IGCSE Chemistry Paper 4 (Theory / Extended) Challenge

Topic 5: Chemical Energetics -- Challenge Paper

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions -- Challenge Paper

This paper is designed to stretch A* candidates with complex multi-step calculations.
Answer all questions in the spaces provided.
Show all working for calculations -- method marks are available even if the final answer is incorrect.
Use appropriate scientific terminology throughout.

Question Navigation

Question 1 -- Multi-Step Bond Energy: Combustion of Butane

Total: 12 marks

Butane (C₄H₁₀) is used as lighter fuel and in portable camping stoves across the Lake District. A student calculates the energy released during its combustion.

(a) [3]

The structural formula of butane is CH₃CH₂CH₂CH₃.

(i) State the number of C–H bonds in one molecule of butane. [1]

(ii) State the number of C–C bonds in one molecule of butane. [1]

(iii) Write the balanced equation for the complete combustion of butane. [1]

Model Answer -- 1(a)

10 C–H bonds [1]

3 C–C bonds [1]

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(g) [1]

⚠ If you missed marks here: If you counted 8 C–H bonds you forgot the two end CH₃ groups have 3 hydrogens each (3+2+2+3 = 10); if you wrote 4 C–C bonds, remember 4 carbons in a chain are joined by only 3 bonds. For the equation, the classic slip is unbalanced oxygen – either write 2C₄H₁₀ + 13O₂ or use 6½O₂ for one mole; check C, H and O atoms all balance before moving on.

1 mark -- ten C-H bonds 1 mark -- three C-C bonds 1 mark -- balanced equation butane combustion

(b) [6]

Using the equation for one mole of butane:

C₄H₁₀(g) + 6½O₂(g) → 4CO₂(g) + 5H₂O(g)

Use the bond energy data to calculate ΔH.

Bond	Bond energy / kJ mol⁻¹
C–H	413
C–C	347
O=O	498
C=O	805
O–H	464

Show all working clearly.

Model Answer -- 1(b)

Bonds broken: 10(C–H) + 3(C–C) + 6.5(O=O) [1]

= (10×413) + (3×347) + (6.5×498) = 4130 + 1041 + 3237 = 8408 kJ [1]

Bonds formed: 8(C=O) + 10(O–H) [1]

= (8×805) + (10×464) = 6440 + 4640 = 11080 kJ [1]

ΔH = 8408 − 11080 = −2672 kJ mol⁻¹ [1]

The large negative value confirms this is a highly exothermic reaction [1]

⚠ If you missed marks here: If you got +2672 you did formed − broken instead of ΔH = bonds broken − bonds formed – the sign matters and loses the mark. Other classic slips: forgetting the 6.5 O=O bonds (your broken total comes out 3237 too low), or using 4 C=O and 5 O–H instead of 8 and 10 – each CO₂ has TWO C=O bonds and each H₂O has TWO O–H bonds. Method marks are available, so always show the 8408 and 11080 subtotals.

1 mark -- correct bonds broken identified ten C-H three C-C six-and-half O=O 1 mark -- bonds broken total 8408 1 mark -- correct bonds formed eight C=O ten O-H 1 mark -- bonds formed total 11080 1 mark -- answer negative 2672 1 mark -- highly exothermic confirmed

The enthalpy changes for the combustion of three alkanes are shown below.

Alkane	ΔH / kJ mol⁻¹	M_r
Methane (CH₄)	−818	16
Propane (C₃H₈)	−2054	44
Butane (C₄H₁₀)	−2672	58

(i) Describe the trend in ΔH as the number of carbon atoms increases. [1]

(ii) Calculate the energy released per gram for methane and butane. State which is the better fuel per gram. [2]

Model Answer -- 1(c)

As the number of carbon atoms increases, the magnitude of ΔH increases / more energy is released per mole [1]

Methane: 818 / 16 = 51.1 kJ g⁻¹; Butane: 2672 / 58 = 46.1 kJ g⁻¹ [1]

Methane releases more energy per gram (51.1 vs 46.1 kJ g⁻¹), so methane is the better fuel per gram [1]

⚠ If you missed marks here: For the trend, saying "ΔH increases" alone is ambiguous because the numbers get MORE negative – say the magnitude increases or more energy is released per mole. If you picked butane as the better fuel, you compared energy per MOLE (−2672 vs −818) instead of dividing by M_r to get energy per GRAM – 818 ÷ 16 = 51.1 beats 2672 ÷ 58 = 46.1.

