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Chapter 17 Spontaneity, Entropy, and Free Energy Chapter 17 Chapter 17 2 First Law of Thermodynamics First law of thermodynamics : energy cannot be created or destroyed. ΔE universe = 0 = ΔE system + ΔE surroundings The total energy of the universe cannot change (though you can transfer it from one place to another). Chapter 17 3 First Law of Thermodynamics conservation of energy exothermic rxn : releases heat to the surroundings (system surroundings) two ways energy “lost” from a system: converted to heat, q and/or used to do work, w Energy conservation requires that the energy change in the system equal the heat released plus the work done. ΔE = q + w ΔE = ΔH + pΔV Chapter 17 4 First Law of Thermodynamics ΔE is a state function : internal energy change independent of how it was done
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122_Ch-17_Thermodyn_NOTES_ZUMD_March_29_14.pptxbe created or destroyed.
ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings
(though you can transfer it from one place to
another).
(system è surroundings)
two ways energy “lost” from a system: converted to heat, q and/or used to do work, w
Energy conservation requires that the energy change in the system equal the heat released plus the work done.
ΔE = q + w ΔE = ΔH + pΔV
Chapter 17 4
First Law of Thermodynamics
ΔE is a state function: internal energy change independent of how it was done
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Nonspontaneous processes require energy input to go.
Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction
If the system after the rxn has less potential energy than before the rxn, the rxn is thermodynamically favorable.
spontaneity ≠ fast or slow
Chapter 17 6
The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.
Comparing Potential Energy
Chapter 17 7
Reversibility of Process
Any spontaneous process is irreversible, i.e., it will proceed in only one direction.
A reversible process will proceed back and forth between the two end conditions.
- equilibrium rxns (they result in no change in free energy;)
If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction.
Chapter 17 8
Thermodynamics and Spontaneity
Thermodynamics and Spontaneity
diamond vs. graphite
Factors Affecting Whether a Rxn is Spontaneous
Reactions are spontaneous in the direction of lower chemical potential energy.
There are two factors that determine the thermodynamic favorability: the enthalpy change and the entropy change.
The enthalpy change, ΔH, is the difference in the sum of the internal energy and pV work energy of the reactants compared to the products.
The entropy change, ΔS, is the difference in randomness of the reactants compared to the products.
Chapter 17 11
Enthalpy (review)
ΔH is generally measured in kJ/mol. - stronger bonds = more stable molecules
exothermic rxn: if the bonds in the products are stronger than the bonds in the reactants
ΔHrxn < 0: exothermic rxn; ΔHrxn > 0: endothermic rxn
endothermic rxn: the bonds in the products are weaker than the bonds in the reactant;
ΔΗrxn = Σ (ΔH°f products) − Σ (ΔH°f reactants)
Chapter 17 12
Entropy, S, is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components (degree of freedom) increases. S is generally measured in J/mol. Random systems require less energy than ordered systems. (read section 17.1)
Entropy
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Changes in Entropy, ΔS
Entropy change is favorable when the result is a more random system (ΔS = Sfinal – Sinit. > 0 )
ΔS > 0 if: - products are in a more random state - going from solid to less ordered liquid to less ordered gas
- rxns that have larger # of product molecules than reactant molecules
- increase in temperature
Chapter 17 14
Chapter 17 15
Changes in Entropy, ΔS
ΔSsystem > 0 for a process in which the final condition is more random than the initial condition (favorable entropy) ΔSsystem < 0 for a process in which the final condition is more orderly than the initial condition (unfavorable entropy)
ΔSsystem = ΔSreaction = Σ (S°prod) − Σ (S°react)
Chapter 17 16
Entropy Change and State Changes When a material changes physical state, the number of macrostates it can have changes as well.
