Edexcel International A Level Chemistry: Energetics (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers Topic 6: Energetics of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 6.1 to 6.11 including Core Practical 2. Topic 6 is IAS content in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. Learn it first from the energetics study guide and the energetics revision notes. Track your progress on the printable checklist, find more on the course hub, and check your gaps with the free diagnostics.

Data for this set: Ar values H = 1.0, C = 12.0, O = 16.0, Ca = 40.1, Fe = 55.8. Specific heat capacity of water and dilute solutions = 4.18 J g⁻¹ °C⁻¹. Density of dilute solutions = 1.00 g cm⁻³.

Questions

1. This question is about standard enthalpy changes.

(a) State the standard conditions used for standard enthalpy changes. [2] (b) State the standard enthalpy change of formation of oxygen gas, O₂(g). [1]

2. Write equations, with state symbols, for the reactions whose enthalpy changes are:

(a) the standard enthalpy change of atomisation of iodine [1] (b) the standard enthalpy change of combustion of but-1-ene, C₄H₈(g) [1] (c) the standard enthalpy change of formation of sodium hydrogencarbonate, NaHCO₃(s). [1]

3. Potassium nitrate dissolves in water: KNO₃(s) + aq → K⁺(aq) + NO₃⁻(aq), ΔH = +35 kJ mol⁻¹. Describe the enthalpy level diagram for this change. [3]

4. 40.0 cm³ of 0.500 mol dm⁻³ copper(II) sulfate solution is placed in a polystyrene cup. Its temperature is recorded each minute. At 3.0 minutes, 2.0 g of iron powder is added and the mixture is stirred.

Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

Time / min	0.0	1.0	2.0	4.0	5.0	6.0	7.0	8.0	9.0
Temperature / °C	20.5	20.5	20.5	34.6	37.0	36.6	36.2	35.8	35.4

(a) The readings from 5.0 to 9.0 minutes lie on a straight line. Use them to find the corrected temperature at 3.0 minutes, and hence the corrected temperature change. [2] (b) Calculate the energy transferred to the solution, in J. [1] (c) Calculate the enthalpy change of the reaction, in kJ mol⁻¹. [2] (d) Show that the iron is in excess. [1] (e) Explain why the corrected temperature change is used rather than the highest reading. [1]

5. A spirit burner containing pentan-1-ol (Mr = 88.0) heats 250 g of water in a copper can. The burner loses 0.660 g and the water temperature rises by 20.0 °C.

(a) Calculate the enthalpy change of combustion of pentan-1-ol. [3] (b) Use the value −3330 kJ mol⁻¹ as the accepted standard enthalpy change of combustion. Give two reasons why your answer is less exothermic. [2] (c) Suggest one change to the apparatus that would improve the result. [1]

6. Calculate ΔrH for Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO(g). Use these values of ΔfH (kJ mol⁻¹): Fe₂O₃(s) −824, CO(g) −111. [3]

7. Calculate the standard enthalpy change of formation of benzene, C₆H₆(l), using these standard enthalpy changes of combustion (kJ mol⁻¹): H₂(g) −286, C₆H₆(l) −3268, C(s) −394. [4]

8. A student uses Core Practical 2 to find ΔH for CaCO₃(s) → CaO(s) + CO₂(g). In each run, a weighed solid is tipped into a polystyrene cup holding 50.0 cm³ of hydrochloric acid, concentration 2.00 mol dm⁻³ (excess).

Run A: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l). 2.50 g of CaCO₃ gives a temperature rise of 1.8 °C. Run B: CaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l). 1.40 g of CaO gives a temperature rise of 22.4 °C.

