Chapter 17 Flashcards

Chapter 17: Entropy, Gibbs Energy, and Equilibrium

First Law of Thermodynamics

• You can't win!
• Energy cannot be created or destroyed.
• The total energy of the universe remains constant.
• Energy can be transferred between locations.
• \Delta E{universe} = 0 = \Delta E{system} + \Delta E_{surroundings}
• For exothermic reactions, the heat lost by the system is gained by the surroundings.
• Energy is converted to heat (q) and used to do work (w).

Spontaneous Processes

• Thermodynamics predicts whether a process will occur under specific conditions.
• Spontaneity is determined by comparing the system's free energy before and after a reaction.
• A reaction is thermodynamically favorable if the system has less free energy after the reaction.
• Processes spontaneous in one direction do not occur spontaneously in the opposite direction under the same conditions.
• Example: Sugar dissolving in tea.

Spontaneity: Kinetics vs. Thermodynamics

• Thermodynamics deals with initial and final states, determining spontaneity.
• Kinetics deals with intermediate states and the speed of the reaction.
• Example: Diamond converting to graphite is spontaneous but slow.

Factors Determining Spontaneity

• Enthalpy ($\Delta H$): Compares bond energies of reactants and products.
  • Bond energy: Energy needed to break a bond.
• Entropy ($\Delta S$): Relates to the randomness/orderliness of a system.
• Enthalpy is generally more important than entropy in determining spontaneity.

Enthalpy ($\Delta H$)

• Related to internal energy.
• Stronger bonds = more stable molecules.
• Exothermic: Energy is released ($\Delta H$ = negative).
• Endothermic: Energy is absorbed ($\Delta H$ = positive).

Entropy (S)

• Thermodynamic function that increases with the number of energetically equivalent arrangements.
• S = k \ln W
  • k = Boltzmann Constant = 1.38 x 10^{-23} J/K
  • W = number of energetically equivalent ways (microstates).
• Random systems require less energy than ordered systems.

Entropy and Microstates (W)

• Macrostate A and B can only be achieved through one possible arrangement of particles.
• State C has 6 possible arrangements.
• Therefore State C has higher entropy than either State A or B

Changes in Entropy ($\Delta S$)

• Entropy change is favorable when resulting system is more random ($\Delta S$ is positive).
• Examples of entropy increases:
  • Reactions with products in a more disordered state (solid < liquid < gas).
  • Reactions with more product molecules than reactant molecules.
  • Solids dissolving into ions (though not always positive).
  • Increasing temperature.

Entropy and Changes of State

• The number of microstates changes when materials change state.
• Entropy order: solid < liquid < gas
• This is because the degrees of freedom of motion increases.

Entropy and Molecular Motion

• Molecules can rotate, vibrate, and have translational energy.
• These motions increase with temperature, leading to higher entropy.

Entropy and Dissolution

• Dissolution of a salt generally increases entropy.
• NaCl(s) \rightarrow NaCl(aq)

Entropy and Dissolution (cont.)

• More particles in solution lead to greater possible arrangements (W).
• Hydration of ions can order water molecules, resulting in a negative entropy change.

Predicting the Sign of Entropy Change ($\Delta S$)

• Examples:
  • CO2(g) \rightarrow CO2(l): Negative (gas to liquid)
  • Solid iodine sublimes: Positive (solid to gas)
  • 2 N2O(g) \rightarrow 2 N2(g) + O_2(g): Positive (2 gas molecules to 3 gas molecules)
  • Dissolving sugar in water: Generally positive (solid to aqueous)
  • Cooling nitrogen gas from 80°C to 20°C: Negative (decreasing temperature)

The 2nd Law of Thermodynamics

• The total entropy change of the universe must be positive for a spontaneous process.
  • Reversible process: \Delta S_{universe} = 0
  • Irreversible (spontaneous) process: \Delta S_{universe} > 0
  • \Delta S{universe} = \Delta S{system} + \Delta S_{surroundings}
• If the system's entropy decreases, the surroundings' entropy must increase by a larger amount.
• When \Delta S{system} is negative, \Delta S{surroundings} is positive.

