CC

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.