(a) Draw a Hess cycle linking the decomposition to runs A and B, and write an expression for ΔH of the decomposition. [2] (b) Calculate ΔH for run A, per mole of CaCO₃. [2] (c) Calculate ΔH for run B, per mole of CaO. [2] (d) Calculate ΔH for the decomposition of calcium carbonate. [1] (e) Each temperature reading is ±0.1 °C. Calculate the percentage uncertainty in the temperature change for run A, and suggest how to reduce it. [2]

9. Use these mean bond enthalpies (kJ mol⁻¹): O–H 463, C=O 804, C≡C 839, C–H 412, O=O 496.

C₂H₂(g) + 2½O₂(g) → 2CO₂(g) + H₂O(g)

(a) Calculate the enthalpy change for this reaction. [3] (b) The standard enthalpy change of combustion of ethyne is −1301 kJ mol⁻¹. Give two reasons for the difference. [2]

10. For N₂(g) + 3F₂(g) → 2NF₃(g), ΔH = −264 kJ mol⁻¹. Use N≡N 945 kJ mol⁻¹ and F–F 158 kJ mol⁻¹ to calculate the mean N–F bond enthalpy in NF₃. [3]

11. This question is about bond enthalpies and reactivity.

(a) Use these values (kJ mol⁻¹): C–F 484, C–Cl 338, C–Br 276. Explain why fluoroalkanes are hydrolysed far more slowly than bromoalkanes. [2] (b) The combustion of methane is very exothermic, yet methane and air can be mixed at room temperature without reacting. Explain this using C–H 413 kJ mol⁻¹ and O=O 498 kJ mol⁻¹. [2]

Answers

1. (a) A pressure of 100 kPa [1]; a stated temperature, usually 298 K [1]. (b) 0 kJ mol⁻¹, because it is an element in its standard state [1]. Examiner insight: Old values such as 101 kPa or 1 atm do not match this specification; quote 100 kPa.

2. (a) ½I₂(s) → I(g) [1] (b) C₄H₈(g) + 6O₂(g) → 4CO₂(g) + 4H₂O(l) [1] (c) Na(s) + ½H₂(g) + C(s) + 1½O₂(g) → NaHCO₃(s) [1] Examiner insight: Iodine is a solid in its standard state; writing I₂(g) or making two moles of atoms loses the mark.

3. Products, K⁺(aq) + NO₃⁻(aq), drawn above the reactants, KNO₃(s) + aq [1]. Arrow pointing upwards from reactants to products [1]. Arrow labelled ΔH = +35 kJ mol⁻¹, with both levels labelled with formulae and state symbols [1]. Examiner insight: A diagram with a hump is a reaction profile, not an enthalpy level diagram; keep to two levels and one arrow.

4. (a) The line falls 0.4 °C per minute, so at 3.0 min it reads 36.6 + 3 × 0.4 = 37.8 °C [1]. ΔT = 37.8 − 20.5 = 17.3 °C [1]. (b) Energy = 40.0 × 4.18 × 17.3 = 2893 J [1]. (c) n(CuSO₄) = 0.500 × 40.0 ÷ 1000 = 0.0200 mol [1]. ΔH = −2.893 ÷ 0.0200 = −145 kJ mol⁻¹ [1]. (d) n(Fe) = 2.0 ÷ 55.8 = 0.036 mol, more than the 0.0200 mol of CuSO₄ needed [1]. (e) Heat is lost while the reaction proceeds, so the highest reading is too low; extrapolating back to the time of mixing allows for this loss [1]. Examiner insight: Divide by moles of copper(II) sulfate, the limiting reagent; using moles of iron gives a value that is too small.

5. (a) Energy = 250 × 4.18 × 20.0 = 20 900 J [1]. n = 0.660 ÷ 88.0 = 0.00750 mol [1]. ΔcH = −20.9 ÷ 0.00750 = −2790 kJ mol⁻¹ [1]. (b) Any two: heat lost to the air and the can [1]; incomplete combustion [1]; evaporation of fuel from the wick; non-standard conditions. (c) Any one: a draught shield, a lid on the can, or a smaller gap between flame and can [1]. Examiner insight: “Heat loss” alone is often too vague for credit; say where the heat goes.

6. Products: 3(−111) = −333, with Fe(s) zero [1]. Reactants: −824, with C(s) zero [1]. ΔrH = −333 − (−824) = +491 kJ mol⁻¹ [1]. Examiner insight: Iron and carbon are elements in their standard states, so they add nothing to either total; giving them a value loses credit.

