Chapter 19: Thermodynamics and Equilibrium

1
Chapter 19 Thermodynamics and Equilibrium
Chemistry 1062 Principles of Chemistry II Andy
Aspaas, Instructor
2
First Law of Thermodynamics

• First Law conservation of energy applied to
  thermodynamic systems
• Internal Energy (U) sum of kinetic and potential
  energies in a system (energy of motion, and
  energy contained in chemical bonds and
  intermolecular forces)
• U is a state function
• Changing from one state to another will give you
  the change in internal energy, and this is
  independent of path (?U Uf Ui is independent
  of path)
• Changes of U are caused by exchanges of energy
  between system and surroundings
• As either heat (due to temp change) or work (due
  to a force moving an object a certain distance)

3
Enthalpy

• Change in enthalpy (?H) heat of reaction at
  constant pressure (qp)
• More specifically, H U PV
• ?H for a reaction can be calculated by summing
  standard ?Hf values for products and subtracting
  summed ?Hf values for reactants

4
Entropy

• Entropy (S) measure of how dispersed the energy
  in a system is among all possible ways a system
  can contain energy
• As a coffee cup cools, it heats the surroundings
  and disperses its energy. S increases
• If a gas is allowed to enter an empty chamber,
  the kinetic energy of the gas molecules
  distributes through the whole volume S increases.

5
Entropy changes

• ?S Sf Si
• For H2O(s) ? H2O(l), ?S 22 J/K
• Second Law of Thermodynamics
• Total entropy for a system and surroundings
  increases in a spontaneous process
• Entropy associated with heat flow q / T
• Total entropy change for the system entropy
  created plus entropy associated with heat flow
  (?S Screated q / T)
• Since entropy must be created in a spontaneous
  process, if ?S gt q / T the process must be
  spontaneous
• At equilibrium, no new entropy is being created,
  so any entropy change is just due to heat flow
• ?S q / T in a system at equilibrium

6
Enthalpy, entropy, and spontaneity

• Since ?H qp, ?S gt ?H / T for any spontaneous
  process (at constant temp and pressure)
• ?H T?S lt 0 for a spontaneous process
• gt 0 for a nonspontaneous process
• 0 for a system at equilibrium

7
3rd Law of Thermodynamics

• 3rd Law a perfectly crystalline solid at 0 K has
  entropy of zero
• When heat is transferred, ?S q / T
• So Standard entropy
• Calculated by changing the temperature slightly
  and measuring the heat absorbed

8
Predicting changes in entropy

• Three occurrences will generally increase entropy
• A single molecule is broken into 2 or more
  molecules
• The number of moles of gas increases
• Phase changes (s) ? (l) or (g) or (l) ? (g)
• An increase in entropy means the sign of ?S is
  ()
• The opposite of any of the above corresponds with
  a decrease in entropy, ?S (-)

9
Predicting the sign of ?S

• CS2 (l) ? CS2 (g)
• 2Hg (l) O2 (g) ? 2HgO (s)
• CaCO3 (s) ? CaO (s) CO2 (g)

10
Calculating ?So

• ?So ?nSo(products) - ?mSo(reactants)
• So CS2 (l) 151.3 J/K
• So CS2 (g) 237.9 J/K

11
Free energy

• Gibbs Free Energy, G H - TS
• Free energy change, ?G ?H - T?S
• ?G (-) for a spontaneous process
• ?G () for a nonspontaneous process
• ?G 0 for a process at equilibrium

12
Standard free energy change

• ?Go ?Ho - T?So
• Calculate ?Go for the following reaction, and
  predict its spontaneity
• CH4 (g) 2O2 (g) ? CO2 (g) 2H2O (g)
• ?Hfo -74.87 0 -393.5 -241.8 (kJ)
• So 186.1 205.0 213.7 188.7 (J/K)

13
Standard free energies of formation

• ?Gfo ?G for production of 1 mol of a compound
  under standard conditions from elements in their
  standard forms
• ?Go ? n ?Gfo (products) - ? m ?Gfo (reactants)

14
More precise predictions of spontaneity

• If ?Go lt -10 kJ, the reaction is spontaneous as
  written, and reactants transform nearly entirely
  to products when equilibrium is reached
• If ?Go gt 10 kJ, the reaction is nonspontaneous
  as written, and the reactants will not give a
  significant amount of products at equilibrium
• If ?Go is between -10 kJ and 10 kJ, an
  equilibrium mixture of reactants and products
  will be obtained, with significant amounts of each

15
Relating ?Go to the equilibrium constant

• ?Go can be converted to ?G (for nonstandard
  temperatures)
• ?G ?Go RT ln Q
• (where Q is the reaction quotient)
• At equilibrium, ?G 0, and Q K
• ?Go -RT ln K
• (this equation relates standard free energy
  change for a reaction with that reactions
  equilibrium constant)

16
Calculating K from ?Go

• 2NH3(g) CO2(g) NH2CONH2(aq) H2O(l)
• ?Go -13.6 kJ
• R 8.31 J/(K mol)
• ?Go -RT ln K
• When K gt 1, ?Go lt 0
• When K lt 1, ?Go gt 0

17
Change of free energy with temperature

• ?GTo ?Ho -T?So (assuming ?Ho and ?So are
  constant with respect to temperature)

?Ho ?So ?Go Description
- - Spont. at all T
- Nonspont at all T
- - or - Spont at low T, nonspont at high T
or - Nonspont at low T, spont at high T
18
Example

• Ba(OH)2 8H2O(s) 2NH4NO3(s) ? Ba(NO3)2(aq)
  2NH3(g) 10H2O(l)
• Predict sign of ?S
• ?Ho
• ?So
• ?Go at room temperature
• At what temperature does this reaction switch
  from being spontaneous to nonspontaneous?