·°C)).
The reaction’s heat is the opposite: $q_\text{reaction} = -q_\text{solution}$ (heat lost by one is gained by the other).
Divide by the moles that reacted: $\Delta H = \dfrac{q_\text{reaction}}{n}$ .

Think of it like this

Think of the reaction as a heater placed in a bath. You can’t see the heat directly, but the bath’s thermometer tells you: if the water warms, the “heater” gave heat out (exothermic); if the water cools, the reaction took heat in (endothermic).

More precisely

Enthalpy, $H$ , is defined as $H = U + PV$ ; at constant pressure, the heat transferred equals $\Delta H$ . Coffee-cup calorimeters work at constant (atmospheric) pressure, so they measure $\Delta H$ directly. A bomb calorimeter works at constant volume, measures $\Delta U$ , and is used for combustion. Simple calorimeters lose some heat to the surroundings and the cups, so measured values are usually slightly smaller in size than accepted ones.

Visualise it

ΔH is negative when the products are lower in enthalpy than the reactants.

A coffee-cup calorimeter: two nested polystyrene cups with a lid, holding the reaction mixture, which is mostly water. A thermometer and a stirrer pass through the lid. The heat of reaction changes the temperature of the solution; q of the solution equals m times c times delta T, and q of the reaction is minus q of the solution. — The solution's temperature change measures the heat of the reaction.

Worked example

Worked example: Heating water

Question: How much heat is needed to warm 250. g of water from 20.0 °C to 80.0 °C?

$\Delta T = 80.0\ °\text{C} - 20.0\ °\text{C} = 60.0\ °\text{C}$
Substitute:
$\begin{aligned} &q = m c \Delta T \\[4pt] &= (250.\ \text{g}) \\[4pt] &\quad \times \left(4.184\ \tfrac{\text{J}}{\text{g·°C}}\right) \\[4pt] &\quad \times (60.0\ °\text{C}) \\[4pt] &= 6.28 \times 10^{4}\ \text{J} = 62.8\ \text{kJ} \end{aligned}$
The units of g and °C cancel, leaving J.

Worked example: Enthalpy of neutralization

Question: 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH, both at 21.0 °C, are mixed in a coffee-cup calorimeter. The temperature rises to 27.8 °C. Calculate ΔH per mole of water formed.

Mass of solution: 100.0 mL × 1.00 g/mL = 100. g; $\Delta T = 27.8\ °\text{C} - 21.0\ °\text{C} = 6.8\ °\text{C}$
Heat gained by the solution:
$\begin{aligned} &q_\text{soln} = (100.\ \text{g}) \\[4pt] &\quad \times \left(4.184\ \tfrac{\text{J}}{\text{g·°C}}\right) \\[4pt] &\quad \times (6.8\ °\text{C}) \\[4pt] &= 2845\ \text{J} \end{aligned}$
$q_\text{reaction} = -2845$ J (the reaction released this heat).
Moles of water formed: $n = 1.00\ \text{mol/L} \times 0.0500\ \text{L} = 0.0500$ mol
Divide:
$\begin{aligned} &\Delta H = \frac{-2845\ \text{J}}{0.0500\ \text{mol}} \\[4pt] &= -5.7 \times 10^{4}\ \text{J/mol} \\[4pt] &= -57\ \text{kJ/mol} \end{aligned}$
ΔT (6.8 °C) has 2 significant figures, so the answer has 2. It is negative: the reaction is exothermic.

Common mistake

Common mistake: Getting the sign wrong

If the solution warms up, the reaction gave out heat: $q_\text{reaction}$ and $\Delta H$ are negative, even though $q_\text{solution}$ is positive.

Common mistake: Using only the mass of the solid

In $q = mc\Delta T$ , $m$ is the mass of the solution that changes temperature (mostly water), not the mass of the substance that reacted.

Common mistake: Forgetting to divide by moles

$q$ is the heat for this experiment (in J). $\Delta H$ is per mole (kJ/mol), so divide by the moles that reacted and convert J to kJ.

Notation note

$\Delta H°$ (with a degree sign) is a standard enthalpy change: 1 bar and a stated temperature, usually 25 °C.
Units: c in J/(g·°C) = J/(g·K) = J g⁻¹ K⁻¹; ΔH in kJ/mol = kJ mol⁻¹.
Some textbooks write $q = mc\Delta T$ as $q = m c_\text{s} \Delta T$ , with $c_\text{s}$ for specific heat.

Remember this

Exothermic: heat released, ΔH negative. Endothermic: heat absorbed, ΔH positive.
$q = mc\Delta T$ ; for water $c = 4.184$ J/(g·°C); ΔT in °C equals ΔT in K.
Calorimetry: $q_\text{reaction} = -q_\text{solution}$ ; $\Delta H = q_\text{reaction}/n$ , in kJ/mol.
Doubling an equation doubles ΔH; reversing it changes the sign.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/enthalpy/