Chemical Energetics

0. What Examiners Want

1. Exothermic, Endothermic and Profile Diagrams

• Exothermic: energy is transferred from the system to the surroundings, the temperature of the surroundings rises and ΔH is negative (e.g. combustion, neutralisation).
• Endothermic: energy is taken in from the surroundings, the temperature falls and ΔH is positive (e.g. thermal decomposition, dissolving ammonium nitrate).
• Enthalpy change, ΔH: the heat energy change measured at constant pressure.
• Activation energy, E_a: the minimum energy that colliding particles must have for a reaction to occur.

2. Standard Conditions and Definitions

Standard conditions: a pressure of 101 kPa, a stated temperature of 298 K, solutions at 1 mol dm⁻³ where relevant, and every substance in its normal physical state under those conditions. The symbol ^⊖ on ΔH shows that standard conditions apply.

• ΔH_neut for a strong acid with a strong alkali is about −57 kJ mol⁻¹ because the reaction is always H⁺(aq) + OH⁻(aq) → H₂O(l).
• The equation for ΔH_f often has fractions, e.g. Mg(s) + ½O₂(g) → MgO(s), because it must make exactly one mole of the compound.

3. Calorimetry

The heat released or absorbed by a reaction in solution changes the temperature of the water present. For an aqueous solution treat 1 cm³ as 1 g and use the specific heat capacity of water, 4.18 J g⁻¹ K⁻¹.

q = mcΔT   then   ΔH = −q ÷ n (in kJ mol⁻¹)

• m is the mass of the solution heated (in g), not the mass of the reagent that dissolved. ΔT is the temperature change, not the final temperature.
• n is the amount (mol) of the limiting reactant.
• Convert q from J to kJ before dividing by n. A temperature rise gives a negative ΔH; a fall gives a positive ΔH.

Worked example 1: an endothermic dissolving. 1.60 g of ammonium nitrate (M_r = 80.0) is dissolved in 50.0 cm³ of water. The temperature falls from 22.0 °C to 19.4 °C.

1. ΔT = 2.6 K; q = 50.0 × 4.18 × 2.6 = 543 J.
2. n = 1.60 ÷ 80.0 = 0.0200 mol.
3. ΔH = +543 ÷ 0.0200 = +27 200 J mol⁻¹ = +27.2 kJ mol⁻¹ (positive because the temperature fell).

Sources of error (why the measured value is less exothermic than the true one): heat lost to the surroundings, heat absorbed by the calorimeter and thermometer, incomplete reaction or combustion, evaporation, and non-standard conditions. Use a lid and insulation (a polystyrene cup) to reduce loss.

Paper 3: a slow reaction. If the reaction takes minutes, the highest temperature reading is too low because heat is lost while the temperature is still rising. Record the temperature at regular intervals for a few minutes before mixing, add the reagent at a noted time, keep stirring and continue recording at regular intervals. Plot temperature against time, draw lines of best fit through the before and after readings, and extrapolate the cooling line back to the time of mixing to find the corrected temperature change.

4. Hess's Law and Enthalpy Cycles

Hess's law: the enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. It lets us find a ΔH that cannot be measured directly.

Route 1: from enthalpies of formation. Both routes start from elements, so the cycle arrows point up from the elements.

Worked example 2. Calculate ΔH_r for 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). ΔH_f / kJ mol⁻¹: NH₃ −46, NO +90, H₂O(g) −242.

1. Products: 4(+90) + 6(−242) = 360 − 1452 = −1092.
2. Reactants: 4(−46) + 5(0) = −184 (oxygen is an element, so zero).
3. ΔH_r = (−1092) − (−184) = −908 kJ mol⁻¹.

Route 2: from enthalpies of combustion. Now the reactants and the products are both burned, so the arrows point down and the formula is reversed.

Worked example 3. Find ΔH_f of methanol, CH₃OH(l), given ΔH_c / kJ mol⁻¹: C(graphite) −394, H₂(g) −286, CH₃OH(l) −726.

1. Equation: C(s) + 2H₂(g) + ½O₂(g) → CH₃OH(l).
2. ΔH_f = ΣΔH_c(reactants) − ΔH_c(product) = [(−394) + 2(−286)] − (−726) = −966 + 726 = −240 kJ mol⁻¹.

5. Bond Energies

Bond energy: the enthalpy change when one mole of a covalent bond is broken in the gas phase (averaged over different molecules for most bonds). Breaking bonds is endothermic (+); making bonds is exothermic (−).

ΔH_r = Σ(bond energies broken) − Σ(bond energies formed)

Worked example 4. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). Bond energies / kJ mol⁻¹: C–H 410, O=O 496, C=O (in CO₂) 805, O–H 460.

1. Broken: 4(410) + 2(496) = 1640 + 992 = 2632.
2. Formed: 2(805) + 4(460) = 1610 + 1840 = 3450.
3. ΔH_r = 2632 − 3450 = −818 kJ mol⁻¹.

• Bond energies of most bonds are average values, so a result from bond energies is only approximate. The values for diatomic molecules (H–H, Cl–Cl) are exact because they refer to a single, definite bond.
• The calculation applies to gaseous species. For liquids you must also include the enthalpy change of vaporisation.
• More shared pairs mean a stronger bond: C≡C > C=C > C–C, and N≡N (very strong) explains why nitrogen is unreactive.

6. Quick Sheet and Checklist

• Exothermic (ΔH −) and endothermic (+); E_a and ΔH arrows on a profile diagram.
• Standard conditions and the four definitions, with the “one mole” wording.
• q = mcΔT; ΔH = −q ÷ n; m is the solution; convert J to kJ; sign from the temperature change.
• Slow reactions: a temperature–time graph, extrapolated to the time of mixing.
• Three routes: ΣΔH_f (P − R), ΣΔH_c (R − P), bonds (broken − formed).
• Elements have ΔH_f = 0; bond energies are averages.

Before you leave this topic, can you:

1. Write the definition of the standard enthalpy change of formation?
2. Calculate ΔH for a reaction from ΔH_f values, including the coefficients and the sign?
3. Explain how to improve the accuracy of a calorimetry result for a slow reaction?
4. Say why a value from average bond energies is only an approximation?

If yes to all four, attempt the ten exam-style questions with the mark schemes covered, then open the flashcards.