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29 Cards in this Set
- Front
- Back
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Spontaneous (product-favored) reaction
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a reaction that takes place on its own without an external force and entropy of universe increases
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A Spontaneous reaction is _____-favored
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A Spontaneous reaction is _product_-favored
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Energy becomes more dispersed when a system consisting of atoms or molecules ________ to occupy a _____ volume
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Energy becomes more dispersed when a system consisting of atoms or molecules _expands_ to occupy a _larger_ volume
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For a process that takes place at constant temerature and pressure, the entropy change can be calculated by . . .
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dividing the thermal energy transferred, q(reverse), by the absolute temperature, T. ΔS = S(final) - S(initial) = q(rev)/T
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For a constant-pressure process, thermal energy transfer, q(reverse), is the same as . . .
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. . . the enthalpy change, ΔHº
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Entropies of gases are ____ than liquids
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Entropies of gases are _larger_ than liquids
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Entropies of liquids are ____ than solids
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Entropies of liquids are _larger_ than solids
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Entropies of more complex molecules are _____ than those of simpler molecules
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Entropies of more complex molecules are _larger_ than those of simpler molecules
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Entropies of ionic solids that have similar formulas are ______ when the attractions among the ions are ______
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Entropies of ionic solids that have similar formulas are _larger_ when the attractions among the ions are _weaker_
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Entropy _____ when a pure liquid or solid dissolves in a solvent
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Entropy _increases_ when a pure liquid or solid dissolves in a solvent
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Entropy ________ when a gas dissolves in a liquid
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Entropy _decreases_ when a gas dissolves in a liquid
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Entropy Change - formula
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ΔSº = Σ {(moles product) x Sº(product)} - Σ {(moles reactant) x Sº(reactant)}
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Second law of thermodynamics
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The total entropy of the universe (a system plus its surroundings) is continually increasing
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Steps to predicting whether a reaction is product-favored:
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1. Calculate how much entropy changes as a result of transfer of energy between system and surroundings (ΔSsurroundings)
2. Calculate how much entropy changes as a result of dispersal of energy within the system (ΔSsystem) 3. Add these two results to get ΔSuniverse = ΔSsystem + ΔSsurroundings |
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ΔSºuniverse =
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-ΔHºsystem/T + ΔSºsystem
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A reaction is certain to be product-favored is it is ___thermic and the entropy of the products is ______ than the reactants
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A reaction is certain to be product-favored is it is _exo_thermic and the entropy of the products is _greater_ than the reactants
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A reaction is certainly not product-favored is it is ___thermic and the entropy of the products is ______ than the reactants
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A reaction is certainly not product-favored is it is _endo_thermic and the entropy of the products is _less_ than the reactants
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ΔHsys | ΔSsys - product favored?
Negative | Positive |
Yes (ΔGº = ΔHº-ΔSº*T)
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ΔHsys | ΔSsys - product favored?
Negative | Negative |
Yes, for low T (ΔGº = ΔHº-ΔSº*T)
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ΔHsys | ΔSsys - product favored?
Positive | Positive |
Yes, for large T (ΔGº = ΔHº-ΔSº*T)
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ΔHsys | ΔSsys - product favored?
Positive | Negative |
No (ΔGº = ΔHº-ΔSº*T)
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ΔSº =
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-ΔHº / T
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Gibbs Free Energy - equation
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ΔGsystem = -T * ΔSuniverse
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Gibbs Free Energy - meaning
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Energy available to do work. A decrease in Gibbs Free Energy in a system is characteristic of a process that is product-favored at constant temperature & Pressure
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Free Energy
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Energy available to do work
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ΔGº =
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Σ {(moles product) x ΔGfº(product)} - Σ {(moles reactant) x ΔGfº(reactant)}
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When ΔGº = 0, the system is . . .
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. . . at equilibrium
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3rd law of thermodynamics
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In a perfect crystal at 0 K, entropy is 0
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(non-standard state) ΔG =
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ΔGº + RT ln(Q)
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