R1.1 Exam Practice
Test your knowledge on Measuring Enthalpy Changes
Section B: Data Analysis (Paper 1B Style)
Calculator and Data Booklet permitted. Show all working clearly.
Question 1: Calorimetry Data Calculate
5 marksA student mixes 50.0 cm³ of 1.00 mol dm⁻³ HCl(aq) with 50.0 cm³ of 1.00 mol dm⁻³ NaOH(aq) in a polystyrene cup calorimeter.
| Measurement | Value |
|---|---|
| Volume of HCl | 50.0 cm³ |
| Volume of NaOH | 50.0 cm³ |
| Initial temperature | 21.0 °C |
| Final temperature | 27.8 °C |
| Specific heat capacity | 4.18 J g⁻¹ K⁻¹ |
| Density of solution | 1.00 g cm⁻³ |
(a) Calculate the enthalpy change of neutralisation in kJ mol⁻¹. [2]
(b) Explain why the experimental value differs from the accepted literature value of −57.1 kJ mol⁻¹. [2]
(c) Suggest one improvement to increase accuracy. [1]
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(a)
- q = mcΔ T = 100.0 × 4.18 × 6.8 = 2842 J [1]
- n = 0.0500 × 1.00 = 0.0500 mol; Δ H = -2842/0.0500 = -56.8 kJ mol⁻¹ [1]
(b) Heat loss to the surroundings / polystyrene cup is not a perfect insulator [1]; assumption that specific heat capacity and density equal pure water [1]
(c) Add a lid to the calorimeter / use a vacuum flask / use graphical extrapolation [1]
Question 2: Bond Enthalpy Calculation Calculate
4 marksMethane undergoes complete combustion: CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(g)
| Bond | Bond enthalpy / kJ mol⁻¹ |
|---|---|
| C–H | 414 |
| O=O | 498 |
| C=O | 804 |
| O–H | 463 |
(a) Calculate the enthalpy change of combustion using the bond enthalpies above. [2]
(b) Explain why bond enthalpy values are described as "averages". [1]
(c) Deduce whether this reaction is exothermic or endothermic. [1]
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(a)
- Bonds broken: (4 × 414) + (2 × 498) = +2652 kJ [1]
- Bonds formed: (2 × 804) + (4 × 463) = -3460 kJ; Δ H = 2652 - 3460 = -808 kJ mol⁻¹ [1]
(b) Bond enthalpies are averages taken across many different compounds; the actual energy depends on the molecular environment [1]
(c) Exothermic: ΔH is negative / more energy released forming bonds than absorbed breaking them [1]
Section C: Structured Questions (Paper 2 Style)
Show all working. State answers with appropriate significant figures and units.
Question 3: Standard Enthalpy & Hess's Law Define
4 marks(a) Define the term standard enthalpy of combustion. [1]
(b) State the standard conditions for thermodynamic measurements. [1]
(c) Calculate the standard enthalpy of combustion of methane using Hess's Law and the following Δ Hf^\ominus data: CH₄(g) = −74.8, CO₂(g) = −393.5, H₂O(l) = −285.8 kJ mol⁻¹. [2]
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(a) The enthalpy change when one mole of a substance undergoes complete combustion in excess O₂ under standard conditions [1]
(b) 100 kPa pressure; all substances in their standard states [1]
(c)
- Δ H = \SigmaΔ Hf^\ominus(products) - \SigmaΔ Hf^\ominus(reactants) [1]
- = [(-393.5) + 2(-285.8)] - [(-74.8) + 0] = -965.1 + 74.8 = -890.3 kJ mol⁻¹ [1]
Question 4: Energy Profile Diagrams Explain
4 marks(a) State the meaning of the term activation energy (Ea). [1]
(b) Explain the difference between exothermic and endothermic reactions using energy profile diagrams. [2]
(c) Suggest the effect of a catalyst on an energy profile diagram. [1]
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(a) The minimum energy required for a reaction to occur / energy difference between reactants and transition state [1]
(b)
- Exothermic: products lower in energy than reactants (ΔH < 0) [1]
- Endothermic: products higher in energy than reactants (ΔH > 0) [1]
(c) A catalyst lowers the activation energy / provides an alternative pathway with a lower Ea; ΔH remains unchanged [1]
Question 5: Dissolution Calorimetry Determine
5 marksA student dissolves 0.0500 mol of ammonium nitrate in 100.0 cm³ of water. The temperature decreases by 4.5 °C. Assume c = 4.18 J g⁻¹ K⁻¹ and density = 1.00 g cm⁻³.
(a) Calculate the heat transferred (q) in joules. [1]
(b) Determine the molar enthalpy change (ΔH) in kJ mol⁻¹, including the correct sign. [2]
(c) Explain two sources of error in this experiment. [2]
Show Mark Scheme
(a) q = 100.0 × 4.18 × 4.5 = 1881 J [1]
(b)
- Temperature decreases → endothermic → ΔH is positive [1]
- Δ H = +1881/0.0500 = +37620 J mol⁻¹ = +37.6 kJ mol⁻¹ [1]
(c)
- Heat is absorbed from the surroundings / container is not a perfect insulator [1]
- Assuming specific heat capacity and density of the solution equal that of pure water [1]
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