1 mark -- more carbon atoms magnitude delta increases more energy released per mole 1 mark -- methane 51.1 per gram butane 46.1 per gram calculated 1 mark -- methane better fuel per gram releases more energy

Question 2 -- Bond Energy Calculation: Hydrogenation of Ethene

Total: 12 marks

The hydrogenation of ethene is used in the food industry to convert vegetable oils into margarine. A student in Ahmedabad investigates the energy change for this reaction.

(a) [4]

The equation for the hydrogenation of ethene is:

CH₂=CH₂(g) + H₂(g) → CH₃CH₃(g)

(i) List the bonds broken in the reactants. [2]

(ii) List the bonds formed in the products. [2]

Model Answer -- 2(a)

Bonds broken in ethene: 4 × C–H + 1 × C=C [1]

Bonds broken in hydrogen: 1 × H–H [1]

Bonds formed in ethane: 6 × C–H [1]

Plus 1 × C–C (the double bond becomes a single bond, and two new C–H bonds form) [1]

⚠ If you missed marks here: At IGCSE you treat ALL bonds as broken and ALL bonds as re-formed – students lose marks by only listing the bonds that actually change (1 C=C, 1 H–H broken; 1 C–C, 2 C–H formed) and leaving out the 4 C–H in ethene and the full 6 C–H in ethane. The other common miss is forgetting the H–H bond in hydrogen entirely, or forgetting that the C=C becomes a C–C single bond in ethane.

1 mark -- bonds broken ethene four C-H one C=C 1 mark -- bonds broken hydrogen one H-H 1 mark -- bonds formed ethane six C-H 1 mark -- bonds formed one C-C single bond

(b) [5]

Use the bond energy data to calculate ΔH for the hydrogenation of ethene.

Bond	Bond energy / kJ mol⁻¹
C–H	413
C=C	614
C–C	347
H–H	436

Show all working.

Model Answer -- 2(b)

Bonds broken: 4(C–H) + 1(C=C) + 1(H–H) = (4×413) + 614 + 436 = 1652 + 614 + 436 = 2702 kJ [1]

Bonds formed: 6(C–H) + 1(C–C) = (6×413) + 347 = 2478 + 347 = 2825 kJ [1]

ΔH = bonds broken − bonds formed [1]

ΔH = 2702 − 2825 = −123 kJ mol⁻¹ [1]

The reaction is exothermic [1]

⚠ If you missed marks here: An answer of +123 means you flipped the formula – it is always bonds broken (2702) minus bonds formed (2825), and breaking needs energy IN while forming gives energy OUT. If your broken total was 2266 you forgot the H–H bond (436); if your formed total was wrong, check you used C–C = 347 for the new single bond, not the C=C value of 614. A negative ΔH means exothermic – state that explicitly for the final mark.

1 mark -- bonds broken 2702 1 mark -- bonds formed 2825 1 mark -- correct formula bonds broken minus formed 1 mark -- answer negative 123 1 mark -- reaction exothermic

The hydrogenation reaction uses a nickel catalyst at 150 °C.

(i) On an energy level diagram for this reaction, sketch two curves: one without the catalyst and one with the catalyst. [2]

(ii) Explain why a catalyst is needed even though the reaction is exothermic. [1]

Model Answer -- 2(c)

Reactants at higher energy, products at lower energy (ΔH = −123 kJ) with two humps: taller for uncatalysed, shorter for catalysed [1]

ΔH the same for both curves / only the activation energy peak differs [1]

The activation energy for breaking the C=C and H–H bonds is high / the catalyst provides an alternative pathway with lower Eₐ, allowing the reaction to proceed at a practical rate [1]

⚠ If you missed marks here: Diagram marks are usually lost by drawing the catalysed curve ending at a different product level – both curves must start and finish at the SAME energies, with ΔH unchanged; only the peak height differs. For (ii), "the catalyst speeds it up" scores nothing – you must say it provides an alternative pathway with a lower activation energy; exothermic only tells you about the overall energy change, not how hard it is to get the reaction started.

