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How do we measure and represent the energy change of a reaction?

Define enthalpy change, classify reactions as exothermic or endothermic, and use calorimetry data with energy profile diagrams.

Exothermic and endothermic reactions, enthalpy change, energy profile diagrams, and calorimetry calculations using q = mcdeltaT, with fully worked TASC-style examples.

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What this dot point is asking

TASC expects you to define enthalpy change, classify and represent reactions on energy profiles, and calculate molar enthalpy changes from calorimetry data.

Exothermic and endothermic reactions

Every reaction breaks bonds (absorbing energy) and makes bonds (releasing energy); the net result sets the energy change.

  • In an exothermic reaction, more energy is released forming new bonds than is absorbed breaking old ones, so heat flows out and the temperature rises. ΔH\Delta H is negative.
  • In an endothermic reaction, more energy is absorbed breaking bonds than is released, so heat flows in and the temperature falls. ΔH\Delta H is positive.

Energy profile diagrams

An energy profile plots enthalpy against reaction progress. Reactants and products sit at different enthalpy levels, and the activation energy EaE_a is the height of the peak above the reactants. For an exothermic reaction the products lie below the reactants (ΔH<0\Delta H < 0); for an endothermic reaction the products lie above (ΔH>0\Delta H > 0). The vertical gap between reactants and products is ΔH\Delta H.

Calorimetry

We measure enthalpy changes by capturing the heat in a known mass of water and recording the temperature change:

Standard enthalpy changes

Chemists define standard enthalpy changes so values can be compared fairly. The standard enthalpy of combustion (ΔHc\Delta H_c^\circ) is the heat released when one mole of a substance burns completely in oxygen under standard conditions. The standard enthalpy of formation (ΔHf\Delta H_f^\circ) is the heat change when one mole of a compound forms from its elements in their standard states; elements in their standard state are defined as zero. The standard enthalpy of neutralisation is the heat released when an acid and base react to form one mole of water, and for strong acid and strong base it is almost constant (about 57 kJ mol1-57\ \text{kJ mol}^{-1}) because the reaction is essentially H++OHH2O\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} in every case.

Sources of experimental error in calorimetry

A school calorimetry result is almost always less exothermic than the data-book value, for predictable reasons: heat is lost to the surroundings and to the apparatus rather than all going into the water; combustion may be incomplete, leaving soot and producing less energy; and some fuel evaporates without burning. Recognising these systematic errors, and stating that they make the measured ΔH\Delta H less negative, is frequently worth a mark in TASC questions.

In the exam, show q=mcΔTq = mc\Delta T with the correct mass and specific heat, divide by moles for a molar value, attach the sign that matches an exothermic or endothermic process, and explain experimental losses when comparing with a data-book value.

Exam-style practice questions

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

TCE 20224 marksA class determined the enthalpy of combustion of butan-1-ol by calorimetry. Mass of water heated =100.0 g= 100.0\ \text{g}; initial temperature 25.0 C25.0\ ^\circ\text{C}; final temperature 31.5 C31.5\ ^\circ\text{C}; mass of butan-1-ol burned =0.152 g= 0.152\ \text{g}; M(butan-1-ol)=74.12 g mol1M(\text{butan-1-ol}) = 74.12\ \text{g mol}^{-1}; c(water)=4.18 J g1C1c(\text{water}) = 4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}. Calculate the experimental molar heat of combustion in kJ mol1\text{kJ mol}^{-1}.
Show worked answer →

Heat absorbed by the water: q=mcΔT=100.0×4.18×(31.525.0)=2717 J=2.717 kJq = mc\Delta T = 100.0 \times 4.18 \times (31.5 - 25.0) = 2717\ \text{J} = 2.717\ \text{kJ}. (2 marks)

Moles of butan-1-ol burned: n=0.152/74.12=2.05×103 moln = 0.152 / 74.12 = 2.05 \times 10^{-3}\ \text{mol}. (1 mark)

Molar heat of combustion: ΔH=q/n=2.717/(2.05×103)=1.33×103 kJ mol1\Delta H = -q/n = -2.717 / (2.05 \times 10^{-3}) = -1.33 \times 10^{3}\ \text{kJ mol}^{-1}. The negative sign confirms combustion is exothermic; the value is well below the data-book figure (about 2676 kJ mol1-2676\ \text{kJ mol}^{-1}) because of heat loss to the surroundings and incomplete combustion. (1 mark)

TCE 20213 marksA student measured the heat of combustion of ethanol with a spirit burner. Mass of water heated =200.0 g= 200.0\ \text{g}; mass of ethanol burned =1.75 g= 1.75\ \text{g}; initial water temperature =20.2 C= 20.2\ ^\circ\text{C}; final =73.5 C= 73.5\ ^\circ\text{C}; M(ethanol)=46.07 g mol1M(\text{ethanol}) = 46.07\ \text{g mol}^{-1}. Calculate the experimental molar heat of combustion and explain why it differs from the data-book value of 1367 kJ mol1-1367\ \text{kJ mol}^{-1}.
Show worked answer →

Heat gained by the water: q=mcΔT=200.0×4.18×(73.520.2)=4.456×104 J=44.56 kJq = mc\Delta T = 200.0 \times 4.18 \times (73.5 - 20.2) = 4.456 \times 10^{4}\ \text{J} = 44.56\ \text{kJ}. (1 mark)

Moles of ethanol: n=1.75/46.07=0.03799 moln = 1.75 / 46.07 = 0.03799\ \text{mol}. (1 mark)

ΔH=q/n=44.56/0.03799=1173 kJ mol1\Delta H = -q/n = -44.56 / 0.03799 = -1173\ \text{kJ mol}^{-1}. It is less negative than 1367 kJ mol1-1367\ \text{kJ mol}^{-1} because heat is lost to the surroundings and apparatus and combustion may be incomplete. (1 mark)

TCE 20235 marksCaCl2(s)Ca2+(aq)+2Cl(aq)\text{CaCl}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq), ΔH=+82.8 kJ mol1\Delta H = +82.8\ \text{kJ mol}^{-1}, and KOH(s)K+(aq)+OH(aq)\text{KOH}(s) \rightarrow \text{K}^+(aq) + \text{OH}^-(aq), ΔH=56.6 kJ mol1\Delta H = -56.6\ \text{kJ mol}^{-1}. One mole of each is dissolved in separate, equal volumes of water. (a) Compare the temperature change of the water in each beaker. (b) Explain the difference in terms of lattice and hydration energy.
Show worked answer →

(a) Dissolving CaCl2\text{CaCl}_2 is endothermic (ΔH>0\Delta H > 0), so that water cools. Dissolving KOH\text{KOH} is exothermic (ΔH<0\Delta H < 0), so that water warms. (1 mark)

(b) When an ionic solid dissolves, energy is absorbed to break the lattice and is released when the ions are hydrated (ion-dipole attractions form). The sign of ΔH\Delta H is the net of these. For CaCl2\text{CaCl}_2 the lattice energy needed exceeds the hydration energy released, so the process is endothermic and absorbs heat from the water. For KOH\text{KOH} the hydration energy released exceeds the lattice energy needed, so it is exothermic and releases heat to the water. (4 marks)

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