Topic 6
Topic 6 on a page
Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.
Definitions to learn word for word
- Enthalpy change, ΔH: the heat energy change measured at constant pressure (page 1).
- Standard conditions: 100 kPa and a specified temperature, usually 298 K.
- Exothermic: ΔH negative, heat given out, temperature of the surroundings rises. Endothermic: ΔH positive, heat taken in, temperature falls.
- Hess's Law: the enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same (page 9).
- Bond enthalpy: the enthalpy change when one mole of a particular covalent bond is broken, in gaseous molecules. Mean bond enthalpy: the average of this over a range of compounds, in the gaseous state (page 13).
The five standard enthalpy changes
| Enthalpy change | Definition (one mole of ...) | Equation features | Sign |
|---|---|---|---|
| reaction, | the molar quantities in the equation, as written | any balanced equation | + or − |
| formation, | a compound formed from its elements in standard states | one mole of compound; fractions on elements | + or − |
| combustion, | a substance completely burned in oxygen | one mole of fuel; CO₂(g), H₂O(l) | always − |
| neutralisation, | water formed from an acid and an alkali | H⁺(aq) + OH⁻(aq) → H₂O(l), about −57 | always − |
| atomisation, | gaseous atoms formed from the element in its standard state | ½Cl₂(g) → Cl(g) | always + |
All five are "under standard conditions". Always give the equation with state symbols. of an element in its standard state is zero (page 3 and page 4). Level diagrams: the arrow starts at the reactants; down is negative, up is positive (page 2).
From a thermometer to kJ mol⁻¹
- ΔT = final − initial (°C).
- q = m × c × ΔT in J, with c = 4.18 J g⁻¹ °C⁻¹ and m = mass of water or solution (1.00 g cm⁻³), never the reactant (page 5).
- ÷ 1000 for kJ.
- ÷ moles of the limiting reactant (or of the fuel burned, or of water formed).
- Sign: negative if the temperature rose, positive if it fell (page 6).
Cooling-curve correction: plot temperature against time, extend the cooling line back to the time of mixing, and take ΔT from there (page 8).
Evaluating an experiment
- Error: name the source, say which way it moves ΔH, suggest an improvement. Heat loss and incomplete combustion make ΔH less exothermic; they are systematic.
- Uncertainty: ± half the smallest division per reading; ΔT uses two readings, so the uncertainties add. Percentage uncertainty = uncertainty ÷ measured value × 100.
- Assumptions: all the energy goes into the water; the solution has the c and density of water; the apparatus absorbs no heat; the reaction goes to completion (page 7).
Hess cycles
- Follow an arrow: add. Go against it: subtract.
- Formation data: ΔrH = ΣΔfH(products) − ΣΔfH(reactants). Arrows point up from the elements.
- Combustion data: ΔrH = ΣΔcH(reactants) − ΣΔcH(products). Arrows point down to the products of burning.
- Multiply each value by its coefficient. Elements are zero (page 10).
- Core Practical 2: ΔfH(MgO) = ΔH₁ + ΔfH(H₂O) − ΔH₂ = −465.5 − 285.8 + 145.3 = −606 kJ mol⁻¹, against about −602 in a data book (pages 11 and 12).
Bond enthalpies
- ΔH = Σ(bonds broken) − Σ(bonds made), counting bonds for the moles in the equation.
- Only an estimate: mean values, gases only, intermolecular forces ignored (page 14).
- Reverse it to find a mean bond enthalpy: divide by the number of bonds (six N–H in 2NH₃).
- A weaker bond (lower bond enthalpy) tends to break first and the reaction tends to go faster at room temperature: an indication only (page 15).
The ten traps
- Giving ΔH without a sign or without units, or calling a negative ΔH endothermic.
- Drawing the level-diagram arrow from the wrong line.
- A formation equation with atoms or compounds on the left, or the compound not at one mole.
- A combustion equation with H₂O(g), CO, or more than one mole of fuel.
- Using ΔH per mole of acid when the definition is per mole of water.
- Using the mass of the reactant instead of the mass of the water or solution in q = mcΔT.
- Leaving q in joules when dividing by moles in kJ.
- Taking ΔT from the highest reading rather than the extended cooling line.