Heat Flow, Entropy, and the 2nd Law

• Heat must flow from water to ice for the universe's entropy to increase.
• The increase in \Delta S_{surroundings} often comes from the heat released in an exothermic reaction.

Temperature Dependence of \Delta S_{surroundings}

• Exothermic process (negative q_{system}) adds heat to surroundings, increasing their entropy.
• Endothermic process takes heat from surroundings, decreasing their entropy.
• The change in surroundings' entropy is proportional to the magnitude of q_{system}.

Temperature Dependence of \Delta S_{surroundings} (cont.)

• Under constant pressure: q{system} = \Delta H{system}
• Therefore: \Delta S{surroundings} = -\frac{\Delta H{sys}}{T}
• As temperature increases, a given negative enthalpy produces a smaller positive \Delta S_{surroundings}.

Calculating Entropy Changes in Surroundings

• Reaction: C3H8(g) + 5 O2(g) \rightarrow 3 CO2(g) + 4 H_2O(g)
• \Delta H_{rxn} = -2044 kJ at 25°C
• Calculate the entropy change of the surroundings: 6860 J/K

Gibbs Free Energy

• G = maximum energy from the system available to do work on the surroundings
• G = H – T(S)
• \Delta G{sys} = \Delta H{sys} - T \Delta S_{sys}
• \Delta G{sys} = \sum n \Delta G{products} - \sum n \Delta G_{reactants}
• When \Delta G < 0, there is a decrease in free energy released into the surroundings; therefore a process will be spontaneous when \Delta G is negative.

Free Energy Change and Spontaneity

• Free energy determines the direction of spontaneous change.
• Example: N2(g) + 3H2(g) \rightleftharpoons 2 NH_3(g)
  • Spontaneous forward reaction when Q < K.
  • Spontaneous reverse reaction when Q > K.
  • Equilibrium when Q = K.

Gibbs Free Energy ($\Delta G$) and Spontaneity

• \Delta G will be negative (reaction spontaneous) when:
  • \Delta H is negative and \Delta S is positive (exothermic and more random).
  • \Delta H is negative and large, and \Delta S is negative and small.
  • \Delta H is positive and small, and \Delta S is positive and large, or at high temperatures.
• \Delta G will be positive when endothermic and \Delta S is negative.
• When \Delta G = 0, the reaction is at equilibrium.

Gibbs Free Energy: The Effect of \Delta H, \Delta S, and T on Spontaneity

• Table summarizing the spontaneity based on signs of \Delta H and \Delta S and temperature.
• Examples:
  • 2 N2O(g) \rightarrow 2N2(g) + O_2(g): Spontaneous at all temperatures.
  • 3O2(g) \rightarrow 2O3(g): Nonspontaneous at all temperatures.
  • H2O(l) \rightarrow H2O(s): Spontaneous at low temperatures, nonspontaneous at high temperatures.
  • H2O(l) \rightarrow H2O(g): Nonspontaneous at low temperatures, spontaneous at high temperatures.

Gibbs Free Energy: Example Calculation

• Reaction: CCl4(g) \rightarrow C(s, graphite) + 2 Cl2(g)
• \Delta H = 95.7 kJ and \Delta S = 142.2 J/K at 25°C.
• Calculate \Delta G and determine if it is spontaneous.
• Solution: 53.3 kJ (non-spontaneous)

Gibbs Free Energy: Temperature Dependence Example

• Reaction: CCl4(g) \rightarrow C(s, graphite) + 2 Cl2(g)
• \Delta H = 95.7 kJ and \Delta S = 142.2 J/K at 25°C.
• Calculate the minimum temperature at which the reaction will be spontaneous: 673K

The 3rd Law of Thermodynamics

• The absolute entropy of a substance is the amount of energy it has due to the dispersion of energy through its particles.
• The 3rd law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol·K.
• Every substance that is not a perfect crystal at absolute zero has some energy from entropy.
• Therefore, the absolute entropy of substances is always positive.

Standard Entropies (S^o)

• Extensive property
• Entropies for 1 mole at 298K for a particular state, allotrope, molecular complexity, molar mass, and degree of dissolution.

