Chapter 20: Spontaneity, Entropy, and Gibbs Free Energy

Learning Objectives

• Define spontaneity and entropy; predict the sign of $\,Delta S$ associated with phase changes and chemical reactions.
• Calculate entropy changes, $\,Delta S$, for reactions using standard entropies or combining entropies for other reactions.
• Calculate entropy change in a reaction, $\,Delta S$, using stoichiometry.
• Define Gibbs Free Energy by relating it to work and spontaneity.
• Predict the sign of Gibbs Free Energy, $\,Delta G$, for a reaction and calculate $\,Delta G$ from $\,Delta S$ and $\,Delta H$.
• Calculate the temperature at which a reaction switches spontaneity.
• Calculate Gibbs Free Energy, $\,Delta G$, from standard Gibbs Free Energies of formation and by combining Gibbs Free Energies of other reactions.
• Calculate the Gibbs Free Energy of a reaction, $\,Delta G$, 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: $\,S = k_B \,ln W$
    • $\,k_B = 1.38 \times 10^{-23}$
    • $\,W$ = # of microstates
  • Microstates: Possible locations of atoms or molecules; different locations = different ways to spread out kinetic energy (KE).
• Entropy Change ($\,\Delta S$): Quantifies the dispersion of energy.
  • Formula: $\,\Delta S = \frac{q_{rev}}{T}$
  • $\,\Delta S_{system} > 0$ 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: $\,\Delta S<em>{universe} = \Delta S</em>{system} + \Delta S_{surroundings}$
  • $\,\Delta S_{universe} > 0$ for spontaneous processes.

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

• For $\,s \,\,\rightarrow \,\, l \,\,\rightarrow \,\, g$, $\,\Delta S_{sys} > 0$ (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 $\,\Delta S$.
  • $\,S<em>{gas} >> S</em>{sol}$ and $\,S_{liq}$ 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 $\,\Delta S$: 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 $\,\Delta S$ for Phase Changes

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

Third Law of Thermodynamics

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

Calculating $\,\Delta S_{sys}^o$

• Can use Hess’s Law with $\,\Delta S$ from several equations:
  • $\,\Delta S<em>{rxn}^o = \,\Sigma \,n</em>i \,S^o (products)<em>i - \,\Sigma \,m</em>j \,S^o (reactants)_j$
  • 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:
  • $\,S^o$ 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)
  • $\,\Delta S_{rxn}^o = \,\Sigma \,n \,S^o (products) - \,\Sigma \,m \,S^o (reactants)$
  • $\,= [2(192.5)] – [191.6 + 3(130.7)] = -198.7 \,J/K$
  • Stoichiometry: If the question said you have 0.232 moles of NH3 and asked what the entropy change was, you need to use stoichiometry: $\,0.232 \,mol NH3 * (\frac{-198.7\,J/K}{2 \,mol NH3})$ bc NH3 has 2 coeffic.
• $\,\Delta S$ depends on stoichiometry (proportionality).

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

• $\,\Delta S<em>{universe} = \,\Delta S</em>{system} + \,\Delta S_{surroundings}$
• @ Const. P, $\, -q<em>{sys} = -\,\Delta H</em>{sys}$
• If $\,\Delta S<em>{surroundings} \,\propto \,-q</em>{sys} \,\propto \frac{1}{T}$
• Then $\,\Delta S<em>{universe} = \,\Delta S</em>{system} - \frac{\Delta H_{sys}}{T}$
• Rearranging: $\,-T\,\Delta S<em>{universe} = -T\,\Delta S</em>{sys} + \,\Delta H_{sys}$
• Gibbs Free Energy ($\,\Delta G<em>{sys}$): $\,\Delta G</em>{sys} = \,\Delta H<em>{sys} - T\,\Delta S</em>{sys}$
  • Another state function

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

• $\,\Delta G < 0$ (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 ($\,\Delta G$) and work: $\,\Delta G = -w_{max}$
  • Maximum work: $\,\Delta G$ 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

• $\,\Delta G^o = \,\Delta H^o + (-T\,\Delta S^o)$
  

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

Gibbs Free Energy: Reference Tables

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