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{universe} = \Delta S{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{gas} >> S{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{CO2} 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{rxn}^o = \,\Sigma \,ni \,S^o (products)i - \,\Sigma \,mj \,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{universe} = \,\Delta S{system} + \,\Delta S_{surroundings}
- @ Const. P, \, -q{sys} = -\,\Delta H{sys}
- If \,\Delta S{surroundings} \,\propto \,-q{sys} \,\propto \frac{1}{T}
- Then \,\Delta S{universe} = \,\Delta S{system} - \frac{\Delta H_{sys}}{T}
- Rearranging: \,-T\,\Delta S{universe} = -T\,\Delta S{sys} + \,\Delta H_{sys}
- Gibbs Free Energy (\,\Delta G{sys}): \,\Delta G{sys} = \,\Delta H{sys} - T\,\Delta S{sys}
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 H^o | \,\Delta S^o | \,-T\,\Delta S^o | \,(\Delta H^o) + (-T\,\Delta S^o) = \,\Delta G^o |
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| + | - | + | (+) + (+) = (+) always (+) → non-spontaneous @ any \,T |
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| + | + | - | (+) + (-) = (more (-) @ high \,T) Spont. @ high \,T, not @ low \,T |
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| - | - | + | (-) + (+) = (more (-) @ low \,T) Spont. @ low \,T, not @ high \,T |
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| - | + | - | (-) + (-) = (-) always (-) → spontaneous @ all \,T | |
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Gibbs Free Energy: Crossover Temperature | | | | |
- \,\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 Hf’s (\Delta Gf^o = 0 for elements)
- refer to formation reactions
- same standard state idea (1 atm(g), 1 M(aq))
- \Delta G{rxn}^o = \,\Sigma \,ni \,\Delta Gf^o (prod)i - \,\Sigma \,nj \,\Delta Gf^o (reactant)_j