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3.1.2Unit 3. Practical Skills in Chemistry I

Determining Enthalpy Change of Reaction — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 4 of 10 Unit quiz

📚 IAL Edexcel Linear (Chemistry) Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Section 1 of 6Calorimetry

Calorimetry measures the energy transferred to or from a solution during a reaction.

Apparatus
  • Apparatus A polystyrene cup with a lid is a cheap, well-insulated calorimeter that keeps heat losses small.
Equation
  • Equation The energy transferred is q = mcΔT, where m is the mass of solution, c the specific heat capacity and ΔT the temperature change.
Value of c
  • Value of c The specific heat capacity of water, 4.18 J g⁻¹ K⁻¹, is used for the solution as an approximation.
Sign of ΔH
  • Sign of ΔH A temperature rise shows an exothermic reaction, so ΔH has a negative sign;
    • a fall gives a positive ΔH.

Section 2 of 6Method and Assumptions

Examplezinc powder displacing copper from copper(II) sulfate solution.

1
Use a measuring cylinder to place 25 cm³ of copper(II) sulfate solution into the polystyrene cup.
2
Record the temperature every 30 seconds for about 2.5 minutes while stirring, to establish the starting trend.
3
At exactly three minutes add an excess of zinc powder, without recording a temperature at that moment.
4
Continue stirring and recording the temperature each half-minute for a further six minutes.
5
Assume the solution has the density and heat capacity of water, there is no heat loss, and the reaction goes to completion.

Section 3 of 6Temperature Correction Graph

Slow reactions lose heat before the highest temperature is reached.

Temperature Correction Graph
Plot
  • Plot Plot temperature against time for the whole experiment, including the readings before the reactant was added.
Extrapolate
  • Extrapolate Extend the cooling part of the line back to the time when the reactant was added.
Find ΔT
  • Find ΔT Measure the vertical distance between the extrapolated cooling line and the line of the starting temperature at that time.
Benefit
  • Benefit This estimates the temperature rise that would have occurred if the reaction were instantaneous with no cooling.

Section 4 of 6Calculating ΔH

Step 1
  • Step 1 Calculate q = mcΔT in joules, using the mass of the solution and the corrected temperature change.
Step 2
  • Step 2 Convert q from joules to kilojoules by dividing by 1000.
Step 3
  • Step 3 Find the moles of the limiting reagent, which is the substance that is not in excess.
Step 4
  • Step 4 Divide q by the moles and add the sign: ΔH = –q ÷ n for a temperature rise, in kJ mol⁻¹.

Section 5 of 6Worked Example

Question

  • 25.0 cm³ of 0.500 mol dm⁻³ copper(II) sulfate solution reacts with excess zinc powder.
  • The corrected temperature rise is 26.0 K.
  • Calculate the enthalpy change for the reaction per mole of copper(II) sulfate (assume 1 g cm⁻³ and c = 4.18 J g⁻¹ K⁻¹).

  1. m = 25.0 g, so q = 25.0 × 4.18 × 26.0 = 2717 J = 2.717 kJ.
  2. n(CuSO₄) = 0.500 × 25.0 ÷ 1000 = 0.0125 mol, which is the limiting reagent because zinc is in excess.
  3. ΔH = 2.717 ÷ 0.0125 = 217 kJ mol⁻¹, and the temperature rose, so the sign is negative.

Answer ΔH = –217 kJ mol⁻¹ (3 significant figures)

Section 6 of 6Exam Checklist & Tips

3.1.2 – Key Points for the Exam

  • Calorimeter An insulated container used to measure heat energy transferred in a reaction.
  • Specific heat capacity The energy needed to raise the temperature of 1 g of a substance by 1 K.
  • Limiting reagent The reactant that is completely used up and so determines the amount of product.
  • Extrapolation Extending a line on a graph beyond the measured data to estimate a value.

Common Mistakes

  • Using the mass of the solid added in q = mcΔT instead of the mass of the solution being heated.
  • Leaving q in joules while using ΔH in kJ mol⁻¹, which makes the answer 1000 times too large.
  • Using the moles of the excess reagent (zinc) rather than those of the limiting reagent.
  • Forgetting the negative sign for an exothermic reaction.

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