the writing of thermochemical equations to represent the energy released or absorbed in physical and chemical changes, including the sign convention for ΔH for exothermic and endothermic reactions, and the use of ΔH values with mole ratios to calculate the energy released or absorbed

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.

Worked example

The molar enthalpy of combustion of propane (C3H8) is -2220 kJ mol^-1. A camp stove burns 44.0 g of propane during a meal. Calculate the energy released and write the thermochemical equation.

Step 1. Thermochemical equation:

C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l) ΔH = -2220 kJ mol^-1

Step 2. Moles of propane: n = 44.0 / 44.0 = 1.00 mol

Step 3. Energy released = 1.00 × 2220 = 2220 kJ

The thermochemical equation provides both the chemistry and the energy bookkeeping in one line.

Common traps

Forgetting states. Markers deduct for missing or wrong state symbols. Combustion uses (l) for water by VCE convention.

Sign-flipping ΔH after halving coefficients. Halving the equation halves the magnitude of ΔH but keeps the sign. Only reversing the equation flips the sign.

Calling Ea "the enthalpy change". Activation energy is the barrier from reactants to transition state. ΔH is the net difference between reactants and products. They are independent: a reaction can be exothermic with a high Ea or endothermic with a low Ea.

Quoting ΔH per gram. Molar enthalpy is always per mole of the species (usually the fuel for combustion). Energy per gram is a different quantity (energy content).

Writing ΔH = 890 kJ for an exothermic reaction. The sign must be negative for the system. The 890 kJ is the magnitude only.

In one sentence

A thermochemical equation is a balanced equation with states and a ΔH value that is negative for exothermic reactions and positive for endothermic ones; you scale ΔH by the mole ratio in the equation to find the energy for any amount.