- Getting a Hess sum the wrong way round, or forgetting a multiplier.
- Counting bonds in one molecule instead of in the equation.
Mixed test
Try these with no notes, then check the answers.
- Define the standard enthalpy change of formation, and write the equation for the formation of methanol, CH₃OH(l).
- Burning 0.230 g of ethanol ( = 46.0) heats 100 g of water by 10.0 °C. Calculate and give one reason why it is less negative than the data-book value of −1367 kJ mol⁻¹.
- 50.0 cm³ of 0.500 mol dm⁻³ HCl is mixed with 50.0 cm³ of 0.500 mol dm⁻³ NaOH, and the temperature rises by 3.3 °C. Calculate . (c = 4.18 J g⁻¹ °C⁻¹)
- The thermometer in question 3 reads to ±0.05 °C. Calculate the percentage uncertainty in ΔT, and say how you could reduce it.
- (kJ mol⁻¹): C(s) −393.5, H₂(g) −285.8, CH₄(g) −890. Calculate of methane.
- Calculate ΔH for from (kJ mol⁻¹): NH₃(g) −46.1, NO(g) +90.4, H₂O(g) −241.8.
- Calculate ΔH for using C=C 612, H–Cl 432, C–C 347, C–H 413 and C–Cl 346 kJ mol⁻¹.
- has ΔH = −185 kJ mol⁻¹. Using H–H 436 and Cl–Cl 243 kJ mol⁻¹, find the mean H–Cl bond enthalpy.
- Give two reasons why a value from mean bond enthalpies differs from a value from formation data.
- Explain why nitrogen is much less reactive than fluorine at room temperature.
- In a mixing experiment, a student extends the cooling line back to the time of the highest reading. State what is wrong and the effect on the result.
Answers cover these until you have tried the questions
- The enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions.
- q = 100 × 4.18 × 10.0 = 4180 J = 4.18 kJ. n = 0.230 ÷ 46.0 = 0.00500 mol. 4.18 ÷ 0.00500 = 836, and the water got hotter, so = −836 kJ mol⁻¹. It is less negative because heat is lost to the surroundings (or combustion is incomplete), so the water gains less energy than the fuel releases.
- m = 100 g. q = 100 × 4.18 × 3.3 = 1379.4 J = 1.3794 kJ. n(H₂O) = 0.0500 × 0.500 = 0.0250 mol. 1.3794 ÷ 0.0250 = 55.2, and the temperature rose, so = −55.2 kJ mol⁻¹.
- Two readings: ±(0.05 + 0.05) = ±0.1 °C. Percentage = 0.1 ÷ 3.3 × 100 = 3.0%. Reduce it by making the temperature rise larger (more concentrated solutions) or by using a thermometer with finer divisions.
- Equation: . Burning the elements: −393.5 + 2(−285.8) = −965.1. Burning methane: −890. = reactants − products = −965.1 − (−890) = −75.1 kJ mol⁻¹.
- Products: 4(+90.4) + 6(−241.8) = 361.6 − 1450.8 = −1089.2. Reactants: 4(−46.1) + 5(0) = −184.4. ΔH = −1089.2 − (−184.4) = −904.8 kJ mol⁻¹ (about −905).
- Broken: C=C + H–Cl = 612 + 432 = 1044. Made: C–C + C–H + C–Cl = 347 + 413 + 346 = 1106. ΔH = 1044 − 1106 = −62 kJ mol⁻¹.
- Broken: 436 + 243 = 679. −185 = 679 − made, so made = 864. Two H–Cl bonds, so mean H–Cl = 864 ÷ 2 = 432 kJ mol⁻¹.
- Any two: bond enthalpies are mean values, averaged over many compounds; the method assumes gaseous species, so changes of state are ignored; intermolecular forces are ignored.
- The N≡N bond is very strong (945 kJ mol⁻¹), so a great deal of energy is needed to break it. The F–F bond is weak (158 kJ mol⁻¹) and breaks easily. The weaker bond breaks more easily, so fluorine reacts faster at room temperature.
- The cooling line should be extended back to the time of mixing, not the time of the highest reading. The extrapolated temperature would be too low, so ΔT is too small and ΔH is less exothermic than it should be.