These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Use these average bond enthalpies, in kJ mol⁻¹, for every question:
| Bond | kJ mol⁻¹ | Bond | kJ mol⁻¹ | Bond | kJ mol⁻¹ |
|---|---|---|---|---|---|
| H–H | 432 | C–C | 347 | O=O | 495 |
| H–Br | 363 | C≡C | 839 | N–H | 391 |
| Br–Br | 193 | C–O | 358 | N–N | 160 |
| C–H | 413 | C=O (in CO₂) | 799 | N≡N | 941 |
| O–H | 467 |
Molar masses: H 1.008, C 12.01, O 16.00 g mol⁻¹. All substances are gases unless a state is given.
Question 1 (multiple choice · foundation)
Which statement best explains why a reaction is exothermic?
- (A) Energy is released when the bonds in the reactants break.
- (B) The energy released forming the bonds in the products is greater than the energy needed to break the bonds in the reactants.
- (C) The products contain a larger number of bonds than the reactants.
- (D) The bonds in the reactants are stronger, in total, than the bonds in the products.
Answer and explanation
Answer: (B). ΔH ≈ Σ(broken) − Σ(formed). When the energy released by forming bonds is larger, ΔH is negative, so the reaction is exothermic.
- (A) Breaking a bond always needs energy; it never releases it.
- (C) The number of bonds is not what matters. Bond strengths differ, so fewer strong bonds can release more energy than many weak ones.
- (D) This is the condition for an endothermic reaction: strong reactant bonds cost more to break than the product bonds give back.
Question 2 (multiple choice · core)
Estimate ΔH for H₂(g) + Br₂(g) → 2HBr(g).
- (A) −101 kJ mol⁻¹
- (B) +101 kJ mol⁻¹
- (C) +262 kJ mol⁻¹
- (D) −1351 kJ mol⁻¹
Answer and explanation
Answer: (A). Broken: H–H + Br–Br = 432 + 193 = 625 kJ. Formed: 2 H–Br = 2 × 363 = 726 kJ. ΔH ≈ 625 − 726 = −101 kJ mol⁻¹.
- (B) uses formed − broken, which reverses the sign.
- (C) counts only one H–Br bond (625 − 363), ignoring the coefficient 2.
- (D) treats breaking bonds as releasing energy too, so it adds the two totals as energy released: −(625 + 726).
Question 3 (multiple choice · core)
Hydrazine, H₂N–NH₂, reacts with oxygen: N₂H₄(g) + O₂(g) → N₂(g) + 2H₂O(g). Estimate ΔH.
- (A) −590 kJ mol⁻¹
- (B) +590 kJ mol⁻¹
- (C) −750 kJ mol⁻¹
- (D) +191 kJ mol⁻¹
Answer and explanation
Answer: (A).
| Bonds | Energy (kJ) | |
|---|---|---|
| Broken | 1 N–N + 4 N–H + 1 O=O | 160 + 1564 + 495 = 2219 |
| Formed | 1 N≡N + 4 O–H | 941 + 1868 = 2809 |
ΔH ≈ 2219 − 2809 = −590 kJ mol⁻¹.
- (B) has the sign reversed (formed − broken).
- (C) forgets to break the N–N bond in hydrazine (2059 − 2809).
- (D) treats the N₂ product as having a single N–N bond (2219 − 2028), but N₂ has a triple bond.
Question 4 (multiple choice · core)
For C₂H₂(g) + 2H₂(g) → C₂H₆(g), a bond-enthalpy estimate differs by about 15 kJ mol⁻¹ from the value calculated from enthalpies of formation. Which is the best explanation?
- (A) Bond enthalpies are negative numbers, so the estimate has the wrong sign.
- (B) The table values are averages taken over many molecules, so they do not exactly match the bonds in these molecules.
- (C) The estimate includes the energy of the intermolecular forces between the gas molecules.
- (D) Bond-enthalpy estimates can only be used for endothermic reactions.
Answer and explanation
Answer: (B). A C–H or C≡C bond has a slightly different strength in different molecules. Using an average value introduces a small error for each bond, and these add up.
- (A) Bond enthalpies are always positive; the method gives the correct sign.
- (C) The method includes no intermolecular forces. That is a reason for error when liquids or solids are involved, but all species here are gases.
- (D) The method works for any gas-phase reaction, exothermic or endothermic.
Question 5 (constructed response · core)
Methanol vapour burns in oxygen: 2CH₃OH(g) + 3O₂(g) → 2CO₂(g) + 4H₂O(g). Methanol has the structure H₃C–O–H.
(a) List the number of each type of bond broken and formed in the equation as written. (b) Estimate ΔH for the equation as written, and then per mole of methanol. (c) Estimate the heat released when 5.00 g of methanol vapour burns completely.
Worked solution and suggested Marlbridge rubric
(a) Each CH₃OH has 3 C–H, 1 C–O and 1 O–H. Each CO₂ has 2 C=O. Each H₂O has 2 O–H.
| Bonds | Energy (kJ) | |
|---|---|---|
| Broken | 6 C–H + 2 C–O + 2 O–H + 3 O=O | 2478 + 716 + 934 + 1485 = 5613 |
| Formed | 4 C=O + 8 O–H | 3196 + 3736 = 6932 |
(b) ΔH ≈ 5613 − 6932 = −1319 kJ for the equation as written (2 mol methanol), so −659.5 kJ mol⁻¹ of methanol (about −660).
