Chemical Energetics

Enthalpy change: a sign tells you the direction of energy flow

Enthalpy, $H$, is the heat energy stored in a system at constant pressure. You can never measure $H$ itself, only how much it changes. A reaction's enthalpy change, $\Delta H$, is the products' enthalpy minus the reactants': $\Delta H = H_{\text{products}} - H_{\text{reactants}}$, usually in $\text{kJ mol}^{-1}$. Its sign tells you which way the energy flows:

• Exothermic: energy is released to the surroundings, so the products have less enthalpy than the reactants. $\Delta H$ is negative, and the surroundings get warmer.
• Endothermic: energy is absorbed from the surroundings, so the products have more enthalpy than the reactants. $\Delta H$ is positive, and the surroundings get cooler.

The sign always describes the reacting chemicals. In an exothermic reaction they lose energy ($\Delta H$ negative), so a thermometer in the surroundings goes up. Students often get this backwards.

Reversing a change reverses the sign of $\Delta H$ with the same size, because the start and end levels simply swap.

Reading and drawing a reaction pathway diagram

A reaction pathway diagram (also called an energy profile) plots enthalpy on the vertical axis against the reaction's progress on the horizontal axis. Three features appear on every one:

• A horizontal line for the reactants, at their own enthalpy level.
• A horizontal line for the products, at their own enthalpy level.
• A curve joining them that rises to a peak in between. The height of the peak above the reactants is the activation energy, $E_A$: the minimum energy colliding particles need before they can react. Every reaction has a positive $E_A$, whether it is exothermic or endothermic.

The reverse reaction on the same diagram

Read the same diagram from right to left and you have the reverse reaction. It starts at the products line and climbs to the same peak, because it goes through the same highest-energy arrangement of atoms. So:

• $\Delta H$ of the reverse reaction has the same size and the opposite sign.
• $E_A$ of the reverse reaction is the height of the peak above the products line.

For an exothermic forward reaction, the products sit lower, so the reverse climb is longer:

Demonstration. A forward reaction has $E_A = +50\text{ kJ mol}^{-1}$ and $\Delta H = -30\text{ kJ mol}^{-1}$.

• The reverse $\Delta H$ is $+30\text{ kJ mol}^{-1}$ (same size, sign flipped).
• The reverse $E_A$ is $50 - (-30) = 50 + 30 = +80\text{ kJ mol}^{-1}$: the $50$ back up from the reactants' level to the peak, plus the $30$ from the products' level up to the reactants' level.

Activation energies are always positive. For an exothermic forward reaction, $E_A$(reverse) is always bigger than $E_A$(forward); for an endothermic forward reaction it is always smaller (but still above zero).

Standard conditions, and the four named enthalpy changes

The same reaction gives out a slightly different amount of heat at different temperatures and pressures, so values are compared under fixed standard conditions: $298\text{ K}$ ($25\,°\text{C}$) and $101\text{ kPa}$. The symbol $\ominus$ (as in $\Delta H^{\ominus}$) means "measured under standard conditions".

Each substance is then in its standard state: its normal physical state at $298\text{ K}$ and $101\text{ kPa}$. For example carbon is $\text{C(s)}$, hydrogen and oxygen are $\text{H}_2\text{(g)}$ and $\text{O}_2\text{(g)}$, bromine is $\text{Br}_2\text{(l)}$, and water and ethanol are liquids.

Four enthalpy changes have their own names. Each one fixes one mole of a particular substance:

Writing the equation that matches a definition

Exam questions often ask which equation "represents" $\Delta H_f^{\ominus}$ or $\Delta H_c^{\ominus}$. Check three things:

1. Exactly one mole of the substance the definition fixes. Fractions such as $\tfrac12\text{O}_2$ are fine on the other side if that is what balancing needs.
2. Elements in their standard states (for $\Delta H_f$): $\text{C(s)}$, not $\text{C(g)}$; $\text{H}_2\text{(g)}$, not $\text{H(g)}$.
3. The right state symbol for the substance itself at $298\text{ K}$.

Demonstration. For $\Delta H_f^{\ominus}$ of water: $\text{H}_2\text{(g)} + \tfrac12\text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O(l)}$. Writing $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$ is balanced but forms two moles, so its $\Delta H$ is $2 \times \Delta H_f$. Writing $\text{H}_2\text{O(g)}$ is the wrong state.

One consequence: forming an element in its standard state from itself involves no change at all, so $\Delta H_f^{\ominus}$ of any element in its standard state is zero (for example $\text{O}_2\text{(g)}$, $\text{N}_2\text{(g)}$, $\text{C(s)}$). You will use this in every Hess's law calculation.