Energy Changes in Reactions

5.1 Exothermic and endothermic reactions

Enthalpy change

The transfer of thermal energy during a reaction is called the enthalpy change, $\Delta H$, of the reaction. It is measured in $\text{kJ/mol}$.

• For an exothermic reaction, $\Delta H$ is negative, because the reacting chemicals lose energy to the surroundings.
• For an endothermic reaction, $\Delta H$ is positive, because the reacting chemicals gain energy from the surroundings.

Activation energy

Even exothermic reactions need some energy to start, which is why a fuel does not burn until it is lit.

Reaction pathway diagrams

A reaction pathway diagram (an energy level diagram) shows the energy of the chemicals as the reaction goes from reactants to products.

To read or draw one:

• The reactants are at the left, and the products at the right, each shown as a horizontal line at its energy level.
• The activation energy, $E_\text{a}$, is the height from the reactants to the top of the curve.
• The enthalpy change, $\Delta H$, is the difference in energy between the reactants and the products, drawn as an arrow between the two levels.
• Exothermic: the products are lower than the reactants, and the arrow points down ($\Delta H$ negative).
• Endothermic: the products are higher than the reactants, and the arrow points up ($\Delta H$ positive).

Bond breaking and bond making

• Breaking bonds is endothermic: energy must be supplied to pull the atoms apart.
• Making bonds is exothermic: energy is released when new bonds form.

$\Delta H = \text{(total energy to break the bonds in the reactants)} - \text{(total energy released making the bonds in the products)}$

• If more energy is released making bonds than is needed to break bonds, the reaction is exothermic ($\Delta H$ negative).
• If more energy is needed to break bonds than is released, the reaction is endothermic ($\Delta H$ positive).

Bonds broken: one $\ce{H-H}$ and one $\ce{Cl-Cl}$: $436 + 242 = 678\ \text{kJ}$. Bonds made: two $\ce{H-Cl}$: $2 \times 431 = 862\ \text{kJ}$.

$\Delta H = 678 - 862 = -184\ \text{kJ/mol}$

The sign is negative, so the reaction is exothermic.

Bonds broken: four $\ce{C-H}$ and two $\ce{O=O}$: $(4 \times 412) + (2 \times 496) = 1648 + 992 = 2640\ \text{kJ}$. Bonds made: two $\ce{C=O}$ and four $\ce{O-H}$: $(2 \times 805) + (4 \times 463) = 1610 + 1852 = 3462\ \text{kJ}$.

$\Delta H = 2640 - 3462 = -822\ \text{kJ/mol}$