(c) M(CH₃OH) = 12.01 + 4(1.008) + 16.00 = 32.04 g mol⁻¹. n = 5.00 ÷ 32.04 = 0.156 mol. Heat released = 0.156 mol × 659.5 kJ mol⁻¹ = 103 kJ.
| Point | What earns it |
|---|---|
| 1 | Correct bonds in the reactants, including 3 O=O (double bonds) |
| 1 | Correct bonds in the products: 4 C=O and 8 O–H |
| 1 | ΔH ≈ −1319 kJ, using broken − formed |
| 1 | −659.5 kJ per mole of methanol (or ΔH correctly scaled in (c)) |
| 1 | About 103 kJ released, with the moles of methanol shown |
Carry forward an error from (a) or (b). The C=O bonds formed here are in CO₂, so use 799 kJ mol⁻¹ from the table; a general C=O value (such as 745 kJ mol⁻¹) would give −551.5 kJ mol⁻¹ and does not earn the ΔH points.
Question 6 (constructed response · core)
For the reaction C₂H₂(g) + 2H₂(g) → C₂H₆(g), the enthalpy change is −311 kJ mol⁻¹. Ethyne is H–C≡C–H.
(a) Use this value and the table to estimate the C≡C bond enthalpy in ethyne. (b) Compare your answer with the table value and explain the difference.
Worked solution and suggested Marlbridge rubric
(a) Using the shortcut, the two C–H bonds in ethyne survive unchanged.
- Broken: 1 C≡C + 2 H–H = x + 864 kJ.
- Formed: 1 C–C + 4 new C–H = 347 + 1652 = 1999 kJ.
- x + 864 − 1999 = −311, so x = −311 − 864 + 1999 = 824 kJ mol⁻¹.
(Counting every bond gives the same result: x + 2(413) + 864 − [347 + 6(413)] = −311.)
(b) The table value is 839 kJ mol⁻¹, 15 kJ mol⁻¹ higher. Table values are averages over many molecules, and the C–H and C–C values used in the calculation are also averages. The real C≡C bond in ethyne, and the bonds in ethane, are not exactly average.
| Point | What earns it |
|---|---|
| 1 | Correct bonds broken and formed, with the unknown C≡C as x |
| 1 | Equation set up as broken − formed = −311 |
| 1 | x = 824 kJ mol⁻¹ |
| 1 | Difference explained by average bond enthalpies (not “experimental error” alone) |
Question 7 (constructed response · stretch)
A student says: “Hydrogen is a good fuel because breaking the bonds in hydrogen releases a lot of energy.” Consider 2H₂(g) + O₂(g) → 2H₂O(g).
(a) Evaluate the student’s statement. (b) Estimate ΔH for the reaction as written, and per mole of water formed. (c) Describe a step diagram for the bond-enthalpy method for this reaction. Give the energy of each level, with the reactants at 0. (d) A value from enthalpies of formation for H₂O(g) is −241.8 kJ mol⁻¹. A classmate wants to use bond enthalpies to estimate ΔH for forming liquid water instead. Explain why the method is less suitable.
Worked solution and suggested Marlbridge rubric
(a) The statement is wrong. Breaking the H–H and O=O bonds needs energy. Hydrogen is a good fuel because forming the strong O–H bonds in water releases more energy than breaking the reactant bonds needs.
(b) Broken: 2 H–H + 1 O=O = 864 + 495 = 1359 kJ. Formed: 4 O–H = 1868 kJ. ΔH ≈ 1359 − 1868 = −509 kJ, which is −254.5 kJ per mole of H₂O (about −255).
(c) Start: 2H₂(g) + O₂(g) at 0. Upward arrow, “break bonds, +1359”, to the level 4H(g) + 2O(g) at +1359. Downward arrow, “form bonds, −1868”, to 2H₂O(g) at −509, below the start. Axis labelled “Energy (kJ mol⁻¹)”.
(d) Bond enthalpies describe molecules in the gas phase. Forming liquid water also releases energy as hydrogen bonds form when the vapour condenses, and no bond enthalpy includes that. The estimate would still be close to the gas-phase value, so it would be further from the true value for liquid water.
| Point | What earns it |
|---|---|
| 1 | Statement rejected: breaking bonds needs energy |
| 1 | Energy comes from forming stronger O–H bonds (energy released > energy needed) |
| 1 | −509 kJ for the equation and about −255 kJ per mole of water |
| 1 | Step diagram levels 0, +1359, −509, with arrows up then down |
| 1 | (d) Method is for gases; condensation energy (intermolecular forces) is not included |
The estimate in (b), −254.5 kJ mol⁻¹, is about 13 kJ more negative than −241.8 kJ mol⁻¹ even for the gas, because of averaging; this comment is not required.
How did you do?
- Q1 or Q7(a) wrong: re-read “Where the energy of a reaction comes from” in the study guide.
- Q2, Q3 or Q5 wrong: check your bond table against “Counting bonds without mistakes” and Worked example 2.
- Q4 or Q6 incomplete: go through Worked example 3 and “Why the answer is only an estimate”.
Then tick off the topic checklist.