Use the Gibbs equation
Calculate Gibbs free-energy change from enthalpy, temperature and entropy using consistent units.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 23. Use the syllabus sign convention for lattice enthalpy (gaseous ions forming the ionic solid); state phases, signs, units and temperature. Written practice is self-checked
Before you begin
- Use AS enthalpy changes and Hess's law, formulae for ions, balanced equations, and temperature in kelvins.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which temperature scale is used in ΔG° = ΔH° − TΔS°?
What you will learn
- State and use ΔG° = ΔH° − TΔS°
- Convert entropy units before substitution
- Use kelvins and retain units for the reaction as written
Entropy is multiplied by absolute temperature
The Gibbs equation is ΔG° = ΔH° − TΔS°. Use temperature in kelvins. The equation combines the enthalpy and entropy terms for the same reaction under the stated conditions.
Check units before calculating: if ΔH is in kJ mol⁻¹ and ΔS is in J K⁻¹ mol⁻¹, convert ΔS to kJ K⁻¹ mol⁻¹ or convert ΔH to joules. Keep the reaction basis consistent.
The entropy term changes with temperature
The contribution TΔS grows in magnitude as temperature increases. Its effect on ΔG depends on the sign of ΔS because the equation subtracts TΔS.
A common error is to enter Celsius values or add TΔS. Convert to kelvins, calculate the signed product, and then subtract it from ΔH.
Pause and explain
If ΔS° is in J K⁻¹ mol⁻¹ and ΔH° is in kJ mol⁻¹, what must you do?
Worked example
Calculate ΔG° at 298 K if ΔH° = +40.0 kJ mol⁻¹ and ΔS° = +100 J K⁻¹ mol⁻¹.
Show the worked solution
- Convert ΔS° to 0.100 kJ K⁻¹ mol⁻¹.
- Calculate TΔS° = 298 × 0.100 = 29.8 kJ mol⁻¹.
- Subtract from ΔH°.
Answer ΔG° = 40.0 − 29.8 = +10.2 kJ mol⁻¹.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate ΔG° at 298 K for ΔH° = −25.0 kJ mol⁻¹ and ΔS° = −50 J K⁻¹ mol⁻¹.
Give me a hint
Convert entropy to kJ K⁻¹ mol⁻¹, multiply by T, then subtract.
Compare my reasoning
- ΔS° = −0.050 kJ K⁻¹ mol⁻¹.
- TΔS° = 298×(−0.050)=−14.9 kJ mol⁻¹.
- ΔG° = −25.0−(−14.9).
ΔG° = −10.1 kJ mol⁻¹.
Look for these in your work
- Unit conversion correct
- Signed TΔS calculated
- Subtraction and result correct
For ΔH° = +30.0 kJ mol⁻¹ and ΔS° = +75 J K⁻¹ mol⁻¹, calculate ΔG° at 400 K.
Give me a hint
Convert entropy to kJ K⁻¹ mol⁻¹ first.
Compare my reasoning
- ΔS° = 0.075 kJ K⁻¹ mol⁻¹.
- TΔS° = 400×0.075 = 30.0 kJ mol⁻¹.
- Subtract from ΔH°.
ΔG° = 30.0−30.0 = 0 kJ mol⁻¹ at 400 K.
Look for these in your work
- Entropy converted
- Temperature product correct
- Zero result interpreted numerically
A student calculates ΔG° = −20 − 298(−40) = +11 900 using ΔH° in kJ mol⁻¹ and ΔS° in J K⁻¹ mol⁻¹. Identify two errors.
Give me a hint
Check both the energy units and the sign of the product before interpreting the magnitude.
Compare my reasoning
- Convert −40 J K⁻¹ mol⁻¹ to −0.040 kJ K⁻¹ mol⁻¹.
- Then TΔS° = −11.92 kJ mol⁻¹.
- Subtracting a negative gives −20−(−11.92), not a value in thousands.
The entropy was not converted from joules to kilojoules, and the subtraction must retain the negative ΔS: ΔG° = −20 − (−11.92) = −8.08 kJ mol⁻¹ at 298 K.
Look for these in your work
- Unit error identified
- Negative entropy retained
- Corrected value and unit stated
Common mistakes
- Using Celsius instead of kelvins
- Combining joules with kilojoules without conversion
- Adding rather than subtracting TΔS
- Dropping the sign of ΔS
What can you explain now?
Which temperature scale is used in ΔG° = ΔH° − TΔS°?
Compare with the explanation
Kelvins
The equation uses absolute temperature.
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Make it stick
Before evaluating ΔG°, how do you check temperature, units and the sign of the TΔS term?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.