5
Chapter 5· Energetics
Numericals Solved Step by Step
Each numerical in the order you write it in the exam: Given Data, To Find, Formula, Calculation, Result.
Worked example (book): formation of water
Calculate ΔH for 2H₂(g) + O₂(g) → 2H₂O(g) from bond energies: H–H 435, O=O 498 and O–H 484 kJ mol⁻¹. Also find ΔH per mole of water.
Given Data
- H–H = 435 kJ mol⁻¹
- O=O = 498 kJ mol⁻¹
- O–H = 484 kJ mol⁻¹
- Bonds broken: 2 H–H, 1 O=O; bonds formed: 4 O–H
To Find
- ΔH for 2 mol and for 1 mol of H₂O(g)
Formula
ΔH = Σ(bonds broken) − Σ(bonds formed)
Calculation
- Energy absorbed = 2(435) + 498 = 1368 kJ.
- Energy released = 4(484) = 1936 kJ.
- ΔH = 1368 − 1936 = −568 kJ for 2 mol.
- Per mole: −568 ÷ 2 = −284 kJ mol⁻¹.
ResultΔH = −568 kJ (2 mol); −284 kJ mol⁻¹
Note: Measured value for H₂O(g) is −241.8 kJ mol⁻¹; the book's O–H value (484) is higher than the usual 463.
In-text exercise 1
Calculate the enthalpy change of the reaction H₂(g) + I₂(g) → 2HI(g). Bond energies of H₂, I₂ and HI are 436, 151 and 299 kJ mol⁻¹ respectively.
Given Data
- H–H = 436 kJ mol⁻¹
- I–I = 151 kJ mol⁻¹
- H–I = 299 kJ mol⁻¹
Formula
ΔH = Σ(bonds broken) − Σ(bonds formed)
Calculation
- Bonds broken: 436 + 151 = 587 kJ.
- Bonds formed: 2 × 299 = 598 kJ.
- ΔH = 587 − 598 = −11 kJ.
ResultΔH = −11 kJ (slightly exothermic, for gaseous iodine)
Note: The book gives +53.08 kJ for solid iodine; turning I₂(s) into gas takes about 62 kJ, which makes the overall change endothermic.
In-text exercise 2
Calculate the enthalpy change for the formation of one mole of liquid water, given 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −571.6 kJ.
Given Data
- ΔH for 2 mol H₂O(l) = −571.6 kJ
Formula
ΔH per mole = ΔH ÷ moles formed
Calculation
- 2 mol of liquid water: −571.6 kJ.
- 1 mol: −571.6 ÷ 2 = −285.8 kJ.
ResultΔH = −285.8 kJ mol⁻¹
Exercise 4(i)
Find the enthalpy change of N₂ + O₂ → 2NO using: bond dissociation energy of N₂ = 958.38 kJ mol⁻¹, of O₂ = 498 kJ mol⁻¹, bond formation energy of NO = −626 kJ mol⁻¹.
Given Data
- N≡N = 958.38 kJ mol⁻¹
- O=O = 498 kJ mol⁻¹
- N=O formed = −626 kJ mol⁻¹ (2 bonds)
Formula
ΔH = Σ(bonds broken) + Σ(bonds formed, negative)
Calculation
- Energy absorbed = 958.38 + 498 = 1456.38 kJ.
- Energy released = 2 × 626 = 1252 kJ.
- ΔH = 1456.38 − 1252 = +204.38 kJ.
ResultΔH = +204.38 kJ (endothermic)
Note: The measured value quoted in the chapter is +180.6 kJ; bond energies are averages, so the answers differ a little.
Exercise 1(ix)
Average bond energies for O–O and O=O are 146 and 496 kJ mol⁻¹. Find ΔH for H–O–O–H(g) → H–O–H(g) + ½O₂(g).
Given Data
- O–O = 146 kJ mol⁻¹
- O=O = 496 kJ mol⁻¹
- O–H bonds: 2 on each side (cancel)
Formula
ΔH = Σ(bonds broken) − Σ(bonds formed)
Calculation
- Broken: one O–O = 146 kJ.
- Formed: ½ O=O = 248 kJ.
- ΔH = 146 − 248 = −102 kJ.
