Chapter 20: Spontaneity, Entropy, and Gibbs Free Energy

Learning Objectives

  • Define spontaneity and entropy; predict the sign of DeltaS\,Delta S associated with phase changes and chemical reactions.
  • Calculate entropy changes, DeltaS\,Delta S, for reactions using standard entropies or combining entropies for other reactions.
  • Calculate entropy change in a reaction, DeltaS\,Delta S, using stoichiometry.
  • Define Gibbs Free Energy by relating it to work and spontaneity.
  • Predict the sign of Gibbs Free Energy, DeltaG\,Delta G, for a reaction and calculate DeltaG\,Delta G from DeltaS\,Delta S and DeltaH\,Delta H.
  • Calculate the temperature at which a reaction switches spontaneity.
  • Calculate Gibbs Free Energy, DeltaG\,Delta G, from standard Gibbs Free Energies of formation and by combining Gibbs Free Energies of other reactions.
  • Calculate the Gibbs Free Energy of a reaction, DeltaG\,Delta G, by proportionality (using stoichiometry).

Spontaneity

  • Spontaneous Process: Occurs without ongoing external intervention; can be fast or slow.
  • Exothermic Reactions: Many, but not all, spontaneous reactions are exothermic; some are endothermic.
  • Temperature Dependence: Spontaneity can depend on temperature.
    • Example: Ice melting at room temperature is spontaneous but endothermic.
  • Recrystallization: Used to purify reaction mixtures by exploiting temperature-dependent solubility.
  • Phase Changes:
    • Melting: Not spontaneous at \,T < 0
    • Freezing: Not spontaneous at \,T > 0

Energy, Spontaneity, and Entropy

  • Entropy (S): Dispersal of energy through available microstates at a given temperature.
    • Formula: S=kBlnW\,S = k_B \,ln W
      • kB=1.38×1023\,k_B = 1.38 \times 10^{-23}
      • W\,W = # of microstates
    • Microstates: Possible locations of atoms or molecules; different locations = different ways to spread out kinetic energy (KE).
  • Entropy Change (ΔS\,\Delta S): Quantifies the dispersion of energy.
    • Formula: ΔS=qrevT\,\Delta S = \frac{q_{rev}}{T}
    • \,\Delta S_{system} > 0 indicates increasing molecular disorder.
  • Second Law of Thermodynamics: The entropy of the universe is always increasing.
    • More precisely, processes occur spontaneously when \,\Delta S_{universe} > 0.
    • Formula: ΔS<em>universe=ΔS</em>system+ΔSsurroundings\,\Delta S<em>{universe} = \Delta S</em>{system} + \Delta S_{surroundings}
    • \,\Delta S_{universe} > 0 for spontaneous processes.

Predicting the Sign of ΔS\,\Delta S: States of Matter

  • For slg\,s \,\,\rightarrow \,\, l \,\,\rightarrow \,\, g, \,\Delta S_{sys} > 0 (molecular disorder increases).
    • Liquids are more disordered than solids.
    • Gases are much more disordered due to high kinetic energy.
    • Changes in gas moles in a reaction indicate the sign of ΔS\,\Delta S.
    • S<em>gas>>S</em>sol\,S<em>{gas} >> S</em>{sol} and Sliq\,S_{liq} because gases have high KE.

Entropy and Disorder: A Macroscopic View

  • \,\Delta S_{system} > 0: Molecular disorder is increasing.
  • Dissolution: More disorder as solvated ions than in a well-ordered solid or liquid.
    • Putting more things in solution increases entropy.

Predicting the Sign of ΔS\,\Delta S: Molecular Structure

  • \,\Delta S_{system} > 0: Molecular disorder is increasing.
  • Molecular Complexity: More complex molecules (more atoms) have more potential for disorder (more possible configurations or microstates).
    • Propane has more configurations.
    • More atoms = more rotations or vibrations.
  • Atomic Structure: Atoms with more electrons usually have higher entropy (more E transitions, e.g., \,Na \,S < K \,S).

Entropy Factors

  • Moles of Gas: Increasing moles of gas increases entropy (for reaction in the forward direction).
  • Temperature: Increasing T = increasing S.
  • Volume: Increasing V = increasing S. Increasing volume allows more degrees of freedom/motion for the particles (more possibilities for different locations of particles = more miscrostates), which results in increasing entropy. Incr. T also incr. entropy because faster molecules move through more microstates

Entropy Practice

  • Gas entropy > solid entropy (more motion).
  • More moles of gas = more entropy (more possibilities of where the molecules can be = more disorder).

Predicting the Sign of ΔS\,\Delta S for Phase Changes

  • For the sublimation of dry ice at room temperature:
    • ΔS<em>CO</em>2\,\Delta S<em>{CO</em>2} is (+) because entropy increases when sg\,s \,\,\rightarrow g
    • ΔSuniverse\,\Delta S_{universe} is (+) because sublimation is spontaneous at room temp.
    • Below the sublimation temperature of dry ice, ΔSuniverse\,\Delta S_{universe} is (-) because sublimation is not spontaneous.

Third Law of Thermodynamics

  • @ 0 K, no molecular motion = no disorder.
    • Only one configuration if no motion.
    • Microstates W=1\,W = 1 and S=kBlnW\,S = k_B \,ln W
    • ln1=0\,ln 1 = 0, so S=0\,S = 0.

