Enthalpy Changes and Calorimetry – Energy Changes and Energy Profile Diagrams for Chemical Reactions (Senior 4 Chemistry) | Ganzaa

Energy Changes and Energy Profile Diagrams for Chemical Reactions

Lessons

Senior 4 · Chemistry

Energy Changes and Energy Profile Diagrams for Chemical Reactions

4 lessons

Lesson 3 of 4

Enthalpy Changes and Calorimetry

30 min

How much heat is in a spoonful of fuel?

Burn a little ethanol in a spirit lamp under a can of water and the water warms up. Measure the mass of fuel burned and the rise in temperature, and you can work out how much energy one mole of ethanol releases. A data book gives −1367 kJ for every mole burned; a school experiment usually finds far less. Where did the rest go? This lesson defines the enthalpy changes chemists measure, shows how to measure them by calorimetry, and explains why the answers come out as they do.

Definition

Enthalpy change, ΔH: the heat transferred between a system and its surroundings at constant pressure, per mole of reaction as written. Unit: kJ mol⁻¹. Negative for exothermic, positive for endothermic reactions.

Standard conditions (symbol ⦵): 298 K (25 °C), 100 kPa, all solutions 1 mol dm⁻³, all substances in their standard states.

Standard enthalpy of combustion, $\Delta_c H^\circ$ : the enthalpy change when 1 mol of a substance burns completely in oxygen under standard conditions. Example: ethanol, −1367 kJ mol⁻¹.

Enthalpy of neutralisation: the enthalpy change when an acid and a base react to form 1 mol of water. For strong acids with strong bases it is about −57 kJ mol⁻¹.

Specific heat capacity, c: the energy needed to raise the temperature of 1 g of a substance by 1 °C (1 K). For water, 4.18 J g⁻¹ K⁻¹.

Figure 17.3: Enthalpy level diagrams, each drawn to its own scale, from standard enthalpies of formation. An enthalpy level diagram shows only the start and the end; Lesson 4 adds the energy barrier between them.

Where the energy comes from: bonds

A chemical reaction breaks bonds in the reactants and makes new bonds in the products. Breaking a bond always takes in energy (endothermic); making a bond always gives out energy (exothermic). $\Delta H \approx (\text{energy taken in to break bonds}) - (\text{energy given out when new bonds form})$ When methane burns, the bonds formed in $\mathrm{CO_2}$ and $\mathrm{H_2O}$ are stronger than those broken in $\mathrm{CH_4}$ and $\mathrm{O_2}$ , so the reaction is exothermic. When limestone decomposes, more energy is needed to break the bonds than is returned, so heat must be supplied continuously, as in a lime kiln.

Measuring ΔH by calorimetry

A calorimeter is an insulated container in which a reaction heats (or cools) a known mass of water or solution. The energy transferred is $q = mc\Delta T$ where m is the mass of the water or solution (g), c its specific heat capacity (J g⁻¹ K⁻¹) and ΔT the temperature change (K or °C). Dividing by the amount of the limiting reactant gives ΔH: $\Delta H = -\frac{q}{n}$ The minus sign is there because the heat gained by the surroundings was lost by the system. Simple calorimetry assumes that dilute solutions have the density (1.00 g cm⁻³) and specific heat capacity of water, that no heat is lost, and that the container itself absorbs no heat.

Worked example 1: enthalpy of neutralisation

Illustrative results: 50.0 cm³ of 1.00 mol dm⁻³ HCl and 50.0 cm³ of 1.00 mol dm⁻³ NaOH, both at 21.0 °C, are mixed in a polystyrene cup. The highest temperature is 27.7 °C. Calculate the enthalpy of neutralisation.

