Predict feasibility from Gibbs energy and temperature
Use the sign of Gibbs energy to assess feasibility under stated conditions and predict how temperature changes the result.
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
What does ΔG < 0 indicate for the forward reaction under the stated conditions?
What you will learn
- Use the sign of ΔG to assess feasibility under stated conditions
- Predict temperature effects from the signs of ΔH and ΔS
- Distinguish feasibility from reaction rate and observed completion
The sign of ΔG indicates feasibility, not speed
For the conditions considered, ΔG < 0 indicates a feasible change, ΔG > 0 indicates it is not feasible in the forward direction, and ΔG = 0 indicates equilibrium. This thermodynamic test does not predict how fast the reaction occurs.
A feasible reaction may be slow, and a reaction can appear incomplete because it reaches equilibrium. Do not use ΔG alone to claim a reaction is rapid, quantitative or safe.
Combine enthalpy and entropy signs to predict temperature effects
If ΔH < 0 and ΔS > 0, both terms favour feasibility at all temperatures. If ΔH > 0 and ΔS < 0, neither favours it. If ΔH < 0 and ΔS < 0, lower temperature favours feasibility; if ΔH > 0 and ΔS > 0, higher temperature favours feasibility.
These patterns follow ΔG = ΔH − TΔS. At a temperature where ΔG = 0, the system is at the boundary; use consistent units if calculating that temperature.
Pause and explain
When ΔH > 0 and ΔS > 0, which temperature direction can favour feasibility?
Worked example
A reaction has ΔH° = +40 kJ mol⁻¹ and ΔS° = +100 J K⁻¹ mol⁻¹. Is it feasible at 298 K and at 500 K?
Show the worked solution
- Convert ΔS° to +0.100 kJ K⁻¹ mol⁻¹.
- At 298 K, ΔG° = 40−29.8=+10.2.
- At 500 K, ΔG° = 40−50=−10.
Answer It is not feasible in the forward direction at 298 K (ΔG° = +10.2 kJ mol⁻¹) but is feasible at 500 K (ΔG° = −10 kJ mol⁻¹), under standard conditions.
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Test this topic →From guided practice to a new situation
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.
A reaction has ΔH° = −55 kJ mol⁻¹ and ΔS° = −120 J K⁻¹ mol⁻¹. Predict whether lower or higher temperature favours feasibility.
Give me a hint
Use ΔG = ΔH−TΔS and notice the sign of the entropy term.
Compare my reasoning
- ΔH is negative and favours the reaction.
- ΔS is negative, so −TΔS is positive.
- Increasing T makes the unfavourable positive term larger; lower T favours feasibility.
Lower temperature favours feasibility because it reduces the positive −TΔS term while ΔH remains negative.
Look for these in your work
- Both signs identified
- TΔS sign interpreted
- Temperature direction correct
A reaction has ΔH° = +25 kJ mol⁻¹ and ΔS° = +80 J K⁻¹ mol⁻¹. Calculate the crossover temperature at which ΔG° = 0.
Give me a hint
Set ΔH° = TΔS° and convert entropy to kJ K⁻¹ mol⁻¹.
Compare my reasoning
- ΔS° = 0.080 kJ K⁻¹ mol⁻¹.
- T = ΔH°/ΔS° = 25/0.080.
- Report kelvins and state the limiting condition.
T = 312.5 K. At this temperature ΔG° = 0; above it, ΔG° becomes negative for these values.
Look for these in your work
- Equation set to zero
- Units made consistent
- Temperature and interpretation correct
A learner says a reaction with ΔG° < 0 is guaranteed to finish instantly. Correct the claim.
Give me a hint
Separate thermodynamic direction from reaction rate and equilibrium extent.
Compare my reasoning
- Negative ΔG° indicates thermodynamic feasibility under the stated conditions.
- It does not give an activation barrier or reaction rate.
- A reaction can be slow or establish equilibrium before complete conversion.
ΔG° < 0 indicates that the forward change is thermodynamically feasible under standard conditions. It does not imply an instant reaction or guarantee complete conversion; kinetics and equilibrium matter.
Look for these in your work
- Feasibility stated
- Rate distinction made
- Completion claim limited
Common mistakes
- Equating feasibility with reaction rate
- Calling a positive ΔG feasible in the forward direction
- Ignoring the sign of ΔS when predicting temperature effects
- Using Celsius in the Gibbs equation
What can you explain now?
What does ΔG < 0 indicate for the forward reaction under the stated conditions?
Compare with the explanation
It is thermodynamically feasible
A negative Gibbs change favours the forward change.
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Make it stick
How does the sign of ΔG establish feasibility, and how do ΔH, ΔS and temperature determine the balance?
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.