Calculating ΔSsyso\,\Delta S_{sys}^o

  • Can use Hess’s Law with ΔS\,\Delta S from several equations:
    • ΔS<em>rxno=Σn</em>iSo(products)<em>iΣm</em>jSo(reactants)j\,\Delta S<em>{rxn}^o = \,\Sigma \,n</em>i \,S^o (products)<em>i - \,\Sigma \,m</em>j \,S^o (reactants)_j
    • n and m are the stoichiometric coefficients for each reactant/product from the balanced chemical equation.
    • Degree symbol (naught) means standard conditions (1 atm gas, 1 M soln).
  • Absolute Entropies:
    • So\,S^o is the entropy gained by a substance as it is heated from 0 K to 298 K.
    • These are absolute entropies (the entropy gained by heating from 0 K). Since we can’t get colder (less ordered) than 0 K, these are not changes in entropy but absolute because they go from no entropy to a known amt.

Entropy Calculation Example

  • N2(g) + 3H2(g) → 2NH3(g) (4 gas moles → 2 gas moles, entropy decreases)
    • ΔSrxno=ΣnSo(products)ΣmSo(reactants)\,\Delta S_{rxn}^o = \,\Sigma \,n \,S^o (products) - \,\Sigma \,m \,S^o (reactants)
    • =[2(192.5)][191.6+3(130.7)]=198.7J/K\,= [2(192.5)] – [191.6 + 3(130.7)] = -198.7 \,J/K
    • Stoichiometry: If the question said you have 0.232 moles of NH3 and asked what the entropy change was, you need to use stoichiometry: 0.232molNH3(198.7J/K2molNH3)\,0.232 \,mol NH3 * (\frac{-198.7\,J/K}{2 \,mol NH3}) bc NH3 has 2 coeffic.
  • ΔS\,\Delta S depends on stoichiometry (proportionality).

Determining Spontaneity: Calculating ΔSuniverse\,\Delta S_{universe}

  • ΔS<em>universe=ΔS</em>system+ΔSsurroundings\,\Delta S<em>{universe} = \,\Delta S</em>{system} + \,\Delta S_{surroundings}
  • @ Const. P, q<em>sys=ΔH</em>sys\, -q<em>{sys} = -\,\Delta H</em>{sys}
  • If ΔS<em>surroundingsq</em>sys1T\,\Delta S<em>{surroundings} \,\propto \,-q</em>{sys} \,\propto \frac{1}{T}
  • Then ΔS<em>universe=ΔS</em>systemΔHsysT\,\Delta S<em>{universe} = \,\Delta S</em>{system} - \frac{\Delta H_{sys}}{T}
  • Rearranging: TΔS<em>universe=TΔS</em>sys+ΔHsys\,-T\,\Delta S<em>{universe} = -T\,\Delta S</em>{sys} + \,\Delta H_{sys}
  • Gibbs Free Energy (ΔG<em>sys\,\Delta G<em>{sys}): ΔG</em>sys=ΔH<em>sysTΔS</em>sys\,\Delta G</em>{sys} = \,\Delta H<em>{sys} - T\,\Delta S</em>{sys}
    • Another state function

Gibbs Free Energy (ΔG=ΔHTΔS\,\Delta G = \,\Delta H - T\,\Delta S)

  • \,\Delta G < 0 (negative) → spontaneous and product favored (in forward direction) → work done by system (engines), work = product
  • \,\Delta G > 0 (positive) → non-spontaneous and reactant favored (in forward direction) → no work done by system, work would need to be a reactant to make this happen

Spontaneity and Gibbs Free Energy

  • Gibbs Free Energy (ΔG\,\Delta G) and work: ΔG=wmax\,\Delta G = -w_{max}
    • Maximum work: ΔG\,\Delta G tells us how much work the system could theoretically do (for a spontaneous reaction).
    • Work is a product for a spontaneous reaction.
    • Work is a reactant for a non-spontaneous reaction (needs input of work).
  • Carnot cycle/engine: max efficiency process
    • Not possible in real reactions
    • All heat cannot be converted to work due to entropy

Gibbs Free Energy: Temperature Dependence


  • ΔGo=ΔHo+(TΔSo)\,\Delta G^o = \,\Delta H^o + (-T\,\Delta S^o)

ΔHo\,\Delta H^oΔSo\,\Delta S^oTΔSo\,-T\,\Delta S^o(ΔHo)+(TΔSo)=ΔGo\,(\Delta H^o) + (-T\,\Delta S^o) = \,\Delta G^o
+-+(+) + (+) = (+) always (+) → non-spontaneous @ any T\,T
++-(+) + (-) = (more (-) @ high T\,T) Spont. @ high T\,T, not @ low T\,T
--+(-) + (+) = (more (-) @ low T\,T) Spont. @ low T\,T, not @ high T\,T
-+-(-) + (-) = (-) always (-) → spontaneous @ all T\,T

Gibbs Free Energy: Crossover Temperature

  • \,\Delta G < 0 (neg) = spont
  • \,\Delta G > 0 (pos) = nonspont
  • at equilibrium (not spont or nonspont), ΔG=0\,\Delta G = 0 → crossover temp between spont and non.
  • ΔG=0\,\Delta G = 0 (equilibrium) examples: Boiling point, Melting point, Triple point, etc.
  • Crossover T=ΔHΔS\,T = \frac{\Delta H}{\Delta S}

Gibbs Free Energy: Reference Tables

  • Can also do Hess’s Law with ΔG\,\Delta G from several equations
  • Tables list ΔGf\,\Delta G_f values
  • Defined & used like ΔH<em>f\,\Delta H<em>f’s (ΔG</em>fo=0\Delta G</em>f^o = 0 for elements)
    • refer to formation reactions
    • same standard state idea (1 atm(g), 1 M(aq))
  • ΔG<em>rxno=Σn</em>iΔG<em>fo(prod)</em>iΣn<em>jΔG</em>fo(reactant)j\Delta G<em>{rxn}^o = \,\Sigma \,n</em>i \,\Delta G<em>f^o (prod)</em>i - \,\Sigma \,n<em>j \,\Delta G</em>f^o (reactant)_j