Solution. Mass of solution $= 100\ \mathrm{cm^3} \times 1.00\ \mathrm{g\,cm^{-3}} = 100\ \mathrm{g}$ ; $\Delta T = 6.7\ \mathrm{^\circ C}$ . $q = mc\Delta T = 100 \times 4.18 \times 6.7 = 2.80 \times 10^{3}\ \mathrm{J}$ $\mathrm{H^+ + OH^- \rightarrow H_2O}$ : $n(\mathrm{H_2O}) = 1.00 \times 0.0500 = 0.0500\ \mathrm{mol}$ . $\Delta H = -\frac{2.80\ \mathrm{kJ}}{0.0500\ \mathrm{mol}} = -56.0\ \mathrm{kJ\,mol^{-1}}$ close to the accepted −57.6 kJ mol⁻¹; the small difference comes mainly from heat lost to the cup and the air.

Worked example 2: why the spirit lamp disappoints

Illustrative results: burning 0.920 g of ethanol ( $M = 46.07\ \mathrm{g\,mol^{-1}}$ ) raises the temperature of 200 g of water in a copper can by 23.0 °C. Calculate $\Delta_c H$ and compare it with the value from enthalpies of formation, $\mathrm{C_2H_5OH(l)}$ −277.0, $\mathrm{CO_2(g)}$ −393.5, $\mathrm{H_2O(l)}$ −285.8 kJ mol⁻¹.

Solution. $q = 200 \times 4.18 \times 23.0 = 1.92 \times 10^{4}\ \mathrm{J} \qquad n = \frac{0.920}{46.07} = 0.0200\ \mathrm{mol} \qquad \Delta_c H = -\frac{19.2\ \mathrm{kJ}}{0.0200\ \mathrm{mol}} = -963\ \mathrm{kJ\,mol^{-1}}$ Data value for $\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}$ : $\Delta_c H^\circ = [2(-393.5) + 3(-285.8)] - [-277.0] = -1367.4\ \mathrm{kJ\,mol^{-1}}$ The experiment captured only $\frac{963}{1367} \times 100 = 70\%$ of the energy. The rest heated the air, the can and the lamp; some ethanol evaporated without burning; and incomplete combustion (soot on the can) releases less energy.

Warning

Common mistakes. (1) Using the mass of the reactants instead of the mass of the water or solution in $q = mc\Delta T$ . (2) Forgetting the sign: a temperature rise means an exothermic reaction and a negative ΔH. (3) Dividing by the wrong amount: divide by the moles of the limiting reactant, or of water formed in neutralisation. (4) Mixing J and kJ: $mc\Delta T$ gives joules; ΔH is usually quoted in kJ mol⁻¹.

Safety. Ethanol is highly flammable: keep the stock bottle closed and away from the burner, refill the lamp only when it is out and cool, and stand it on a heat-proof mat. Sodium hydroxide and hydrochloric acid, even at 1 mol dm⁻³, harm the eyes: wear goggles, and rinse any splash with plenty of water.

Tip

"Mass of water, not of fuel." q = mcΔT uses the mass of what is heated. Then ΔH = −q ÷ n, with n the moles that reacted. A rise in temperature means ΔH is negative.

Malmesbury Education: Enthalpy Change of Neutralisation, an A-level Practical (11 min). Watch how the temperature is measured as acid is added to sodium hydroxide, and how the results are processed.

✦

Practice

Define standard enthalpy of combustion and state the standard conditions.
Explain, in terms of bonds, why combustion reactions are exothermic.
25.0 cm³ of 2.00 mol dm⁻³ HNO₃ is mixed with 25.0 cm³ of 2.00 mol dm⁻³ KOH. The temperature rises by 13.4 °C. Calculate the enthalpy of neutralisation.
When 2.00 g of ammonium nitrate (M = 80.05) dissolves in 50.0 g of water, the temperature falls by 2.9 °C. Calculate ΔH of solution in kJ mol⁻¹.
0.300 g of methanol (M = 32.05) heats 150 g of water by 9.5 °C. Calculate the experimental $\Delta_c H$ of methanol.
List three reasons why values of $\Delta_c H$ from a spirit-lamp experiment are less exothermic than data-book values.
Calculate $\Delta_c H^\circ$ of methane from enthalpies of formation: $\mathrm{CH_4}$ −74.8, $\mathrm{CO_2}$ −393.5, $\mathrm{H_2O(l)}$ −285.8 kJ mol⁻¹.
Sketch (in words) the enthalpy level diagram for the decomposition of calcium carbonate, ΔH = +178.3 kJ mol⁻¹, and say what the diagram tells you about the stability of $\mathrm{CaCO_3}$ .

