AP Chemistry · Lesson 1

Entropy and Gibbs free energy

Know Know it

Entropy measures how spread out a system’s matter and energy are. At the particle level, entropy rises when particles get more freedom to move or more ways to arrange themselves: a solid melts or dissolves, a liquid boils, a gas expands, or a reaction makes more moles of gas. Gases have far more entropy than liquids or solids, so the count of gas moles usually decides the sign of ΔS∘\Delta S^\circ.

To calculate it, use standard molar entropies: ΔS∘=∑S∘(products)−∑S∘(reactants)\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants}), each multiplied by its coefficient. Entropies are listed in J/(mol·K), so the answer comes out in J/(mol·K).

Gibbs free energy combines enthalpy and entropy: ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. If ΔG∘\Delta G^\circ is negative, the process is thermodynamically favorable. Watch the units: change ΔS∘\Delta S^\circ to kJ/(mol·K) (divide by 1000) before you combine it with ΔH∘\Delta H^\circ in kJ/mol, and use T in kelvins.

How the signs of enthalpy and entropy decide favorability
ΔH∘\Delta H^\circΔS∘\Delta S^\circSign of ΔG∘\Delta G^\circFavorable when
negativepositivealways negativeat every temperature
positivenegativealways positivenever
negativenegativenegative at low Tbelow about ΔH∘/ΔS∘\Delta H^\circ/\Delta S^\circ
positivepositivenegative at high Tabove about ΔH∘/ΔS∘\Delta H^\circ/\Delta S^\circ

Calculator On the exam

On the exam you may use a scientific or graphing calculator on both sections (a four-function calculator is allowed but is a poor choice). You also get a reference booklet of chemistry equations and constants. Check that ΔS∘\Delta S^\circ is in kJ/(mol·K) before you multiply it by T, and use exe^x for K from ΔG∘\Delta G^\circ.

Worked example Easy

Which process has a negative entropy change, ΔS∘<0\Delta S^\circ < 0?

  1. Solid ammonium nitrate dissolving in water
  2. 2NH3(g)→N2(g)+3H2(g)2\text{NH}_{3}(g) \rightarrow \text{N}_{2}(g) + 3\text{H}_{2}(g)
  3. CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_{3}(s) \rightarrow \text{CaO}(s) + \text{CO}_{2}(g)
  4. 2SO2(g)+O2(g)→2SO3(g)2\text{SO}_{2}(g) + \text{O}_{2}(g) \rightarrow 2\text{SO}_{3}(g)Answer

How to solve it

  1. Count the moles of gas on each side: the side with more gas has more entropy.
  2. For the sulfur trioxide reaction, 2 + 1 = 3 mol of gas becomes 2 mol, so ΔS∘\Delta S^\circ is negative.
  3. Dissolving a solid and making a gas from a solid both raise the entropy.

Why each choice is right or wrong

  • A. Incorrect. Dissolving a crystal frees its ions to move through the solution, so the entropy rises.
  • B. Incorrect. This reaction goes from 2 mol of gas to 4 mol of gas, so the entropy rises.
  • C. Incorrect. This reaction makes a gas, carbon dioxide, from a solid, so the entropy rises.
  • D. Correct. This reaction goes from 3 mol of gas to 2 mol of gas, so the particles have fewer ways to arrange themselves and the entropy falls.

Trap Favorable does not mean fast

A negative ΔG∘\Delta G^\circ says a process can happen on its own, not how quickly. A favorable reaction with a high activation energy can sit for years without visible change. Speed is kinetics (Unit 5); favorability is thermodynamics.

Explain Explain it

Task verbs: calculate, justify

When ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ have the same sign, they pull in opposite directions, and the TΔS∘T\Delta S^\circ term grows with temperature. So the temperature decides: an exothermic reaction that loses entropy is favorable only at low temperatures, and an endothermic reaction that gains entropy is favorable only at high temperatures. The crossover is near T=ΔH∘/ΔS∘T = \Delta H^\circ/\Delta S^\circ (with matching units).

Free energy also sets the equilibrium constant: ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K, with R = 8.314 J/(mol·K) and ΔG∘\Delta G^\circ in joules. A negative ΔG∘\Delta G^\circ gives K greater than 1 (products favored at equilibrium); a positive one gives K less than 1.

Worked example Medium

A reaction has ΔH∘=−120\Delta H^\circ = -120 kJ/mol and ΔS∘=−150\Delta S^\circ = -150 J/(mol·K). What is ΔG∘\Delta G^\circ at 298 K, and is the reaction thermodynamically favorable at that temperature?

  1. -165 kJ/mol; favorable
  2. -75.3 kJ/mol; favorableAnswer
  3. +44,580 kJ/mol; not favorable
  4. +75.3 kJ/mol; not favorable

How to solve it

  1. Convert: ΔS∘=−150\Delta S^\circ = -150 J/(mol·K) = -0.150 kJ/(mol·K).
  2. TΔS∘=(298)(−0.150)=−44.7T\Delta S^\circ = (298)(-0.150) = -44.7 kJ/mol.
  3. ΔG∘=−120−(−44.7)=−75.3\Delta G^\circ = -120 - (-44.7) = -75.3 kJ/mol, negative, so the reaction is favorable at 298 K.

Why each choice is right or wrong

  • A. Incorrect. -165 kJ/mol subtracts TΔS∘T\Delta S^\circ with the wrong sign; minus a negative TΔS∘T\Delta S^\circ adds 44.7 kJ/mol to -120.
  • B. Correct. ΔG∘=−120−(298)(−0.150)=−120+44.7=−75.3\Delta G^\circ = -120 - (298)(-0.150) = -120 + 44.7 = -75.3 kJ/mol, which is negative, so the reaction is favorable at 298 K.
  • C. Incorrect. +44,580 kJ/mol multiplies T by ΔS∘\Delta S^\circ in J/(mol·K) without converting to kJ, so the units do not match.
  • D. Incorrect. +75.3 kJ/mol has the right size but the wrong sign; -120 + 44.7 is negative.

Justify Your turn

Solid ammonium chloride decomposes: NH4Cl(s)→NH3(g)+HCl(g)\text{NH}_{4}\text{Cl}(s) \rightarrow \text{NH}_{3}(g) + \text{HCl}(g), with ΔH∘=+176\Delta H^\circ = +176 kJ/mol and ΔS∘=+285\Delta S^\circ = +285 J/(mol·K). Say whether the reaction is thermodynamically favorable at 298 K, supporting your answer with a calculation, and then say whether raising the temperature makes it more or less favorable.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Calculates ΔG∘\Delta G^\circ at 298 K, about +91 kJ/mol
  • Says the reaction is not favorable at 298 K because ΔG∘\Delta G^\circ is positive
  • Says that raising the temperature makes the reaction more favorable

Model answer

ΔG∘=176−(298)(0.285)=176−84.9=+91.1\Delta G^\circ = 176 - (298)(0.285) = 176 - 84.9 = +91.1 kJ/mol. The value is positive, so the reaction is not thermodynamically favorable at 298 K. Because ΔS∘\Delta S^\circ is positive, the −TΔS∘-T\Delta S^\circ term becomes more negative as the temperature rises, so the reaction becomes more favorable at higher temperature; it turns favorable above about 176/0.285=618176/0.285 = 618 K.

Apply Apply it

Practice this topic. AP Practice questions on exactly what this lesson covers, with an explanation for every answer.

Practice this topic