Gen Chem II: Chapter 17

First Law of Thermodynamics

• You can’t win!
• Energy cannot be Created or Destroyed
  • The total energy of the universe cannot change
  • Energy can transfer from one place to another
• \Delta E{universe} = 0 = \Delta E{system} + \Delta E_{surroundings}
  • For an exothermic reaction, “lost” heat from the system goes into the surrounds
  • Converted to heat (q)
  • Used to do work (w)

Spontaneous Processes

• Thermodynamics predicts whether a process will proceed under the given conditions
• Spontaneity is determined by comparing the free energy of the system before the reaction with the free energy of the system after the reaction
  • If the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable
• Processes that occur spontaneously in one direction cannot, under the same conditions, also occur spontaneously in the opposite direction
• Examples:
  • A lump of sugar spontaneously dissolves in a cup of tea
• Spontaneity can be fast or slow, determined by kinetics
• Thermodynamics focuses on initial and final states to determine spontaneity

Two Factors Determining Spontaneity

1. \Delta H (Enthalpy): comparison of the bond energy of the reactants to the products
   • Bond energy = amount needed to break a bond.
2. \Delta S (Entropy): relates to the randomness/orderliness of a system

• The enthalpy factor is generally more important than the entropy factor in determining spontaneity

Enthalpy

• Related to the internal energy
• Stronger bonds = more stable molecules
  • If products are more stable than reactants, energy is released - Exothermic \Delta H = negative
  • If reactants are more stable than products, energy is absorbed - Endothermic \Delta H = positive

Entropy

• Entropy (S) is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases
• S = k \ln W
  • k = Boltzmann Constant = 1.38 x 10^{-23} J/K
  • W is the number of energetically equivalent ways
• Random Systems require less energy than ordered systems

Entropy (W)

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

Changes in Entropy, \Delta S

• Entropy change is favorable when the result is a more random system
  • \Delta S is positive
• Examples of increases in entropy
  • Reactions whose products are in a more disordered state (solid < liquid < gas)
  • Reactions which have larger numbers of products molecules than reactant molecules
  • Solids dissociating into ions upon dissolving, although not always positive
  • Increasing temperature
• When materials change state, the number of microstates it can have changes as well
  • For entropy: solid < liquid < gas
  • Because the degrees of freedom of motion increases

Changes in Entropy, \Delta S

• Molecules can rotate as well as vibrate and have translational energy.
• This increases with increasing temperature

Changes in Entropy

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

Changes in Entropy

• More particles dissolved in solution leads to a greater number of possible arrangements (W).
• Also need to consider hydration of ions, which causes water molecules to become ordered around ions.
  • For these cases, the entropy change is negative

Predicting the Sign of Entropy Change

• Predict the sign of \Delta S for each process
  • a) CO2(g) \rightarrow CO2(l)
  • b) Solid iodine sublimes
  • c) 2 N2O(g) \rightarrow 2 N2(g) + O_2(g)
  • d) dissolving sugar in water
  • e) cooling nitrogen gas from 80oC to 20oC

The 2nd Law of Thermodynamics

• The total entropy change of the universe must be positive for a process to be spontaneous
  • For a reversible process \Delta S_{universe} = 0
  • For irreversible (spontaneous) process \Delta S_{universe} > 0
  • \Delta S{universe} = \Delta S{system} + \Delta S_{surroundings}
• If the entropy of the system decreases, then the entropy of the surroundings 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 in order for the entropy of the universe to increase
• The increase in \Delta S_{surroundings} often comes from the heat released in an exothermic reaction

Temperature Dependence of \Delta S_{surroundings}

• When a system process is exothermic (q_{system} negative) it adds heat to the surroundings, increasing the entropy of the surroundings
• When a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings
• The amount of the entropy of the surroundings changes is proportional to the magnitude of q_{system}

Temperature Dependence of \Delta S_{surroundings}

• Under conditions of 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

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

Gibbs Free Energy

• G = maximum amount of 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 of the system that is 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
• N2(g) + 3H2(g) \rightleftharpoons 2 NH_3(g)

Gibbs Free Energy

• \Delta G will be negative when (reaction will be spontanteous 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 reaction: CCl4(g) \rightarrow C(s, graphite) + 2 Cl2(g)
• Has \Delta H = 95.7 kJ and \Delta S = 142.2 J/K at 25oC. Calculate \Delta G and determine if it is spontaneous 53.3 kJ

Gibbs Free Energy

• The reaction: CCl4(g) \rightarrow C(s, graphite) + 2 Cl2(g)
• Has \Delta H = 95.7 kJ and \Delta S = 142.2 J/K at 25oC. Calculate \Delta G and determine if it is spontaneous. Calculate the minimum temperature it will be spontaneous. 673K

The 3rd Law of Thermodynamics

• The absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particle
• 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, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution

Standard Entropy Values

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

Standard Entropy, 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, Allotropes

• The less constrained the structure of an allotrope is, the larger the entropy

• 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, 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)
• 178.8 J/K

Calculating \Delta G^o

• At 25oC: \Delta G^o{reaction} = \sum n\Delta G^o{f(products)} - \sum n\Delta G^o_{f(reactants)}
• At temperatures other than 25oC: assuming the change in \Delta H^o{reaction} and \Delta S^o{reaction} is neglible, \Delta G^o{reaction} = \Delta H^o{reaction} - T\Delta S^o_{reaction}

Calculating \Delta G^o

• Calculate the \Delta G^o reaction at 25oC for the reaction: CH4(g) + 8 O2(g) \rightarrow CO2(g) + 2 H2O(g) + 4 O_3 (g)
• -148.3 kJ

Calculating \Delta G^o

• Calculate the \Delta G^o reaction at 125oC for the reaction: SO2(g) + 0.5 O2(g) \rightarrow SO_2(g)
• \Delta H^o = -98.9 kJ, \Delta S^o = -94.0 J/K
• -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 the reactants and products are in the standard states
  • Their 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, \Delta G^o = -RT\ln K

\Delta G under nonstandard Conditions

• H2O(l) \rightleftharpoons H2O(g)
• Water evaporates when \\Delta G is negative
• Water condenses when \Delta G is positive
• Equilibrium is when \Delta G = 0

Calculating \Delta G under non-standard conditions

• Consider the following reaction at 25oC
• 2 NO (g) + O2 (g) \rightarrow 2 NO2(g)
• \Delta G^o_{rxn} = -71.2 kJ
• Calculate \Delta G under the following conditions
  • P_{NO} = 0.100 atm
  • P{O2} = 0.100 atm
  • P{NO2} = 2.00 atm
• Is the reaction more or less spontaneous under these conditions under standard conditions?

Free Energy and Equilibrium Constant

• A(g) ⇌ B(g) K<1
  • Standard conditions PA=PB=1 atm Q=1. Pure A to Pure B. Reverse reaction spontaneous
• A(g) ⇌ B(g) K=1
  • Standard conditions PA=PB=1 atm Q=1. Pure A to Pure B. At equilibrium
• A(g) ⇌ B(g) K>1
  • Standard conditions PA=PB=1 atm Q=1. Pure A to Pure B. Forward reaction spontaneous

Practice

• Estimate the equilibrium constant and position of the equilibrium for the following reaction at 427oC
• 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 N_2(g) = 191.5 J/K
  • \Delta S^o of H_2(g) = 130.5 J/K
• 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