1 mark -- reactants higher products lower two humps taller uncatalysed shorter catalysed 1 mark -- delta same both curves activation energy peak differs 1 mark -- activation energy high catalyst alternative pathway lower practical rate

Question 3 -- Multi-Step Bond Energy: Combustion of Glucose

Total: 12 marks

Glucose (C₆H₁₂O₆) is the fuel used in respiration. A biochemist in Oxford calculates the enthalpy change for the combustion of glucose using bond energies.

(a) [3]

Glucose has the simplified structural formula showing these bonds: 5 × C–C, 7 × C–H, 5 × C–O, 5 × O–H, and 1 × C=O.

The equation for the combustion of glucose is:

C₆H₁₂O₆(g) + 6O₂(g) → 6CO₂(g) + 6H₂O(g)

(i) List all bonds broken in the reactants (glucose + oxygen). [2]

(ii) List all bonds formed in the products. [1]

Model Answer -- 3(a)

Bonds broken in glucose: 5(C–C) + 7(C–H) + 5(C–O) + 5(O–H) + 1(C=O) [1]

Bonds broken in oxygen: 6(O=O) [1]

Bonds formed in products: 12(C=O) in 6CO₂ + 12(O–H) in 6H₂O [1]

⚠ If you missed marks here: The most common miss is forgetting the 6 O=O bonds in oxygen – O₂ is a reactant too, and its bonds must be broken. On the products side, if you wrote 6 C=O and 6 O–H you counted molecules instead of bonds: each CO₂ contains 2 C=O bonds and each H₂O contains 2 O–H bonds, giving 12 of each.

1 mark -- glucose bonds five C-C seven C-H five C-O five O-H one C=O 1 mark -- oxygen six O=O 1 mark -- products twelve C=O twelve O-H

(b) [6]

Calculate the enthalpy change for the combustion of glucose using the bond energy data.

Bond	Bond energy / kJ mol⁻¹
C–C	347
C–H	413
C–O	358
O–H	464
C=O	805
O=O	498

Show all working.

Model Answer -- 3(b)

Bonds broken in glucose: 5(347) + 7(413) + 5(358) + 5(464) + 1(805) = 1735 + 2891 + 1790 + 2320 + 805 = 9541 kJ [1]

Bonds broken in O₂: 6(498) = 2988 kJ [1]

Total bonds broken: 9541 + 2988 = 12529 kJ [1]

Bonds formed: 12(C=O) + 12(O–H) = 12(805) + 12(464) = 9660 + 5568 = 15228 kJ [1]

ΔH = 12529 − 15228 = −2699 kJ mol⁻¹ [1]

This confirms glucose combustion / respiration is highly exothermic [1]

⚠ If you missed marks here: With this many terms the usual losses are: leaving out the 6(O=O) = 2988 kJ (broken total comes out at 9541 instead of 12529), using 6 C=O and 6 O–H instead of 12 each (formed total halves to 7614), or an arithmetic slip in the long 5(347) + 7(413) + 5(358) + 5(464) + 805 sum. If you got +2699 you subtracted the wrong way round – ΔH = broken − formed. Write each subtotal on its own line so method marks survive a slip.

1 mark -- glucose bonds broken calculated 9541 1 mark -- oxygen bonds broken 2988 1 mark -- total bonds broken 12529 1 mark -- bonds formed 15228 1 mark -- answer negative 2699 1 mark -- glucose combustion respiration highly exothermic

The experimentally measured value of ΔH for glucose combustion is −2803 kJ mol⁻¹.

(i) Calculate the percentage difference between your calculated value and the experimental value. [2]

(ii) Suggest why there is a difference between the two values. [1]

Model Answer -- 3(c)

Difference = |−2699 − (−2803)| = 104 kJ [1]

Percentage difference = (104 / 2803) × 100 = 3.7% [1]

Bond energies are average values taken from many different compounds; the actual bond energies in glucose may differ from these averages / glucose has a complex ring structure not fully represented by simple bond energies [1]

⚠ If you missed marks here: For the percentage, divide the difference (104) by the EXPERIMENTAL value 2803, not your calculated 2699 – and take the difference of the magnitudes, ignoring the minus signs. For (ii), "experimental error" or "heat loss" scores nothing here because the question compares two calculated/measured values – the expected answer is that bond energies are AVERAGES across many compounds, so the actual bond energies in glucose differ slightly.

