First law and Hess's law

MDCAT Chemistry · Chemistry · Thermochemistry and energetics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

The first law of thermodynamics says that energy cannot be created or destroyed, only changed from one form into another. When a system gains energy, the surroundings lose exactly the same amount. The internal energy, E, of a system changes only through heat, q, and work, w.

Internal energy and enthalpy are state functions: their change depends only on the start and the end, not on the route taken. Heat and work do depend on the route.

Hess's law follows from this. ΔH for a reaction is the same whether it happens in one step or in several. So we can calculate enthalpy changes that cannot be measured directly, such as ΔH°_f of CO, or a lattice energy from a Born–Haber cycle.

Rules to remember

ΔE = q + w. q is positive when the system absorbs heat; w is positive when work is done on the system.
Work of a gas at constant pressure: w = −PΔV. Expansion gives a negative w. 1 kPa × 1 dm³ = 1 J.
At constant volume no work is done, so q_v = ΔE. At constant pressure q_p = ΔH, where H = E + PV and ΔH = ΔE + PΔV.
Hess's law: if an equation is reversed, change the sign of ΔH; if it is multiplied, multiply ΔH by the same number; then add.
ΔH° = ΣΔH°_f(products) − ΣΔH°_f(reactants), each multiplied by its coefficient in the equation.
Born–Haber cycle: ΔH°_f = atomisation of the metal + ionisation energy + atomisation of the non-metal + electron affinity + lattice energy.
Lattice energy is the energy released when one mole of an ionic solid forms from its gaseous ions, so it is negative. Atomisation and ionisation are endothermic.

Common mistake

Students mix up the signs in ΔE = q + w. Heat leaving the system and work done by the system are both negative; heat entering and work done on the system are both positive.

Worked example

The standard enthalpies of combustion of C(graphite), H₂(g) and CH₄(g) are −394, −286 and −891 kJ mol⁻¹. Find the standard enthalpy of formation of methane.

Target equation: C(graphite) + 2H₂(g) → CH₄(g).
Route: burn the carbon (−394) and the two moles of hydrogen (2 × −286 = −572), then reverse the combustion of methane (+891).
ΔH°_f = −394 − 572 + 891 = −966 + 891 = −75 kJ mol⁻¹.

Answer: ΔH°_f of CH₄ = −75 kJ mol⁻¹.

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