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How do we measure and calculate the energy changes that accompany chemical reactions?

Enthalpy changes, exothermic and endothermic reactions, standard enthalpy changes of reaction, formation and combustion, calorimetry, Hess's law and mean bond enthalpies.

An Eduqas A-Level Chemistry C2.2 answer on enthalpy changes, standard enthalpies of reaction, formation and combustion, calorimetry, Hess's law and mean bond enthalpies.

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  1. What this topic is asking
  2. Enthalpy changes and their sign
  3. Standard enthalpy changes
  4. Calorimetry
  5. Hess's law
  6. Mean bond enthalpies
  7. Examples in context
  8. Try this

What this topic is asking

Eduqas topic C2.2 covers enthalpy changes and their sign, the standard enthalpies of reaction, formation and combustion, measuring enthalpy changes by calorimetry, using Hess's law to find enthalpy changes indirectly, and estimating enthalpy changes from mean bond enthalpies. The more advanced lattice and Born-Haber treatment comes later in the Physical and Inorganic section.

Enthalpy changes and their sign

Standard enthalpy changes

Eduqas defines several standard enthalpy changes precisely:

Calorimetry

Calorimetry measures the heat released or absorbed using the temperature change of a known mass of water (or solution):

Dividing qq (in kJ\text{kJ}) by the moles reacted gives the molar enthalpy change. Experimental combustion values are always less exothermic than data-book values because of heat loss to the surroundings and incomplete combustion.

Hess's law

Mean bond enthalpies

A mean bond enthalpy is the average energy needed to break one mole of a particular bond, averaged over many compounds. Breaking bonds is endothermic and making bonds is exothermic, so ΔH≈(bonds broken)−(bonds made)\Delta H \approx \text{(bonds broken)} - \text{(bonds made)}. The values are averages, so bond-enthalpy answers are only estimates.

Examples in context

Example 1. Self-heating cans. A self-heating coffee can mixes water with quicklime, an exothermic hydration (CaO+H2O→Ca(OH)2\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2); the released enthalpy warms the drink, a direct use of an exothermic ΔH\Delta H.

Example 2. Comparing fuels. Enthalpies of combustion let engineers compare fuels per gram or per mole; the calorimetry experiment is the practical that introduces this, and Hess cycles reconcile the lower measured values with the true (data-book) values.

Try this

Q1. State whether bond breaking is exothermic or endothermic, and explain why. [1 mark]

  • Cue. Endothermic: energy must be supplied to overcome the attraction holding the bonded atoms together.

Q2. Calculate the heat released when a reaction raises the temperature of 100 g100\ \text{g} of water by 8.0 ∘C8.0\ ^{\circ}\text{C}. (c=4.18 J g−1K−1c = 4.18\ \text{J g}^{-1}\text{K}^{-1}.) [2 marks]

  • Cue. q=mcΔT=100×4.18×8.0=3344 J=3.34 kJq = mc\Delta T = 100 \times 4.18 \times 8.0 = 3344\ \text{J} = 3.34\ \text{kJ}.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20184 marksWhen 1.00 g1.00\ \text{g} of ethanol was burned, it raised the temperature of 200 g200\ \text{g} of water by 25.0 ∘C25.0\ ^{\circ}\text{C}. (a) Calculate the heat released. (b) Calculate the enthalpy of combustion of ethanol in kJ mol−1\text{kJ mol}^{-1}. (c=4.18 J g−1K−1c = 4.18\ \text{J g}^{-1}\text{K}^{-1}, MrM_r of ethanol =46.0= 46.0.)
Show worked answer →

(a) q=mcΔT=200×4.18×25.0=20900 J=20.9 kJq = mc\Delta T = 200 \times 4.18 \times 25.0 = 20900\ \text{J} = 20.9\ \text{kJ} (2).

(b) Moles of ethanol =1.0046.0=0.0217 mol= \dfrac{1.00}{46.0} = 0.0217\ \text{mol}. Enthalpy of combustion =−20.90.0217=−962 kJ mol−1= -\dfrac{20.9}{0.0217} = -962\ \text{kJ mol}^{-1} (2).

Markers reward q=mcΔTq = mc\Delta T, the moles of ethanol and the per-mole value with a negative sign (exothermic). The data-book value is more exothermic because of heat losses.

Eduqas 20214 marksUse the mean bond enthalpies below to calculate the enthalpy change for H2(g)+Cl2(g)→2HCl(g)\text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{HCl}(\text{g}). Mean bond enthalpies in kJ mol−1\text{kJ mol}^{-1}: H-H=436\text{H-H} = 436, Cl-Cl=242\text{Cl-Cl} = 242, H-Cl=431\text{H-Cl} = 431.
Show worked answer →

Bonds broken (reactants) =436+242=678 kJ mol−1= 436 + 242 = 678\ \text{kJ mol}^{-1} (1).

Bonds made (products) =2×431=862 kJ mol−1= 2 \times 431 = 862\ \text{kJ mol}^{-1} (1).

ΔH=bonds broken−bonds made=678−862=−184 kJ mol−1\Delta H = \text{bonds broken} - \text{bonds made} = 678 - 862 = -184\ \text{kJ mol}^{-1} (2).

Markers reward bonds broken, bonds made and the subtraction giving an exothermic value.

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