1 mark -- difference 104 1 mark -- percentage 3.7 percent 1 mark -- bond energies average values different compounds actual differ

Question 4 -- Bond Energy and the Haber Process

Total: 12 marks

The Haber process is used to manufacture ammonia at IFFCO plants in Gujarat, India, and at chemical works in Teesside, England.

(a) [5]

The equation for the Haber process is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Calculate ΔH using bond energy data.

Bond	Bond energy / kJ mol⁻¹
N≡N	945
H–H	436
N–H	391

Model Answer -- 4(a)

Bonds broken: 1(N≡N) + 3(H–H) = 945 + (3×436) = 945 + 1308 = 2253 kJ [1]

Bonds formed: 6(N–H) (two NH₃ molecules, each with 3 N–H bonds) = 6×391 = 2346 kJ [1]

ΔH = 2253 − 2346 = −93 kJ mol⁻¹ [1]

The reaction is exothermic (negative ΔH) [1]

The reaction is reversible -- the reverse reaction (decomposition of ammonia) is endothermic with ΔH = +93 kJ mol⁻¹ [1]

⚠ If you missed marks here: The classic error is counting only 3 N–H bonds formed – there are TWO NH₃ molecules, each with 3 N–H bonds, so 6 × 391 = 2346 kJ; using 3 gives a wildly wrong +1080. Also check you multiplied H–H by 3 (the coefficient in 3H₂), and remember the reverse of an exothermic reaction is endothermic with the same magnitude but opposite sign (+93).

1 mark -- bonds broken 2253 one triple N three H-H 1 mark -- bonds formed 2346 six N-H 1 mark -- delta negative 93 1 mark -- reaction exothermic 1 mark -- reverse reaction endothermic positive 93

(b) [4]

(i) The activation energy for the uncatalysed Haber process is approximately 230 kJ mol⁻¹. The iron catalyst reduces this to approximately 160 kJ mol⁻¹. Draw an energy level diagram showing both the catalysed and uncatalysed reaction profiles. [3]

(ii) Calculate the energy at the top of the uncatalysed curve if the reactant energy level is set at 2253 kJ. [1]

Model Answer -- 4(b)

Reactants (N₂ + 3H₂) at higher energy, products (2NH₃) at lower energy, ΔH = −93 kJ shown [1]

Two curves: uncatalysed with Eₐ = 230 kJ (taller peak), catalysed with Eₐ = 160 kJ (shorter peak) [1]

Both curves start and end at the same energy levels [1]

Energy at top = 2253 + 230 = 2483 kJ [1]

⚠ If you missed marks here: Diagram marks go missing when the two curves do not start and end at the SAME levels, or when the catalysed peak is drawn without labelling Eₐ = 160 vs 230 kJ. For (ii), the peak sits at reactants + activation energy: 2253 + 230 = 2483 kJ – if you used 160 you took the catalysed curve, and if you subtracted 93 you mixed ΔH into a calculation that only needs Eₐ.

1 mark -- reactants higher products lower delta negative 93 shown 1 mark -- two curves uncatalysed 230 taller catalysed 160 shorter 1 mark -- both curves same start end energy levels 1 mark -- energy top 2483

(i) Explain why the N≡N triple bond requires so much energy to break. [1]

(ii) Explain how the iron catalyst helps overcome this problem. [1]

(iii) Despite being exothermic, the Haber process uses a temperature of 450 °C. Explain why a high temperature is used. [1]

Model Answer -- 4(c)

The triple bond consists of three shared pairs of electrons, creating a very strong bond with a high bond energy (945 kJ mol⁻¹) [1]

The iron catalyst provides an alternative pathway with a lower activation energy -- nitrogen molecules adsorb onto the catalyst surface, which weakens the N≡N bond [1]

