Enthalpy changes and thermochemical equations

What an enthalpy change is

• The enthalpy change, $\Delta H$, is the heat energy transferred during a change carried out at constant pressure.

• Exothermic changes release energy to the surroundings, so $\Delta H$ is negative.

• Endothermic changes absorb energy from the surroundings, so $\Delta H$ is positive.

• The sign is part of the answer. An enthalpy change written without its sign is wrong, not incomplete.

The symbols

The standard lists six enthalpy changes by name, and each has a precise definition.

Symbol	Name	Definition	Sign
$\Delta_rH^\circ$	reaction	for the equation exactly as written	either
$\Delta_cH^\circ$	combustion	1 mol of a substance burned completely in oxygen	always negative
$\Delta_fH^\circ$	formation	1 mol made from its elements in their standard states	either
$\Delta_{fus}H^\circ$	fusion	1 mol solid → liquid	always positive
$\Delta_{vap}H^\circ$	vaporisation	1 mol liquid → gas	always positive
$\Delta_{sub}H^\circ$	sublimation	1 mol solid → gas directly	always positive

• The ° symbol means standard conditions.
• Notice the phrase "1 mol" in five of the six definitions. Only $\Delta_rH^\circ$ refers to the equation as written.

Two definitions worth learning word for word

• $\Delta_cH^\circ$ is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.
  • "Completely" matters: carbon must go to $\text{CO}_2$, not $\text{CO}$.
• $\Delta_fH^\circ$ is the enthalpy change when one mole of a substance is formed from its elements in their standard states under standard conditions.
  • Standard state means the form the element takes at 25 °C and 100 kPa — $\text{O}_2(g)$, $\text{C}(s, \text{graphite})$, $\text{Br}_2(l)$.
  • It follows that $\Delta_fH^\circ$ of any element in its standard state is zero. There is no change involved in making an element from itself.

Thermochemical equations

• A thermochemical equation is a balanced equation with its enthalpy change written beside it:

$$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \qquad \Delta_rH^\circ = -572 \text{ kJ mol}^{-1}$$

• State symbols are compulsory. The same reaction producing steam instead of liquid water has a different enthalpy change, because vaporising the water costs energy.
• The quoted value applies to the molar amounts in that equation — here, 2 mol of hydrogen, not 1.

Scaling with amount

• Energy transferred = (amount reacting ÷ amount in the equation) × $\Delta_rH$.

• Double the amount, double the energy. Halve it, halve the energy. The sign never changes.

• Reverse the equation and the sign flips: forming water releases 572 kJ, so splitting it back up absorbs 572 kJ.

• The two quantity relationships you will need are supplied in the resource booklet:

$$n = \frac{m}{M} \qquad\qquad n = cV$$