DPST1031 Chap.11 Entropy, Gibbs Energy and Spontaneity
Entropy, Gibbs Energy and Spontaneity
This chapter answers the question enthalpy cannot: will a change happen by itself? It introduces entropy as the dispersal of energy, states the second and third laws, predicts the sign of ΔS for phase changes, heating and changes in moles of gas, and calculates reaction entropies from standard molar entropies.
Entropy changes of the system and surroundings are combined into ΔS of the universe and then into the Gibbs energy, ΔG = ΔH − TΔS, with its four sign combinations and crossover temperatures. Finally ΔG° is linked to the equilibrium constant, and the van 't Hoff equation estimates K at a new temperature, joining thermodynamics back to equilibrium.
What this chapter covers
- 01
Entropy as energy dispersal
- 02
The second and third laws
- 03
Predicting the sign of ΔS
- 04
Standard reaction entropy from tables
- 05
Entropy of the surroundings and universe
- 06
Gibbs energy and spontaneity
- 07
Crossover temperatures
- 08
ΔG°, K and the van 't Hoff equation
Worked example · free
Temperature limit for ammonia synthesis
- +1Convert ΔS° to kJ: −198.7 J K⁻¹ = −0.1987 kJ K⁻¹.
- +1ΔG°(298) = −92.2 − 298 × (−0.1987) = −92.2 + 59.2 = −33.0 kJ, so it is spontaneous at 25 °C.
- +1Both ΔH° and ΔS° are negative, so ΔG° becomes positive at high temperature.
- +1Crossover: T = ΔH° ÷ ΔS° = −92 200 J ÷ −198.7 J K⁻¹ = 464 K.
Key terms
- Entropy
- A state function S that measures how widely energy is dispersed among a system's particles and arrangements.
- Second law of thermodynamics
- The law that the entropy of the universe increases in every spontaneous process.
- Third law of thermodynamics
- The law that a perfect crystal has zero entropy at absolute zero.
- Standard molar entropy
- The absolute entropy S° of one mole of a substance in its standard state.
- Spontaneous process
- A change that proceeds without continuous outside intervention, whatever its rate.
- Crossover temperature
- The temperature T = ΔH ÷ ΔS at which ΔG changes sign.
Entropy, Gibbs Energy and Spontaneity FAQ
Why can an endothermic reaction be spontaneous?
Spontaneity depends on ΔG = ΔH − TΔS, not on ΔH alone. If an endothermic reaction has a large enough positive ΔS, the −TΔS term outweighs ΔH at high temperature and ΔG becomes negative. Melting ice above 0 °C and decomposing limestone in a kiln are examples.
How do I predict the sign of ΔS without data?
Look first at the moles of gas: more gas molecules among the products usually means a positive ΔS. Melting, boiling, dissolving and heating also increase entropy, while freezing, condensing and forming fewer particles decrease it. Use data only when these clues balance out.
How are ΔG° and K related?
The data sheet gives ΔG° = −RT ln K. A negative ΔG° gives K greater than 1, favouring products; a positive ΔG° gives K less than 1. Use R = 8.314 J K⁻¹ mol⁻¹, so ΔG° must be in joules, not kilojoules, in this equation.
Does a negative ΔG mean the reaction is fast?
No. ΔG says whether a reaction can proceed by itself, not how quickly. Diamond converting to graphite has a negative ΔG° at room temperature but is immeasurably slow. Speed is controlled by the activation barrier, which thermodynamics does not describe.
Why does an element have a non-zero standard entropy?
Standard entropies are absolute values measured from the third-law zero at 0 K. At 298 K every substance, including an element, has energy dispersed among its particles, so its entropy is positive. Formation enthalpies use a different convention, which sets elements in their standard states to zero.
Exam move
Start with sign predictions: write ΔS as positive or negative with a reason for twenty processes before calculating anything. Then calculate ΔS° from tables and combine it with ΔH° to give ΔG° at two temperatures, always converting entropy to kilojoules. Classify reactions into the four sign cases and find crossover temperatures for the two mixed cases.
Finally, practise converting between ΔG° and K in both directions and use the van 't Hoff equation to move K from one temperature to another, checking each answer against Le Chatelier's prediction for an exothermic or endothermic reaction.
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