A high temperature is used to increase the rate of reaction / give particles enough kinetic energy to overcome the activation energy barrier, even though it reduces the yield (since the forward reaction is exothermic) [1]

⚠ If you missed marks here: Saying "the triple bond is strong" restates the question – you need WHY: three shared pairs of electrons give a very high bond energy (945 kJ mol⁻¹). For the catalyst, "it speeds up the reaction" is not enough – state that it lowers the activation energy via an alternative pathway (adsorption on the iron surface). For (iii), the trap is saying high temperature increases yield – it actually LOWERS yield for this exothermic reaction; it is used to increase the RATE.

1 mark -- triple bond three shared pairs electrons strong high bond energy 945 1 mark -- iron catalyst alternative pathway lower activation energy adsorb surface weakens 1 mark -- high temperature increase rate kinetic energy overcome activation energy

Question 5 -- Working Backwards from ΔH to Find Bond Energy

Total: 10 marks

A chemistry teacher in Singapore sets her students a challenging problem: using a known ΔH value to calculate an unknown bond energy.

(a) [6]

The reaction between hydrogen and fluorine is:

H₂(g) + F₂(g) → 2HF(g) ΔH = −542 kJ mol⁻¹

The bond energies of H–H and F–F are known:

Bond	Bond energy / kJ mol⁻¹
H–H	436
F–F	158
H–F	?

(i) Write expressions for the total energy of bonds broken and bonds formed. [2]

(ii) Use the formula ΔH = bonds broken − bonds formed to set up an equation. [1]

(iii) Solve for the H–F bond energy. [2]

(iv) Comment on the strength of the H–F bond compared with H–H and F–F. [1]

Model Answer -- 5(a)

Bonds broken = 1(H–H) + 1(F–F) = 436 + 158 = 594 kJ [1]

Bonds formed = 2(H–F) = 2x (where x is the H–F bond energy) [1]

ΔH = bonds broken − bonds formed: −542 = 594 − 2x [1]

2x = 594 + 542 = 1136 [1]

x = 1136 / 2 = 568 kJ mol⁻¹ [1]

The H–F bond (568 kJ) is significantly stronger than both H–H (436 kJ) and F–F (158 kJ), because the large electronegativity difference between H and F creates a very polar bond with strong electrostatic attraction [1]

⚠ If you missed marks here: Sign handling is where this goes wrong: from −542 = 594 − 2x, adding 542 to both sides gives 2x = 1136, so x = 568. If you got 26 you solved 2x = 594 − 542 (dropped the minus sign on ΔH); if you got 1136 you forgot the reaction makes TWO moles of HF, so bonds formed is 2x not x. For the comment mark, compare 568 with BOTH 436 and 158 – just saying "H–F is strong" is not a comparison.

1 mark -- bonds broken 594 1 mark -- bonds formed expressed 2x 1 mark -- equation negative 542 equals 594 minus 2x 1 mark -- rearranged 2x equals 1136 1 mark -- H-F bond energy 568 1 mark -- H-F stronger than H-H F-F electronegativity polar strong

(b) [4]

Using the H–F bond energy you calculated (568 kJ mol⁻¹), predict the enthalpy change for:

CH₄(g) + 4F₂(g) → CF₄(g) + 4HF(g)

Additional bond energies: C–H = 413 kJ mol⁻¹, C–F = 485 kJ mol⁻¹, F–F = 158 kJ mol⁻¹

Model Answer -- 5(b)

Bonds broken: 4(C–H) + 4(F–F) = (4×413) + (4×158) = 1652 + 632 = 2284 kJ [1]

Bonds formed: 4(C–F) + 4(H–F) = (4×485) + (4×568) = 1940 + 2272 = 4212 kJ [1]

ΔH = 2284 − 4212 = −1928 kJ mol⁻¹ [1]

This is a very highly exothermic reaction [1]

⚠ If you missed marks here: Every bond count here is multiplied by 4 – forgetting a ×4 on any term (4 C–H, 4 F–F, 4 C–F, 4 H–F) throws the whole answer off. Check you used YOUR calculated H–F value of 568, not the ΔH of 542 from part (a), and that you finished with broken (2284) minus formed (4212) = −1928; a +1928 answer means the subtraction was reversed.

