Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- A process with ΔG° > 0 is thermodynamically unfavourable (K < 1). It will not make much product by itself.
- It can still be made to happen if free energy is supplied in one of two ways.
- Outside energy source. Electrical energy drives an electrolytic cell or recharges a battery. Light drives the overall conversion of CO₂ and water into glucose in photosynthesis.
- Coupled reaction. A favourable reaction (large negative ΔG°) is linked to the unfavourable one. The two reactions share a common intermediate, which cancels when the equations are added.
- In living cells, the favourable partner is often the hydrolysis of ATP to ADP (ΔG° about −30 kJ mol⁻¹).
- The unfavourable step keeps its positive ΔG°. Only the overall reaction is favourable.
Key relationships
| Relationship | What it means | Use it to |
|---|---|---|
| ΔG°(overall) = ΔG°₁ + ΔG°₂ | free energy is a state function, so changes add | decide whether a coupled pair works (needs ΔG°(overall) < 0) |
| K(overall) = K₁ × K₂ | adding equations multiplies K | combine equilibrium constants of coupled steps |
| Reverse an equation → −ΔG° | the reverse of a favourable reaction is unfavourable | make the intermediate cancel |
| Multiply an equation by n → n × ΔG° | ΔG° is per mole of reaction as written | e.g. use 2 ATP: 2 × (−30.5) kJ mol⁻¹ |
| K = e^(−ΔG°/RT) | ΔG° in J mol⁻¹, R = 8.314 J mol⁻¹ K⁻¹ | turn an overall ΔG° into a K value |
Assumptions behind the numbers
- ΔG° values are for standard conditions at the stated temperature; real cells, furnaces and leaves are not at standard conditions.
- The reactions are truly coupled: the common intermediate is passed directly from one step to the other.
- A negative overall ΔG° says products are favoured, not that the reaction is fast.
- Values for ATP hydrolysis vary with conditions; questions will give you the value to use.
Mistakes to avoid
- Saying coupling changes ΔG° of the unfavourable step. It does not; it creates a new overall reaction.
- Adding K values. Multiply them.
- Forgetting to change the sign when you reverse an equation.
- Coupling without a link. Two reactions in separate beakers are not coupled; there must be a shared intermediate.
- Calling a catalyst a driving force. Catalysts change the rate, not ΔG°.
- Using kJ in K = e^(−ΔG°/RT) with R in J mol⁻¹ K⁻¹. Convert to J first.
Quick self-check
- Step 1 has ΔG° = +12.0 kJ mol⁻¹ and step 2 has ΔG° = −20.0 kJ mol⁻¹. They share an intermediate. What is ΔG°(overall), and is it favourable? (−8.0 kJ mol⁻¹; yes)
- For the same pair, what is ΔG° for the reverse of the overall reaction? (+8.0 kJ mol⁻¹)
- K₁ = 2.0 × 10⁻³ and K₂ = 4.0 × 10⁴. What is K(overall)? (80)
- Name the outside energy source that drives photosynthesis. (Light)
Next: practice questions.