H2 Chemistry · JC 1 and 2
Chemical Energetics
Chemical energetics deals with the energy changes in reactions and whether reactions can happen at all. In H2 Chemistry you define a set of standard enthalpy changes, measure them by calorimetry, combine them with Hess's law and Born-Haber cycles, and then add entropy and Gibbs free energy to judge whether a reaction is feasible. These ideas connect directly to equilibrium constants and to electrochemistry, where ΔG links to cell potentials.
What the syllabus expects
- Define standard enthalpy changes of formation, combustion, neutralisation, atomisation, hydration and solution, and bond energy
- Calculate enthalpy changes from calorimetry data using q = mcΔT
- Apply Hess's law using energy cycles with enthalpy changes of formation, combustion and bond energies
- Construct Born-Haber cycles and use them to calculate lattice energy
- Explain how ionic charge and ionic radius affect the magnitude of lattice energy
- Predict the sign of ΔS and use ΔG = ΔH − TΔS to predict whether a reaction is feasible
Standard enthalpy changes
Standard conditions are 298 K and 1 bar, with substances in their standard states. The standard enthalpy change of formation, ΔHf, is the enthalpy change when one mole of a substance is formed from its elements in their standard states, so ΔHf of an element is zero. The standard enthalpy change of combustion, ΔHc, is the enthalpy change when one mole of a substance is completely burnt in oxygen. The enthalpy change of neutralisation is for the formation of one mole of water from an acid and a base. The enthalpy change of atomisation is for the formation of one mole of gaseous atoms from the element. Bond energy is the energy needed to break one mole of a covalent bond in the gas phase, and values for bonds in different molecules are usually averages. Hydration enthalpy is for one mole of gaseous ions dissolving in water, and the enthalpy change of solution is for one mole of solute dissolving in water to give an infinitely dilute solution.
Calorimetry
In a calorimetry experiment, the heat change of the reaction is measured from the temperature change of a known mass of water or solution. Heat change q = mcΔT, where m is the mass of the solution (not the solid added), c is the specific heat capacity, usually 4.18 J g⁻¹ K⁻¹, and ΔT is the temperature change. For dilute solutions, take the density as 1.00 g cm⁻³, so the volume in cm³ gives the mass in grams. Then ΔH = −q ÷ n, where n is the number of moles that reacted as specified in the definition. The minus sign is added because a temperature rise means heat was released, so ΔH is negative. Heat losses to the surroundings make experimental values less exothermic than data book values, which is why a temperature-time graph extrapolated back to the time of mixing is used to estimate the true maximum temperature.
Hess's law and energy cycles
Hess's law states that the enthalpy change of a reaction depends only on the initial and final states, not on the route taken. This allows enthalpy changes that cannot be measured directly to be calculated. With formation data, ΔHr = ΣΔHf(products) − ΣΔHf(reactants), since both reactants and products can be formed from the same elements. With combustion data, ΔHr = ΣΔHc(reactants) − ΣΔHc(products), since both can be burnt to the same combustion products. With bond energies, ΔHr = Σ(bonds broken) − Σ(bonds formed), valid only when all species are gases. Because bond energies are averages, this last method gives approximate values. Drawing a clear cycle with every arrow labelled and multiplied by the right coefficients is the most reliable way to avoid sign errors.
Born-Haber cycles and lattice energy
Lattice energy is the energy released when one mole of a solid ionic compound is formed from its gaseous ions, so it is always negative. It cannot be measured directly, so it is calculated with a Born-Haber cycle. For sodium chloride, the steps from the elements to the gaseous ions are atomisation of sodium, the first ionisation energy of sodium, atomisation of chlorine (half the Cl-Cl bond energy) and the first electron affinity of chlorine. By Hess's law, ΔHf(NaCl) = ΔHat(Na) + IE1(Na) + ΔHat(Cl) + EA1(Cl) + LE(NaCl). The magnitude of lattice energy is proportional to the product of the ionic charges divided by the sum of the ionic radii. So MgO has a much more exothermic lattice energy than NaCl, because its ions carry charges of 2+ and 2− and are smaller.
