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Every year, examiner reports highlight the same recurring mistakes. Knowing what these are — and checking for them — can add several marks to your total. This lesson catalogues the most common and costly mistakes across all topic areas, with worked examples showing how to avoid each one.
The most common energetics mistake is getting the sign of ΔH wrong. Remember:
When using q = mcΔT, remember that ΔH = -q/n (the negative sign accounts for the convention that energy released by the system is negative).
Worked Example: A calorimetry experiment measures a temperature rise of 8.5°C when 50.0 cm³ of 1.00 mol dm⁻³ acid reacts with excess alkali.
q = 50.0 × 4.18 × 8.5 = 1776.5 J n = 1.00 × 0.0500 = 0.0500 mol ΔH = **-**1776.5 / 0.0500 = -35530 J mol⁻¹ = -35.5 kJ mol⁻¹
The negative sign is essential — this is an exothermic reaction. Many students write +35.5, which would indicate an endothermic reaction and is incorrect.
When constructing enthalpy cycles, pay attention to the direction of arrows. If you reverse a reaction, you reverse the sign of ΔH. Students frequently forget to reverse the sign when going "against the arrow" in a cycle.
Worked Example: Given: (1) C(s) + O₂(g) → CO₂(g) ΔH₁ = -393 kJ mol⁻¹ (2) H₂(g) + ½O₂(g) → H₂O(l) ΔH₂ = -286 kJ mol⁻¹ (3) CH₃OH(l) + 1½O₂(g) → CO₂(g) + 2H₂O(l) ΔH₃ = -726 kJ mol⁻¹
Calculate the enthalpy of formation of methanol: C(s) + 2H₂(g) + ½O₂(g) → CH₃OH(l)
ΔH_f = ΔH₁ + 2×ΔH₂ - ΔH₃ (reversing reaction 3 reverses its sign) ΔH_f = (-393) + 2×(-286) - (-726) ΔH_f = -393 - 572 + 726 = -239 kJ mol⁻¹
Common error: Forgetting to reverse the sign of ΔH₃ when going backwards through reaction 3. This gives -393 - 572 - 726 = -1691 kJ mol⁻¹, which is completely wrong.
The most frequent Born-Haber mistake is applying electron affinity in the wrong direction. Remember:
Many students get these signs reversed, which cascades through the entire calculation.
Always check that your equation balances for atoms and charge (for ionic equations). An unbalanced equation receives zero marks.
Checklist for balancing:
State symbols (s, l, g, aq) are required in thermochemistry, equilibrium, and redox equations. Missing state symbols can cost 1 mark per equation.
When state symbols are essential:
When writing half-equations for redox:
Worked Example: Write the half-equation for the reduction of dichromate ions in acid solution. Cr₂O₇²⁻ → 2Cr³⁺ (balance Cr) Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (balance O with H₂O) Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O (balance H with H⁺) Charge: left = -2 + 14 = +12; right = +6. Need +6 electrons on left. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O ✓
graph LR
A["Correct: Arrow starts from<br/>lone pair or bond"] --> B["Shows electron<br/>pair movement"]
C["Incorrect: Arrow starts<br/>from an atom"] --> D["Meaningless — atoms<br/>don't move as electrons"]
E["Correct: Fish-hook arrow<br/>for radicals"] --> F["Shows single<br/>electron movement"]
Common error 1: Drawing an arrow from the Br atom to the OH group. Correct: Draw the arrow from the lone pair on OH⁻ to the δ+ carbon.
Common error 2: Drawing the arrow from the C-Br bond to the Br atom. Correct: The arrow goes from the C-Br bond to the Br atom (showing the bonding electrons moving onto bromine as it leaves).
Common error 3: In electrophilic addition to alkenes, the first arrow must come from the C=C π bond, not from the electrophile to the alkene.
Students often give the reagent but forget the conditions:
| Reagent | Solvent | Product | Reaction type |
|---|---|---|---|
| NaOH | Water (aqueous) | Alcohol | Nucleophilic substitution |
| NaOH | Ethanol | Alkene | Elimination |
| NH₃ | Water | No useful reaction with halogenoalkane | — |
| NH₃ | Ethanol (sealed tube) | Amine | Nucleophilic substitution |
This is one of the most commonly tested distinctions. If you write "NaOH" without specifying the solvent, the examiner cannot tell which reaction you mean and cannot award full marks.
This is one of the most common conceptual errors in the entire specification. Learn this table:
| Change | Effect on rate | Shifts equilibrium? | Changes Kc/Kp? |
|---|---|---|---|
| Increase temperature | Increases (both directions) | Yes — favours endothermic | Yes |
| Increase concentration of reactant | Increases forward rate | Yes — shifts right | No |
| Increase pressure (gases) | Increases rate | Yes — favours fewer moles | No |
| Add a catalyst | Increases rate (both equally) | No | No |
The confusion arises because students think "if a catalyst increases the rate, it must make more product." But a catalyst increases both the forward and reverse rates equally, so the equilibrium position is unchanged. The system simply reaches equilibrium faster.
Worked Example: "Explain the effect of a catalyst on the equilibrium yield of ammonia in the Haber process."
Wrong answer: "A catalyst increases the rate, so more ammonia is produced, shifting the equilibrium to the right."
Correct answer: "A catalyst increases the rate of both the forward and reverse reactions equally. It does not change the position of equilibrium or the equilibrium yield of ammonia. It only allows the system to reach equilibrium more quickly."
When temperature increases for an exothermic forward reaction:
Many students correctly state the shift direction but incorrectly say "Kc does not change." Kc only stays constant when temperature is constant.
Example: A student is given data to 3 s.f. and writes their answer as 0.04821974 mol dm⁻³. This is 7 significant figures. The correct answer is 0.0482 mol dm⁻³ (3 s.f.). Note: the leading zeros are not significant — 0.0482 has 3 significant figures.
| Value | Sig figs | Why |
|---|---|---|
| 4320 | 3 (or 4 if trailing zero is intentional) | Ambiguous — use scientific notation |
| 0.00340 | 3 | Leading zeros not significant, trailing zero is |
| 1.00 | 3 | Trailing zeros after decimal point are significant |
| 100 | 1 (or 2 or 3) | Ambiguous — write 1.00 × 10² for clarity |
The most frustrating mistake of all. Check:
Costly example: A question asks "Give one reason why the boiling point of water is higher than that of hydrogen sulfide." A student writes three reasons. The examiner marks only the first. If the first reason is wrong (e.g., "water has a higher molar mass" — incorrect), the student scores zero even though the second reason was correct.
Before moving to the next question, run through this mental checklist:
graph TD
A[Finished writing answer] --> B{Equations balanced?}
B -->|No| C[Balance atoms and charge]
B -->|Yes| D{State symbols included?}
D -->|No| E[Add s, l, g, aq]
D -->|Yes| F{Units on all numerical answers?}
F -->|No| G[Add units]
F -->|Yes| H{Correct sig figs?}
H -->|No| I[Round appropriately]
H -->|Yes| J{Answered the actual question asked?}
J -->|No| K[Re-read question, adjust answer]
J -->|Yes| L[Move to next question]
C --> D
E --> F
G --> H
I --> J
K --> L
This takes 15-20 seconds per question and can prevent the loss of 1-2 marks per question — potentially 10+ marks across a paper.
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