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Chemistry · Ch 5 — Thermodynamics

Thermochemical Equations

5.4(d)

Thermochemical Equations

Thermochemical Equations

A balanced chemical equation together with the value of its ΔrH\Delta_r H is called a thermochemical equation. The physical state (including allotropic state) of each substance must be specified.

For example:

C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l);ΔrH∘=−1367 kJ mol−1\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l); \quad \Delta_r H^\circ = -1367\ \text{kJ mol}^{-1}

This describes the combustion of liquid ethanol at constant temperature and pressure. The negative sign indicates an exothermic reaction.

Conventions for Thermochemical Equations

Convention 1: The coefficients in a balanced thermochemical equation refer to the number of moles (never molecules) of reactants and products involved in the reaction.

Convention 2: The numerical value of ΔrH∘\Delta_r H^\circ refers to the number of moles of substances specified by the equation. Standard enthalpy change ΔrH∘\Delta_r H^\circ has units of kJ mol−1^{-1}.

To illustrate, consider the reaction:

Fe2O3(s)+3H2(g)→2Fe(s)+3H2O(l)\text{Fe}_2\text{O}_3(s) + 3\text{H}_2(g) \rightarrow 2\text{Fe}(s) + 3\text{H}_2\text{O}(l)

From Table 5.2:

  • ΔfH∘(H2O,l)=−285.83 kJ mol−1\Delta_f H^\circ(\text{H}_2\text{O}, l) = -285.83\ \text{kJ mol}^{-1}
  • ΔfH∘(Fe2O3,s)=−824.2 kJ mol−1\Delta_f H^\circ(\text{Fe}_2\text{O}_3, s) = -824.2\ \text{kJ mol}^{-1}
  • ΔfH∘(Fe,s)=0\Delta_f H^\circ(\text{Fe}, s) = 0 and ΔfH∘(H2,g)=0\Delta_f H^\circ(\text{H}_2, g) = 0 (by convention)

ΔrH1∘=3(−285.83 kJ mol−1)−1(−824.2 kJ mol−1)\Delta_r H_1^\circ = 3(-285.83\ \text{kJ mol}^{-1}) - 1(-824.2\ \text{kJ mol}^{-1})

=(−857.5+824.2) kJ mol−1=−33.3 kJ mol−1= (-857.5 + 824.2)\ \text{kJ mol}^{-1} = -33.3\ \text{kJ mol}^{-1}

The coefficients used are pure numbers equal to the stoichiometric coefficients. The unit kJ mol−1^{-1} means per mole of reaction as written.

If we balance the equation differently:

12Fe2O3(s)+32H2(g)→Fe(s)+32H2O(l)\frac{1}{2}\text{Fe}_2\text{O}_3(s) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{Fe}(s) + \frac{3}{2}\text{H}_2\text{O}(l)

Then:

ΔrH2∘=32(−285.83 kJ mol−1)−12(−824.2 kJ mol−1)\Delta_r H_2^\circ = \frac{3}{2}(-285.83\ \text{kJ mol}^{-1}) - \frac{1}{2}(-824.2\ \text{kJ mol}^{-1})

=(−428.7+412.1) kJ mol−1=−16.6 kJ mol−1=12ΔrH1∘= (-428.7 + 412.1)\ \text{kJ mol}^{-1} = -16.6\ \text{kJ mol}^{-1} = \frac{1}{2}\Delta_r H_1^\circ

This shows that enthalpy is an extensive quantity — it depends on the amount of substance.

Convention 3: When a chemical equation is reversed, the value of ΔrH∘\Delta_r H^\circ is reversed in sign. …

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