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Enthalpies of formation, photosynthesis and respiration
Standard states and standard enthalpy changes, standard enthalpy of formation, calculating ΔH° from enthalpies of formation, finding enthalpies of formation from heats of combustion, and the enthalpy changes of photosynthesis and cellular respiration.
A gum tree on a summer afternoon is running an endothermic reaction on an industrial scale, pulling carbon dioxide out of the air and storing sunlight as sugar. You're running the same reaction in reverse right now, burning that sugar slowly to power every heartbeat. With one table of numbers you can calculate exactly how much energy each process moves, without measuring either.
Standard conditions
ΔH depends on temperature, pressure and the states of the substances, so tables quote standard enthalpy changes, ΔH°, with everything in its standard state:
- a pure substance in its most stable form (solid, liquid or gas) at 100 kPa;
- a solute at a concentration of 1 mol L⁻¹;
- at a stated temperature, almost always 298 K (25 °C).
So the standard state of water at 298 K is liquid, of carbon is graphite, and of oxygen is O₂ gas. (Older books use 101.3 kPa, 1 atm, as standard pressure; the difference changes ΔH by far less than the precision of school data.)
Standard enthalpy of formation
The standard enthalpy of formation, ΔfH°, of a compound is the enthalpy change when one mole of it forms from its elements in their standard states, under standard conditions. For example:
$$\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g) \qquad \Delta_f H^\circ = -74.8 \text{ kJ mol}^{-1}$$
$$\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \qquad \Delta_f H^\circ = -285.8 \text{ kJ mol}^{-1}$$
Note the ½O₂: the definition fixes one mole of product, so fractions on the reactant side are normal. By definition, ΔfH° of an element in its standard state is zero, because forming it from itself involves no change.
| Substance | ΔfH° (kJ mol⁻¹) | Substance | ΔfH° (kJ mol⁻¹) |
|---|---|---|---|
| CO₂(g) | −393.5 | CaCO₃(s) | −1206.9 |
| CO(g) | −110.5 | CaO(s) | −635.1 |
| H₂O(l) | −285.8 | Fe₂O₃(s) | −824.2 |
| H₂O(g) | −241.8 | Al₂O₃(s) | −1675.7 |
| CH₄(g) | −74.8 | NH₃(g) | −46.1 |
| C₂H₄(g) | +52.4 | NO(g) | +90.3 |
| C₂H₅OH(l) | −277.6 | NO₂(g) | +33.2 |
| C₆H₁₂O₆(s) | −1273.3 | N₂O₄(g) | +9.2 |
Calculating ΔH° from enthalpies of formation
Imagine taking the reactants apart into their elements and then building the products from those elements. Hess's law says this route gives the same ΔH as the real reaction:
$$\Delta H^\circ = \sum n\,\Delta_f H^\circ(\text{products}) - \sum n\,\Delta_f H^\circ(\text{reactants})$$
where each $n$ is the coefficient in the balanced equation. Products minus reactants: the opposite order to the bond-energy formula.
Calculate ΔH° for Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s).
Calculate ΔH° for 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), the first step in making nitric acid.
Enthalpies of formation from heats of combustion
Most compounds can't be made directly from their elements in a calorimeter (try making glucose from carbon, hydrogen and oxygen). But nearly every carbon compound can be burned. Since ΔfH° of CO₂ and H₂O are known accurately, a measured ΔcH lets you work backwards.
ΔcH°(C₃H₈) = −2220 kJ mol⁻¹. Calculate ΔfH° of propane.
Photosynthesis and respiration
Photosynthesis and cellular respiration are the same chemical change running in opposite directions:
$$6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) \xrightarrow{\text{light, chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g)$$
$$\text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g) \xrightarrow{\text{enzymes}} 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l)$$
Use enthalpies of formation to calculate ΔH° for photosynthesis.
- Photosynthesis is endothermic and only happens while light supplies energy (captured by chlorophyll). The energy ends up stored in the bonds of glucose and oxygen.
- Respiration is exothermic. It releases the same 2803 kJ mol⁻¹ as burning glucose in a bomb calorimeter (Hess's law: same start, same finish), but in about 30 enzyme-catalysed steps at 37 °C. About a third of the energy is captured in ATP to drive muscles and cells; the rest keeps you warm.
- Per gram, glucose provides 2803 ÷ 180.2 = 15.6 kJ g⁻¹, which is why food labels use about 16–17 kJ g⁻¹ for carbohydrate.
- For Methane burns, work out ΔH° yourself from the table, then press Show ΔH. (Answer: −890.3 kJ mol⁻¹.)
- Compare Photosynthesis and Respiration. What do you notice about the two ladders?
- In Thermite, why are Al and Fe drawn at zero?
- For 2NO₂ → N₂O₄, both species sit above the elements. Is the reaction exothermic or endothermic?
Test yourself
Which equation represents the standard enthalpy of formation of ethanol?
One mole of ethanol, formed from elements in their standard states (graphite, H₂ gas, O₂ gas).
Calculate ΔH° for CaCO₃(s) → CaO(s) + CO₂(g).
(−635.1 − 393.5) − (−1206.9) = −1028.6 + 1206.9 = +178.3 kJ mol⁻¹.
The standard enthalpy of formation of O₂(g) is
Forming O₂(g) from O₂(g) involves no change.
How much energy does a plant store when it makes 9.0 g of glucose (M = 180.2 g mol⁻¹)?
n = 9.0 ÷ 180.2 = 0.0499 mol; 0.0499 × 2803 = 140 kJ.
Label the energy diagram for photosynthesis and respiration.
Photosynthesis, respiration, or both?
Standard conditions are a pressure of kPa and solution concentrations of 1 mol L⁻¹, usually at K. The standard enthalpy of formation of an element in its standard state is . ΔH° equals the sum of ΔfH° of the products minus the sum for the .
(a) Use enthalpies of formation to calculate ΔH° for the respiration of glucose: C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l). (3 marks)
(b) Explain, with reference to Hess's law, why the energy released by respiration in the body equals the heat of combustion of glucose, even though the two processes are very different. (3 marks)
(a) Σ products = 6(−393.5) + 6(−285.8) = −2361.0 − 1714.8 = −4075.8 kJ. Σ reactants = −1273.3 + 6(0) = −1273.3 kJ. ΔH° = −4075.8 − (−1273.3) = −2802.5 kJ mol⁻¹.
(b) Hess's law states that the enthalpy change of a reaction depends only on the initial and final states, not on the route taken. Combustion and respiration both start with glucose and oxygen and finish with carbon dioxide and liquid water, so their total ΔH must be identical. Combustion happens in one rapid step at high temperature, while respiration occurs through dozens of enzyme-catalysed steps at 37 °C, but the sum of the enthalpy changes of all those steps is still −2802.5 kJ mol⁻¹. The route only changes how the energy appears: in the body part of it is first captured as chemical energy in ATP before eventually becoming heat.