4.1.4Organic Chemistry
Determining Enthalpy Change of Reaction — Edexcel AS Chemistry Revision Notes & Topical Questions
5 min read 6 parts Lesson 36 of 42
Part 1 of 6Calorimetry Basics
Calorimetry measures the energy transferred as heat by the temperature change of a known mass of water or solution.
Calorimeter
- An insulated container, often a polystyrene cup with a lid, that limits heat exchange with the surroundings.
Specific heat capacity
- The energy needed to raise the temperature of 1 g of a substance by 1 K; for water it is 4.18 J g⁻¹ K⁻¹.
Energy equation
- Q = m × c × ΔT, where m is the mass of solution heated and ΔT is the temperature change.
Assumptions
- The solution is treated as water with density 1.00 g cm⁻³, and heat absorbed by the cup is ignored.
Part 2 of 6Experimental Method
Examplezinc powder displacing copper from copper(II) sulfate solution, Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s).
1
Measure 25.0 cm³ of 1.00 mol dm⁻³ copper(II) sulfate solution into a polystyrene cup and place a thermometer in it.
2
Record the temperature every 30 seconds for a few minutes while stirring, to establish the starting trend.
3
At a fixed time add an excess of zinc powder and stir continuously without recording that moment's temperature.
4
Keep recording the temperature at regular intervals as it rises to a maximum and then slowly falls.
5
Plot temperature against time for all readings, ready for extrapolation to find the true temperature rise.
Part 3 of 6Correcting for Heat Loss
Correcting for Heat Loss
Cooling line
- Extend the downward-sloping line after the maximum back to the time the solid was added.
Starting line
- Extend the horizontal line of initial readings forward to the same time.
Corrected ΔT
- The vertical distance between the two extrapolated lines at the time of mixing is the corrected temperature change.
Why this works
- The reaction is not instant, so the solution begins cooling before it reaches its true maximum, and extrapolation removes that loss.
Part 4 of 6Calculating ΔH
Limiting reagent
- Moles are worked out from the reagent not in excess, here the copper(II) sulfate solution.
Sign of ΔH
- A temperature rise means the reaction is exothermic, so ΔH is negative.
Units
- Convert joules to kilojoules, then divide by moles to give kJ mol⁻¹.
Main errors
- Heat loss, heat absorbed by the cup and thermometer, and the assumption of water's properties all make the result smaller in magnitude.
Part 5 of 6Worked Example
Question
- 25.0 cm³ of 1.00 mol dm⁻³ copper(II) sulfate reacts with excess zinc and the corrected temperature rise is 45.0 K.
- Calculate the enthalpy change in kJ mol⁻¹.
- (c = 4.18 J g⁻¹ K⁻¹, density = 1.00 g cm⁻³)
- Mass of solution = 25.0 g, so Q = 25.0 × 4.18 × 45.0 = 4702.5 J.
- Moles of CuSO₄ = 1.00 × 0.0250 = 0.0250 mol.
- Energy per mole = 4702.5 ÷ 0.0250 = 188 100 J mol⁻¹ = 188.1 kJ mol⁻¹.
- The reaction is exothermic, so the sign is negative.
Answer ΔH = −188 kJ mol⁻¹
Part 6 of 6Exam Checklist & Tips
4.1.4 – Key Points for the Exam
- Enthalpy change The heat energy change measured at constant pressure.
- Calorimeter Insulated container used to measure heat transfer in a reaction.
- Specific heat capacity Energy needed to raise the temperature of 1 g of a substance by 1 K.
- Exothermic A reaction that releases heat to its surroundings, with a negative ΔH.
Common Mistakes
- Using the mass of zinc rather than the mass of solution in Q = mcΔT.
- Forgetting to divide by moles, so the answer is in kJ instead of kJ mol⁻¹.
- Leaving out the negative sign for an exothermic reaction.
- Dividing joules by moles but quoting the answer in kJ mol⁻¹ without converting.
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