1 mark -- bonds broken 2284 1 mark -- bonds formed 4212 1 mark -- delta negative 1928 1 mark -- very highly exothermic

Question 6 -- Advanced Energy Level Diagram Analysis

Total: 12 marks

A chemistry examiner in Cambridge designs questions that test deep understanding of energy level diagrams.

(a) [5]

An energy level diagram for a reaction shows the following energy values:
-- Reactants: 200 kJ
-- Top of uncatalysed curve: 350 kJ
-- Top of catalysed curve: 280 kJ
-- Products: 50 kJ

Calculate:
(i) ΔH for the reaction. [1]
(ii) The activation energy without a catalyst. [1]
(iii) The activation energy with a catalyst. [1]
(iv) The reduction in activation energy caused by the catalyst. [1]
(v) State whether the reaction is exothermic or endothermic. Explain your reasoning. [1]

Model Answer -- 6(a)

ΔH = products − reactants = 50 − 200 = −150 kJ [1]

Eₐ (uncatalysed) = 350 − 200 = 150 kJ [1]

Eₐ (catalysed) = 280 − 200 = 80 kJ [1]

Reduction = 150 − 80 = 70 kJ [1]

Exothermic because ΔH is negative / products are at a lower energy level than reactants [1]

⚠ If you missed marks here: ΔH is always products − reactants: 50 − 200 = −150 kJ – if you wrote +150 you subtracted the other way. Activation energy is measured from the REACTANT level up to the peak (350 − 200 = 150; 280 − 200 = 80), not from zero and not from the products. For the last mark, "exothermic" alone is not enough – give the reason (ΔH negative / products lower than reactants).

1 mark -- delta negative 150 1 mark -- uncatalysed activation energy 150 1 mark -- catalysed activation energy 80 1 mark -- reduction 70 1 mark -- exothermic negative products lower energy

(b) [4]

For the reverse of the reaction in part (a):

(i) State ΔH for the reverse reaction. [1]

(ii) Calculate the activation energy for the reverse uncatalysed reaction. [1]

(iii) Calculate the activation energy for the reverse catalysed reaction. [1]

(iv) State whether the reverse reaction is exothermic or endothermic. [1]

Model Answer -- 6(b)

ΔH (reverse) = +150 kJ (same magnitude, opposite sign) [1]

Eₐ (reverse, uncatalysed) = 350 − 50 = 300 kJ [1]

Eₐ (reverse, catalysed) = 280 − 50 = 230 kJ [1]

Endothermic (because ΔH is positive) [1]

⚠ If you missed marks here: For the reverse reaction the STARTING level is the old products (50 kJ), so Eₐ is peak minus 50: 350 − 50 = 300 and 280 − 50 = 230 – if you got 150 and 80 again you measured from the wrong side. ΔH simply flips sign to +150 kJ (same magnitude), which is why the reverse reaction is endothermic.

1 mark -- delta reverse positive 150 1 mark -- reverse uncatalysed activation 300 1 mark -- reverse catalysed activation 230 1 mark -- endothermic positive

(i) State the mathematical relationship between Eₐ(forward), Eₐ(reverse), and ΔH. [1]

(ii) Verify this relationship using the uncatalysed values from parts (a) and (b). [1]

(iii) Explain why the activation energy for the reverse reaction is always larger than the activation energy for the forward reaction when the forward reaction is exothermic. [1]

Model Answer -- 6(c)

ΔH = Eₐ(forward) − Eₐ(reverse), or equivalently Eₐ(reverse) = Eₐ(forward) − ΔH [1]

Check: Eₐ(forward) − Eₐ(reverse) = 150 − 300 = −150 kJ = ΔH (verified) [1]

For an exothermic reaction, products are at a lower energy level, so the reverse reaction must climb from the lower product level all the way up to the same peak, covering both the activation energy AND the energy difference ΔH [1]

⚠ If you missed marks here: The order matters: ΔH = Eₐ(forward) − Eₐ(reverse) – writing it reversed gives +150 instead of −150 when you verify with 150 − 300. For (iii), just saying "the reverse Eₐ is bigger" repeats the question – the mark is for explaining that the reverse reaction starts from the LOWER product level but must reach the SAME peak, so it climbs Eₐ(forward) plus the extra drop of ΔH.