Show answersHide answers

The enthalpy change when 1 mol of a substance burns completely in oxygen; 298 K, 100 kPa, solutions at 1 mol dm⁻³, substances in standard states.
The bonds formed in the products ( $\mathrm{C{=}O}$ in $\mathrm{CO_2}$ , O–H in water) release more energy than is needed to break the bonds in the fuel and in oxygen.
$q = 50.0 \times 4.18 \times 13.4 = 2801\ \mathrm{J}$ ; $n(\mathrm{H_2O}) = 2.00 \times 0.0250 = 0.0500\ \mathrm{mol}$ ; $\Delta H = -\frac{2.80}{0.0500} = -56.0\ \mathrm{kJ\,mol^{-1}}$ .
$q = 50.0 \times 4.18 \times 2.9 = 606\ \mathrm{J}$ taken in; $n = \frac{2.00}{80.05} = 0.0250\ \mathrm{mol}$ ; $\Delta H = +\frac{0.606}{0.0250} = +24\ \mathrm{kJ\,mol^{-1}}$ (endothermic: the temperature fell).
$q = 150 \times 4.18 \times 9.5 = 5957\ \mathrm{J}$ ; $n = \frac{0.300}{32.05} = 9.36 \times 10^{-3}\ \mathrm{mol}$ ; $\Delta_c H = -\frac{5.96}{9.36 \times 10^{-3}} = -636\ \mathrm{kJ\,mol^{-1}}$ .
Heat lost to the air, the can and the lamp; incomplete combustion; evaporation of the fuel; the heat capacity of the can ignored.
$[-393.5 + 2(-285.8)] - [-74.8] = -890.3\ \mathrm{kJ\,mol^{-1}}$ .
$\mathrm{CaCO_3}$ on a low level, $\mathrm{CaO + CO_2}$ 178.3 kJ higher, with an upward arrow. The products are at higher enthalpy, so $\mathrm{CaCO_3}$ is energetically stable relative to them and decomposes only when heated strongly.

Quick check

Key points

ΔH is the heat transferred at constant pressure per mole of reaction; negative = exothermic, positive = endothermic. Standard conditions: 298 K, 100 kPa, 1 mol dm⁻³.
Bond breaking takes in energy, bond making releases it; ΔH is the balance between them.
Calorimetry: $q = mc\Delta T$ (m = mass of water or solution), then $\Delta H = -\frac{q}{n}$ .
Neutralisation of strong acid by strong base: about −57 kJ mol⁻¹. Ethanol: $\Delta_c H^\circ = -1367\ \mathrm{kJ\,mol^{-1}}$ ; simple experiments find less because heat is lost.

Sources and credits

Image: Aluminium Calorimeter.jpg by Maciej J. Mrowinski, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Enthalpies of formation from Standard enthalpy of formation; neutralisation value from Enthalpy of neutralization; Wikipedia contributors, licensed under CC BY-SA 4.0.
Video: Enthalpy Change of Neutralisation - Chemistry A-level Practical, Malmesbury Education, embedded from YouTube.
The text, diagrams, worked examples and Practice items were written and drawn by Ganzaa. The experimental results in the worked examples are illustrative.
This lesson uses CC BY-SA material and is shared under CC BY-SA 4.0.

1How much heat is in a spoonful of fuel?

2Where the energy comes from: bonds

3Measuring ΔH by calorimetry

How much heat is in a spoonful of fuel?

Where the energy comes from: bonds

Measuring ΔH by calorimetry