Entropy and Gibbs free energy
Entropy, S, is a measure of the disorder of a system, or the number of ways the particles and their energy can be arranged. ΔS is positive when the number of moles of gas increases, when a solid melts or a liquid boils, when substances mix or dissolve, and when temperature rises. Whether a reaction is feasible depends on both ΔH and ΔS, combined in the Gibbs free energy change: ΔG = ΔH − TΔS, with T in kelvin. A reaction is feasible, or spontaneous, if ΔG is negative. Since ΔS is usually given in J K⁻¹ mol⁻¹, divide it by 1000 before combining with ΔH in kJ mol⁻¹. An endothermic reaction with a positive ΔS becomes feasible above T = ΔH ÷ ΔS. A negative ΔG does not mean the reaction is fast, because a high activation energy may make it too slow to observe.
Worked example
Use the data to calculate the lattice energy of sodium chloride. ΔHf(NaCl) = −411 kJ mol⁻¹, ΔHat(Na) = +107 kJ mol⁻¹, first ionisation energy of Na = +496 kJ mol⁻¹, ΔHat(Cl) = +122 kJ mol⁻¹, first electron affinity of Cl = −349 kJ mol⁻¹.
- Write the cycle: Na(s) + ½Cl₂(g) gives Na(g) + Cl(g), then Na⁺(g) + Cl⁻(g), then NaCl(s), and this route must equal ΔHf.
- By Hess's law, ΔHf = ΔHat(Na) + IE1(Na) + ΔHat(Cl) + EA1(Cl) + LE.
- Sum of the known steps = 107 + 496 + 122 + (−349) = +376 kJ mol⁻¹.
- So −411 = 376 + LE, giving LE = −411 − 376 = −787 kJ mol⁻¹.
- The value is negative, as expected for energy released when gaseous ions form a lattice.
Answer: The lattice energy of sodium chloride is −787 kJ mol⁻¹.
Common mistakes
- Using the total mass of reactants, including the solid added, in q = mcΔT. Use the mass of the solution being heated or cooled.
- Leaving ΔH positive when the temperature rose. A temperature rise means the reaction released heat, so ΔH is negative.
- Combining ΔS in J K⁻¹ mol⁻¹ directly with ΔH in kJ mol⁻¹. Convert ΔS to kJ K⁻¹ mol⁻¹ first.
- Using the full Cl-Cl bond energy for the atomisation of chlorine. Atomisation makes one mole of Cl atoms, so it is half the bond energy.
- Using bond energies when a species is a liquid, such as H₂O(l). The bond energy method applies to gaseous species only.
- Assuming a reaction with a negative ΔG will happen quickly. ΔG says nothing about rate, which depends on activation energy.
Exam tips
- Write the definition of each enthalpy change in full when asked, including 'one mole', the states and standard conditions. Definitions are a common source of easy marks lost.
- In Hess's law and Born-Haber questions, draw the cycle before writing any numbers, and write the multiplier beside each arrow.
- When asked to explain feasibility, quote the signs of ΔH and ΔS and say which term, ΔH or TΔS, dominates at the given temperature.
Practice questions
1. Easy: 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH, and the temperature rises by 6.8 °C. Calculate the enthalpy change of neutralisation. Take c = 4.18 J g⁻¹ K⁻¹ and the density of the solution as 1.00 g cm⁻³.
Show answer
−56.8 kJ mol⁻¹. q = 100 × 4.18 × 6.8 = 2842 J. Moles of water formed = 0.0500 mol. ΔH = −2842 ÷ 0.0500 = −56 800 J mol⁻¹ = −56.8 kJ mol⁻¹.
2. Medium: Calculate the standard enthalpy change of formation of propane, 3C(s) + 4H₂(g) gives C₃H₈(g), given ΔHc of C(s) = −394, H₂(g) = −286 and C₃H₈(g) = −2220 kJ mol⁻¹.
Show answer
−106 kJ mol⁻¹. Using ΔHr = ΣΔHc(reactants) − ΣΔHc(products): ΔHf = [3(−394) + 4(−286)] − (−2220) = (−1182 − 1144) + 2220 = −2326 + 2220 = −106 kJ mol⁻¹.
3. Harder: For CaCO₃(s) gives CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Show that the reaction is not feasible at 298 K and find the minimum temperature at which it becomes feasible.
Show answer
At 298 K, ΔG = 178 − 298 × 0.161 = 178 − 48.0 = +130 kJ mol⁻¹, which is positive, so it is not feasible. It becomes feasible when ΔG = 0, at T = ΔH ÷ ΔS = 178 ÷ 0.161 = 1106 K (about 833 °C), and above.
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