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Bond Energy Calculations and Energy Profiles: O-Level Chemistry

10 August 2026 · Dojo Education · 4 min read

Bond Energy Calculations and Energy Profiles: O-Level Chemistry

Energetics is one of the most predictable topics in the O-Level Pure Chemistry paper. The questions look different every year, but the marks are always given for the same three or four things. Once you know exactly what the marker wants, bond energy calculations and energy profile diagrams become some of the fastest marks in Paper 2.

Here is the method we teach at Dojo, written the way we actually did it in the exam hall.

The two rules everything is built on

  1. Breaking bonds absorbs energy. It is endothermic. Energy goes in.
  2. Forming bonds releases energy. It is exothermic. Energy comes out.

Every calculation and every diagram in this topic is just bookkeeping between those two rules.

The enthalpy change of reaction is:

ΔH = (total energy absorbed to break all bonds in the reactants) − (total energy released when all bonds in the products are formed)

A useful shorthand: ΔH = bonds broken − bonds formed.

Getting the sign right

Singapore markers are strict here. A correct number with a missing minus sign usually loses the final mark, and "kJ/mol" is expected as the unit.

Worked example 1: hydrogen and chlorine

H₂(g) + Cl₂(g) → 2HCl(g)

Given bond energies in kJ/mol: H–H = 436, Cl–Cl = 242, H–Cl = 431.

Step 1: list the bonds broken. 1 × H–H = 436 1 × Cl–Cl = 242 Total in = 678 kJ

Step 2: list the bonds formed. 2 × H–Cl = 2 × 431 = 862 Total out = 862 kJ

Step 3: subtract. ΔH = 678 − 862 = −184 kJ/mol

Step 4: state the conclusion. The reaction is exothermic because the energy released when bonds are formed is greater than the energy absorbed when bonds are broken.

That final sentence is the mark-scheme phrasing. Write it out even when the question only says "comment on your answer".

Worked example 2: combustion of methane

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Bond energies: C–H = 410, O=O = 496, C=O = 805, O–H = 460.

Bonds broken: 4 × C–H + 2 × O=O = 1640 + 992 = 2632 kJ Bonds formed: 2 × C=O + 4 × O–H = 1610 + 1840 = 3450 kJ

ΔH = 2632 − 3450 = −818 kJ/mol

Notice where students lose marks: forgetting that CO₂ has two C=O bonds, that each H₂O has two O–H bonds, and that the coefficient 2 in front of O₂ and H₂O multiplies everything. Draw the displayed structures in the margin and count the lines. It takes fifteen seconds and saves the whole question.

Drawing the energy profile diagram

Examiners award marks for specific labelled features, not artistic quality. Include all of these:

For an exothermic reaction the products line sits below the reactants line. For an endothermic reaction it sits above. If your ΔH sign and your diagram disagree, you will lose marks in both parts.

Adding a catalyst

If asked to show the effect of a catalyst, draw a dashed curve with a lower peak starting and ending at the same reactant and product levels. Then write: the catalyst provides an alternative pathway of lower activation energy, so more particles have enough energy to react, but ΔH is unchanged.

Common traps to avoid

A quick self-check routine

Before you move on, ask yourself: is the equation balanced, did I multiply every bond by its coefficient, does my sign match my diagram, and did I write kJ/mol? Four questions, four extra marks.

If energetics still feels like guesswork, come and work through a past-paper question with one of our tutors live. Message us on WhatsApp to book a free trial class: https://wa.link/dsgbkf

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