1. Does a negative ΔG guarantee a fast reaction?
- Yes
- No
- Only at room temperature
Check the answer and reasoning
No
A large activation barrier can make a thermodynamically feasible reaction slow.
CHEM / A2 level
Entropy, Gibbs energy and ionic energetics
Entropy describes the distribution of energy among accessible arrangements. Gibbs energy combines enthalpy and entropy to assess thermodynamic feasibility at a given temperature.
Learning objectives
- Explain entropy, gibbs energy and ionic energetics using a concrete example.
- Apply the worked method and justify each step.
- Recognise the limitation described in the final check.
The big idea
Entropy describes the distribution of energy among accessible arrangements. Gibbs energy combines enthalpy and entropy to assess thermodynamic feasibility at a given temperature.
Born–Haber cycles connect lattice, ionisation and electron-affinity terms. Keep a consistent lattice-enthalpy sign convention and distinguish thermodynamic feasibility from reaction speed.
Equations and relationships
Use kelvin and consistent energy units.
Worked example
ΔH=20 kJ/mol, ΔS=100 J/(mol K), T=300 K. Find ΔG.
Follow the solution
Convert ΔS to 0.100 kJ/(mol K); 20−300×0.100=−10.
Investigate and explain
Predict the worked result before calculating. Change one input, record the new result, and explain which assumptions still hold.
Use the evidence activity below; no exact PhET model is claimed.
Check your understanding
2. ΔH=20 kJ/mol, ΔS=100 J/(mol K), T=300 K. Find ΔG.
- −29980 kJ/mol
- 50 kJ/mol
- −10 kJ/mol
Check the answer and reasoning
−10 kJ/mol
Convert ΔS to 0.100 kJ/(mol K); 20−300×0.100=−10.
3. Why convert joules to kilojoules in the example?
- To keep energy units consistent
- To change the physical result
- Because kelvin is optional
Check the answer and reasoning
To keep energy units consistent
Subtracting unlike energy units gives a numerical error, even when the equation is correct.