1 mark -- delta equals forward activation minus reverse activation formula 1 mark -- verified 150 minus 300 equals negative 150 1 mark -- products lower reverse must climb higher peak covering activation energy plus delta

Question 7 -- Comparing Three Fuels: Hydrogen, Methanol and Octane

Total: 10 marks

An environmental scientist in Kochi compares three fuels -- hydrogen, methanol and octane -- for their suitability as vehicle fuels in India.

(a) [5]

The combustion of methanol is:

2CH₃OH(g) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)

Methanol (CH₃OH) has: 3 × C–H, 1 × C–O, 1 × O–H bonds.

Calculate ΔH per mole of methanol (divide final answer by 2).

Bond	Bond energy / kJ mol⁻¹
C–H	413
C–O	358
O–H	464
O=O	498
C=O	805

Model Answer -- 7(a)

Bonds broken (for 2 moles methanol + 3 O₂): 6(C–H) + 2(C–O) + 2(O–H) + 3(O=O) = (6×413) + (2×358) + (2×464) + (3×498) = 2478 + 716 + 928 + 1494 = 5616 kJ [1]

Bonds formed: 4(C=O) + 8(O–H) = (4×805) + (8×464) = 3220 + 3712 = 6932 kJ [1]

ΔH (for 2 moles) = 5616 − 6932 = −1316 kJ [1]

ΔH per mole of methanol = −1316 / 2 = −658 kJ mol⁻¹ [1]

The reaction is exothermic [1]

⚠ If you missed marks here: The equation shows 2CH₃OH, so every methanol bond is doubled: 6 C–H, 2 C–O, 2 O–H – if you used 3, 1, 1 your broken total came out roughly half. An answer of −1316 means you forgot the final divide by 2 to get per mole of methanol; +658 means the broken/formed subtraction was reversed. Also check the products: 2CO₂ gives 4 C=O and 4H₂O gives 8 O–H.

1 mark -- bonds broken 5616 1 mark -- bonds formed 6932 1 mark -- delta two moles negative 1316 1 mark -- per mole negative 658 1 mark -- exothermic

(b) [5]

The table compares three fuels.

Fuel	ΔH / kJ mol⁻¹	M_r	Products of combustion
Hydrogen (H₂)	−486	2	H₂O only
Methanol (CH₃OH)	−658	32	CO₂ + H₂O
Octane (C₈H₁₈)	−5470	114	CO₂ + H₂O

(i) Calculate the energy released per gram for each fuel. [3]

(ii) Using your answers and the combustion products, evaluate which fuel is best for the environment and which releases the most energy per gram. [2]

Model Answer -- 7(b)

Hydrogen: 486 / 2 = 243.0 kJ g⁻¹ [1]

Methanol: 658 / 32 = 20.6 kJ g⁻¹ [1]

Octane: 5470 / 114 = 48.0 kJ g⁻¹ [1]

Hydrogen releases the most energy per gram (243 kJ g⁻¹) and is best for the environment because it produces only water -- no CO₂ emissions / no contribution to global warming [1]

However, hydrogen is difficult to store and transport / methanol and octane are liquids and easier to handle / octane releases the most energy per mole but not per gram [1]

⚠ If you missed marks here: Energy per gram is ΔH ÷ M_r – dividing the other way round, or forgetting hydrogen's tiny M_r of 2, hides why hydrogen wins at 243 kJ g⁻¹. If you chose octane as "best" you compared per MOLE (−5470) instead of per gram. For the environment mark, "hydrogen is cleaner" is too vague – state that it produces ONLY water and no CO₂, and for the evaluate mark give a balancing drawback such as hydrogen being hard to store and transport.

1 mark -- hydrogen 243 per gram 1 mark -- methanol 20.6 per gram 1 mark -- octane 48.0 per gram 1 mark -- hydrogen most energy per gram best environment only water no CO2 1 mark -- hydrogen difficult store transport octane methanol liquids easier handle

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