Greenhouse gases and climate change (SACE Stage 2 Chemistry)

Syllabus dot point

“Explain the greenhouse effect in terms of the absorption of infrared radiation by greenhouse gas molecules.”

8 min readUpdated 2 Jun 2026

Quick answer

The greenhouse effect occurs because greenhouse gases ( $\text{CO}_2$ , $\text{H}_2\text{O}$ , $\text{CH}_4$ , $\text{N}_2\text{O}$ ) absorb the infrared radiation that the Earth re-emits, then re-radiate it in all directions, warming the lower atmosphere. A gas absorbs IR only if a bond vibration changes its dipole moment, which is why polar-bonded molecules absorb while symmetrical $\text{N}_2$ and $\text{O}_2$ do not. Burning fossil fuels raises $\text{CO}_2$ , enhancing the effect.

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What this dot point is asking
Lead worked calculation
Why some gases absorb infrared and others do not
The enhanced greenhouse effect
Monitoring greenhouse gases
Why it matters for monitoring
Exam-style questions

What this dot point is asking

SACE expects you to explain absorption in terms of bond vibrations and dipole change, distinguish greenhouse from non-greenhouse gases, and link combustion to rising $\text{CO}_2$ .

Lead worked calculation

Comparing the CO2 from two fuels per mole

Compare the $\text{CO}_2$ released per mole burned for methane and octane. Methane: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ . Octane: $2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}$ .

Read the carbon-to-CO2 ratio

Methane gives $1\ \text{mol}$ of $\text{CO}_2$ per mole burned. Octane gives $\tfrac{16}{2} = 8\ \text{mol}$ of $\text{CO}_2$ per mole burned.

Per gram comparison (more meaningful)

For methane, $\dfrac{1 \times 44.0}{16.0} = 2.75\ \text{g CO}_2$ per gram of fuel. For octane, $\dfrac{8 \times 44.0}{114.0} = 3.09\ \text{g CO}_2$ per gram.

Interpret

Methane releases less $\text{CO}_2$ per gram because it has the highest hydrogen-to-carbon ratio of common fuels, so a larger share of its energy comes from forming water rather than $\text{CO}_2$ . This is one reason natural gas is a lower-carbon fuel than petrol.

Why some gases absorb infrared and others do not

The Earth absorbs visible sunlight and re-emits energy as infrared. Whether a gas captures this IR depends on its molecular structure.

The infrared absorption rule

A molecule absorbs infrared radiation only when a vibration (bond stretch or bend) produces a change in dipole moment. The vibrating polar bonds act like an oscillating charge that couples to the oscillating electric field of the IR photon, allowing energy transfer. If a vibration leaves the dipole unchanged, no absorption occurs.

$\text{CO}_2$ : although linear and symmetric overall, its bending vibration and asymmetric stretch create a temporary changing dipole, so it absorbs IR strongly.
$\text{H}_2\text{O}$ : bent and polar; its stretches and bend all change the dipole, making it the most significant natural greenhouse gas.
$\text{N}_2$ and $\text{O}_2$ : homonuclear diatomics with non-polar bonds; stretching keeps the dipole at zero, so they are IR-transparent despite being abundant.

The enhanced greenhouse effect

The natural greenhouse effect keeps Earth habitable, around $33\ ^\circ\text{C}$ warmer than it would otherwise be. The enhanced greenhouse effect is the extra warming from human activity raising greenhouse-gas concentrations, chiefly through combustion of fossil fuels: $\text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ . More $\text{CO}_2$ means more IR is absorbed and re-radiated downward, raising average temperatures.

Global warming potential

Gases differ in warming impact per molecule. Methane absorbs IR far more effectively than $\text{CO}_2$ per molecule (a higher global warming potential), though it is less abundant and shorter-lived. Nitrous oxide ( $\text{N}_2\text{O}$ ) is also a potent greenhouse gas. Monitoring tracks all of them, not just $\text{CO}_2$ .

Monitoring greenhouse gases

Atmospheric $\text{CO}_2$ is measured continuously (famously at Mauna Loa) using infrared gas analysers, which exploit the very IR absorption that makes $\text{CO}_2$ a greenhouse gas: the more $\text{CO}_2$ in the sample, the more IR it absorbs. This neatly ties the monitoring technique back to the molecular property being studied.

Saying abundance determines greenhouse behaviour

The most common error is assuming the most abundant atmospheric gases must be the strongest greenhouse gases. $\text{N}_2$ and $\text{O}_2$ make up about $99\%$ of dry air yet absorb no IR, because their non-polar bonds give no changing dipole on vibration. Absorption depends on molecular structure (a dipole change), not abundance. Another slip is forgetting that $\text{CO}_2$ absorbs through its bending and asymmetric vibrations even though it has no permanent dipole.

Why it matters for monitoring

Linking molecular structure to IR absorption explains both why certain trace gases drive climate change and how instruments detect them. It connects combustion chemistry (the source of extra $\text{CO}_2$ ) to the physical mechanism of warming, the foundation of climate monitoring and emissions policy.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2021-style4 marks

Explain, in terms of molecular vibrations and bond polarity, why carbon dioxide and water vapour are greenhouse gases but nitrogen (

\text{N}_2

) and oxygen (

\text{O}_2

) are not, even though

\text{N}_2

and

\text{O}_2

make up most of the atmosphere.

Show worked answer →

A molecule absorbs infrared radiation only if a vibration (a stretching or bending of its bonds) causes a change in its dipole moment. (1 mark)

$\text{CO}_2$ and $\text{H}_2\text{O}$ contain polar bonds; when these bonds vibrate, the distribution of charge changes, creating a changing dipole that can absorb IR photons of matching energy. (1 mark)

$\text{N}_2$ and $\text{O}_2$ are diatomic molecules made of two identical atoms, so their bonds are non-polar; stretching them does not change the dipole (which stays zero). With no changing dipole, they cannot absorb IR. (1 mark)

Therefore, despite being the most abundant gases, $\text{N}_2$ and $\text{O}_2$ are transparent to the IR re-emitted by Earth, while trace $\text{CO}_2$ and $\text{H}_2\text{O}$ trap it. (1 mark)

SACE 2019-style4 marks

The complete combustion of octane is

2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}

. Calculate the mass of

\text{CO}_2

produced when

1.00\ \text{kg}

of octane is burned completely. (

M(\text{C}_8\text{H}_{18}) = 114.0

M(\text{CO}_2) = 44.0\ \text{g mol}^{-1}

Show worked answer →

Step 1: $n(\text{C}_8\text{H}_{18}) = \dfrac{1000}{114.0} = 8.77\ \text{mol}$ . (1 mark)

Step 2: from the $2:16$ (i.e. $1:8$ ) ratio, $n(\text{CO}_2) = 8 \times 8.77 = 70.2\ \text{mol}$ . (1 mark)

Step 3: $m(\text{CO}_2) = nM = 70.2 \times 44.0 = 3.09 \times 10^{3}\ \text{g} = 3.09\ \text{kg}$ . (2 marks)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board