Thermochemical equations and enthalpy changes: VCE Chemistry Unit 3

What this dot point is asking

VCAA wants you to write a thermochemical equation (balanced equation plus ΔH with the correct sign), know the sign convention for exothermic and endothermic reactions, and use the mole ratio in the equation to scale ΔH up or down for a given amount of substance.

The answer

A thermochemical equation

A thermochemical equation is a balanced chemical equation that includes the enthalpy change (ΔH) for the reaction as written. The equation must include the states of all reactants and products because enthalpy changes depend on phase (vapour vs liquid water release different energies, for example).

General form:

reactants -> products ΔH = ± x kJ mol^-1

The "per mole" refers to the reaction as written. ΔH for the reaction 2H2(g) + O2(g) -> 2H2O(l) is double the ΔH for H2(g) + ½O2(g) -> H2O(l).

Sign convention for ΔH

ΔH is the change in enthalpy (H_products minus H_reactants), measured at constant pressure.

• Exothermic reaction: energy is released to the surroundings; H_products < H_reactants; ΔH is negative. Combustion of fuels, neutralisation, freezing, condensation.
• Endothermic reaction: energy is absorbed from the surroundings; H_products > H_reactants; ΔH is positive. Photosynthesis, thermal decomposition of CaCO3, melting, evaporation.

A useful memory aid: "the system loses, so the number is negative."

Reading a thermochemical equation

C(s) + O2(g) -> CO2(g) ΔH = -394 kJ mol^-1

Means: when 1 mole of solid carbon reacts with 1 mole of oxygen gas to form 1 mole of CO2 gas, 394 kJ of energy is released to the surroundings.

Double the equation:

2C(s) + 2O2(g) -> 2CO2(g) ΔH = -788 kJ mol^-1

The coefficients and ΔH both double together.

Reverse the equation:

CO2(g) -> C(s) + O2(g) ΔH = +394 kJ mol^-1

Reversing flips the sign of ΔH.

Scaling energy for a given mass

To find the energy released or absorbed by a non-stoichiometric amount, use the mole ratio.

1. Calculate moles of the reactant or product of interest: n = m / M.
2. Multiply by the ΔH per mole as written.

Example. The molar enthalpy of combustion of ethanol (C2H5OH) is -1367 kJ mol^-1. How much energy is released when 11.5 g of ethanol burns?

• n(C2H5OH) = 11.5 / 46.0 = 0.250 mol
• Energy released = 0.250 × 1367 = 342 kJ

The negative sign is dropped because the question asks for energy released (a positive quantity). The thermochemical equation keeps the negative sign on ΔH.

Energy profile diagrams

Thermochemical equations are visualised as energy profile diagrams showing reactants, products, and the activation energy barrier.

• Exothermic: products below reactants on the energy axis. ΔH measured as products minus reactants (negative).
• Endothermic: products above reactants. ΔH positive.
• The activation energy (Ea) is the height from reactants to the peak (transition state). It is always positive and is not ΔH.

States matter

Always include states because the same reaction releases different ΔH depending on phase. For example:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) ΔH = -890 kJ mol^-1

vs

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) ΔH = -802 kJ mol^-1

The difference (88 kJ mol^-1 less released) is the energy used to convert 2 mol of liquid water into vapour. By VCAA convention, combustion enthalpies are reported with liquid water as the product unless told otherwise.

Examples in context

Example 1. CO$_2$ enthalpy at Latrobe Valley CCS pilot. The CO2CRC pilot at Otway near Warrnambool tests carbon capture and storage. The capture reaction $\text{CO}_2 + \text{MEA(aq)} \to \text{MEA-CO}_2 (\text{aq})$ has $\Delta H = -85 \, \text{kJ/mol}$ (exothermic), so heat is released as CO$_2$ binds to monoethanolamine. To regenerate the solvent, $+85 \, \text{kJ/mol}$ must be added at $120^{\circ}\text{C}$ in the stripper. Capturing $1 \, \text{Mt}$ of CO$_2$ per year requires $\sim 2.27 \times 10^{10}$ mol $\times 85 \, \text{kJ} = 1.93 \, \text{PJ}$ of regeneration heat, supplied by low-pressure steam. The energy penalty of $\sim 25\%$ on the power station is the main reason CCS is so costly.

Example 2. Hess cycle for ethanol synthesis at Sunshot bioethanol. Sunshot engineers verify ethanol's enthalpy of formation via Hess's law. Three reactions: (1) $\text{C(s)} + \text{O}_2 \to \text{CO}_2$, $\Delta H = -393.5$; (2) $\text{H}_2 + 1/2 \text{O}_2 \to \text{H}_2 \text{O}$, $\Delta H = -285.8$; (3) $\text{C}_2 \text{H}_5 \text{OH} + 3 \text{O}_2 \to 2 \text{CO}_2 + 3 \text{H}_2 \text{O}$, $\Delta H = -1367$. By Hess: $\Delta H_f (\text{ethanol}) = 2 \times (-393.5) + 3 \times (-285.8) - (-1367) = -787 - 857.4 + 1367 = -277.4 \, \text{kJ/mol}$. The literature value is $-277.7 \, \text{kJ/mol}$, agreeing within $0.1\%$. This confirms the bomb-calorimeter $\Delta H_c$ for purposes of MEPS reporting.

Try this

Q1. Write the thermochemical equation for the complete combustion of propane gas, $\Delta H_c = -2220 \, \text{kJ/mol}$. State the states of all species. [3 marks]

• Cue. $\text{C}_3 \text{H}_8 (\text{g}) + 5 \text{O}_2 (\text{g}) \to 3 \text{CO}_2 (\text{g}) + 4 \text{H}_2 \text{O} (\text{l})$, $\Delta H = -2220 \, \text{kJ/mol}$.

Q2. Calculate the heat released when $100 \, \text{g}$ of ethanol burns completely. $\Delta H_c (\text{ethanol}) = -1367 \, \text{kJ/mol}$. [3 marks]

• Cue. $n = 100 / 46.0 = 2.17$ mol; heat $= 2.17 \times 1367 = 2972 \, \text{kJ}$.

Q3. Use Hess's law to determine $\Delta H$ for $\text{C(s)} + 2 \text{H}_2 (\text{g}) \to \text{CH}_4 (\text{g})$ from: $\text{C(s)} + \text{O}_2 \to \text{CO}_2$ ($-394$); $\text{H}_2 + 1/2 \text{O}_2 \to \text{H}_2 \text{O}$ ($-286$); $\text{CH}_4 + 2 \text{O}_2 \to \text{CO}_2 + 2 \text{H}_2 \text{O}$ ($-890$). (a) Write equations 1+2 and reverse 3. (b) Add and simplify. (c) State whether the formation of methane is exo- or endothermic. [2+2+2 marks]

• Cue. (a) Sum: $-394 + 2(-286) - (-890)$. (b) $\Delta H = -394 - 572 + 890 = -76 \, \text{kJ/mol}$. (c) Exothermic.