7. Cycle with combustion products at the bottom, or ΔfH = ΣΔcH(reactants) − ΔcH(benzene) [1]. 6(−394) + 3(−286) = −2364 − 858 = −3222 [1]. ΔfH = −3222 − (−3268) [1] = +46 kJ mol⁻¹ [1]. Examiner insight: A positive answer is correct here, so write the + sign; leaving it out can cost the final mark.

8. (a) Cycle: CaCO₃(s) → CaO(s) + CO₂(g) across the top; both sides react with 2HCl(aq) to give CaCl₂(aq) + CO₂(g) + H₂O(l) at the bottom [1]. ΔH = ΔH(A) − ΔH(B) [1]. (b) Energy = 50.0 × 4.18 × 1.8 = 376.2 J; n(CaCO₃) = 2.50 ÷ 100.1 = 0.02498 mol [1]. ΔH(A) = −0.3762 ÷ 0.02498 = −15.1 kJ mol⁻¹ [1]. (c) Energy = 50.0 × 4.18 × 22.4 = 4681.6 J; n(CaO) = 1.40 ÷ 56.1 = 0.02496 mol [1]. ΔH(B) = −4.6816 ÷ 0.02496 = −187.6 kJ mol⁻¹ [1]. (d) ΔH = −15.06 − (−187.6) = +173 kJ mol⁻¹ [1]. (e) 0.2 ÷ 1.8 × 100 = 11% [1]. Use more CaCO₃ (with acid still in excess) or less acid to get a larger temperature rise, or use a more precise thermometer [1]. Examiner insight: For ΔT, add the uncertainties of both readings; using ±0.1 °C alone halves the percentage uncertainty.

9. (a) Broken: 839 + 2(412) + 2.5(496) = 2903 kJ [1]. Made: 4(804) + 2(463) = 4142 kJ [1]. ΔH = 2903 − 4142 = −1239 kJ mol⁻¹ [1]. (b) The values are means taken from many different compounds, so they are not exact for ethyne, CO₂ or water [1]. Water is H₂O(g) here but H₂O(l) under standard conditions, so the enthalpy change of condensing is missing [1]. Examiner insight: Count the bonds in each CO₂ molecule (two C=O each); one C=O per CO₂ is a common slip.

10. Broken − made = ΔH: 945 + 3(158) − 6x = −264 [1]. 6x = 1419 + 264 = 1683 [1]. x = 281 kJ mol⁻¹ [1]. Examiner insight: There are six N–F bonds in two NF₃ molecules; dividing by three gives double the right answer.

11. (a) The C–halogen bond must break during hydrolysis [1]. C–F (484) is much stronger than C–Br (276), so it is far harder to break and the reaction is much slower [1]. (b) C–H and O=O bonds must break before new bonds form; their bond enthalpies are high [1]. So the energy needed to start the reaction is large and very few molecules have it at room temperature, so the rate is negligible [1]. Examiner insight: Link the bond enthalpy to rate; saying only “C–F is strong” without stating the effect on the reaction does not answer the question.

Where marks are usually lost

• Putting the fuel or solid mass into m × c × ΔT in place of the water or solution mass.
• Giving the energy in J as the final answer instead of ΔH in kJ mol⁻¹.
• Missing signs, especially + on endothermic or positive answers.
• Dividing by moles of the excess reagent.
• Reading ΔT from the highest temperature instead of extrapolating.
• Reversing the subtraction in combustion-data cycles.
• Forgetting coefficients when counting bonds or multiplying ΔfH values.
• Giving vague evaluation points such as “human error”.

Next steps

• Revise with the energetics revision notes.
• Reread any weak area in the energetics study guide.
• Browse the Edexcel International A Level Chemistry hub.
• Tick off outcomes on the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Topic 6: Energetics (outcomes 6.1 to 6.11, including Core Practical 2). Published by Pearson Education Limited.