Standard Entropy Values

• For different substances in the same phase, molecular complexity determines which ones have higher entropies.

Standard Entropy and Molar Mass

• Greater the molar mass, the higher the standard entropy.
• Available energy states are more closely spaced, allowing more dispersal of energy through the states.

Standard Entropy and Allotropes

• The less constrained the structure of an allotrope, the larger the entropy.
• Example:
  • C(s, diamond): S^o = 2.4 J/mol·K
  • C(s, graphite): S^o = 5.7 J/mol·K

Standard Entropy and Molecular Complexity

• Larger, more complex molecules generally have larger standard entropy values.
• There is more available energy states, allowing more dispersal of energy through the states.

Standard Entropy and Dissolution

• Dissolved solids generally have larger entropy.
• Distributing particles throughout the mixture.

Standard Entropy Calculation

• Calculate \Delta S^o for the reaction: 4 NH3 (g) + 5 O2(g) \rightarrow 4 NO (g) + 6 H_2O(g)
• Solution: 178.8 J/K.

Calculating \Delta G^o

• At 25°C: \Delta G^o{reaction} = \sum n \Delta G^of(products) - \sum n \Delta G^o_f(reactants)
• At temperatures other than 25°C (assuming \Delta H^o{reaction} and \Delta S^o{reaction} are negligible): \Delta G^o{reaction} = \Delta H^o{reaction} - T \Delta S^o_{reaction}

Calculating \Delta G^o: Example

• Calculate the \Delta G^o{reaction} at 25°C for the reaction: CH4(g) + 8 O2(g) \rightarrow CO2(g) + 2 H2O(g) + 4 O3 (g)
• Result: -148.3 kJ

Calculating \Delta G^o at Non-Standard Temperature: Example

• Calculate the \Delta G^o{reaction} at 125°C for the reaction: SO2(g) + 0.5 O2(g) \rightarrow SO2(g)
• Given: \Delta H^o = -98.9 kJ, \Delta S^o = -94.0 J/K
• Result: -65.1 kJ

Free Energy and Reversible Reactions

• The change in free energy is a theoretical limit as to the amount of work that can be done.
• If the reaction achieves its theoretical limit, it is a reversible reaction.

\Delta G under Nonstandard Conditions

• \Delta G = \Delta G^o only when reactants and products are in their standard states.
  • Normal state at that temperature.
  • Partial pressure of gas = 1 atm.
  • Concentration = 1M.
• Under nonstandard conditions: \Delta G = \Delta G^o + RT\ln Q
• At equilibrium: \Delta G = 0 and \Delta G^o = -RT\ln K

\Delta G under nonstandard Conditions

• Water evaporation example.
• When P{H2O} = 0 atm, \Delta G is negative and water evaporates.
• When P{H2O} = 1 atm, \Delta G is positive and water condenses.
• When P{H2O} = 0.0313 atm, \Delta G = 0 and the system is at equilibrium.

Calculating \Delta G under Non-Standard Conditions: Example

• Reaction at 25°C: 2 NO (g) + O2 (g) \rightarrow 2 NO2(g)
• \Delta G^o_{rxn} = -71.2 kJ
• Conditions: P{NO} = 0.100 atm, P{O2} = 0.100 atm, P{NO_2} = 2.00 atm
• Calculate \Delta G and determine if the reaction is more or less spontaneous under these conditions than under standard conditions.

Free Energy and Equilibrium Constant

• Relationship between free energy and equilibrium constant is explored through graphs.

Practice

• Estimate the equilibrium constant and position of the equilibrium for the following reaction at 427°C
• N2 (g) + 3 H2 (g) \rightarrow 2 NH_3 (g)
• \Delta H^of of NH3(g) = -46.19kJ
• \Delta S^o of NH_3(g) = 192.5 J/K
• \Delta S^o of N2(g) = 191.5 J/K\n* \Delta S^o of H2(g) = 130.5 J/K
• Result: 3.45 x 10^{-4}

Temperature Dependence of K

• For an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant.
• For an endothermic reaction, increasing the temperature increases the value of the equilibrium constant.