- the more degrees of freedom the molecules have, the more macrostates are possible;
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Chapter 17 17
Entropy Change and State Changes - solids have fewer macrostates than liquids, which have fewer macrostates than gases;
Chapter 17 18
•Water vapor condensing
•Dissolving sugar in tea
•2 NH3 (g) → N2 (g) + 3 H2 (g)
•Ag+ (aq) + Cl−(aq) → AgCl(s)
Predict whether ΔSsystem is + or − for each of the following:
Problem A
Chapter 17 19
• The total entropy change of the universe must be positive for a process to be spontaneous. - for reversible process, ΔSuniv = 0. - for irreversible (spontaneous) process, ΔSuniv >0
ΔSuniverse = ΔSsystem + ΔSsurroundings • If the entropy of the system decreases, then the entropy
of the surroundings must increase by a larger amount. - when ΔSsystem < 0 , ΔSsurroundings > 0 .
the 2nd Law of Thermodynamics
Chapter 17 20
Temperature Dependence of ΔSsurroundings
When a system process is exothermic, 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 the entropy of the surroundings changes depends on its initial temperature.
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Temperature dependence of ΔSsurroundings
The higher the original temperature, the less effect addition or removal of heat has.
Chapter 17 22
Calculate the ΔSsurr. at 25ºC for the rxn: C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) ΔHrxn = − 2044 kJ.
Problem B
Chapter 17 23
The rxn below has ΔHrxn = +66.4 kJ at 25°C. 2 O2 (g) + N2 (g) → 2 NO2 (g)
i. Determine ΔSsurroundings.
The rxn below has ΔHrxn = +66.4 kJ at 25°C.
2 O2 (g) + N2 (g) → 2 NO2 (g)
ii. Determine the sign of ΔSsystem.
Problem C
Chapter 17 25
The rxn below has ΔHrxn = + 66.4 kJ at 25°C. iii. Determine whether the process is spontaneous 2 O2 (g) + N2 (g) → 2 NO2 (g)
Problem C
ΔGsys = ΔHsys− TΔSsys
ΔSuniv = ΔSsys + ΔSsurr
ΔSuniv = ΔSsys - ΔHsys
-Τ ΔSuniv = -T ΔSsys + ΔHsys
-Τ ΔSuniv = ΔHsys - Τ ΔSsys
-Τ ΔSuniv = ΔGsys
Gibbs Free Energy and Spontaneity
The Gibbs free energy is the maximum amount of work energy that can be released to the surroundings by a system.
- valid at a constant temperature and pressure - often called the chemical potential energy because it is analogous to the storing of energy in a mechanical system
ΔGsys = ΔHsys− T ΔSsys
- since ΔSuniv determines whether a process is spontaneous, ΔG also determines spontaneity;
-Τ ΔSuniv = ΔGsys Chapter 17 28
Gibbs Free Energy, ΔG
Possible ΔG values: ΔG < 0:
ΔH < 0 and ΔS > 0
ΔGsys = ΔHsys− T ΔSsys
rxn exothermic & more random
or high temperature
-Τ ΔSuniv = ΔGsys
Possible ΔG values: ΔG > 0 : ΔH > 0 and ΔS < 0
ΔGsys = ΔHsys− T ΔSsys
ΔG = 0 : the reaction is at equilibrium
Chapter 17 30
Chapter 17 31
The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. i. Calculate ΔG for the reaction.
Problem D
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Chapter 17 33
The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. i. Calculate ΔG for the reaction.
Problem E
Chapter 17 35
The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. Calculate the minimum temperature for it to be spontaneous.
Problem F
Chapter 17 37
The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. Calculate the maximum temperature for it to be spontaneous.
Problem G
Chapter 17 41
Changes in Entropy
the standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions;
ΔSºreaction = ( ∑ np Sºproducts) − ( ∑ nr Sºreactants)
Chapter 17 42
Calculate ΔS° for the rxn 4 NH3 (g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g)
Problem H
Substance S°, J/mol⋅K NH3 (g) 192.8 O2 (g) 205.2 NO(g) 210.8 H2O(g) 188.8
Chapter 17 43
Problem H continued
Chapter 17 44
Problem H continued
Chapter 17 45
Substance S°, J/mol⋅K H2 (g) 130.7 O2 (g) 205.2 H2O(g) 188.8
Calculate ΔS° for the rxn 2 H2 (g) + O2 (g) → 2 H2O(g)
Problem I
at 25 °C at temperatures other than 25 °C:
assuming the change in ΔHo reaction and
ΔSo reaction negligible
1.
2.
3.
CO2 (g) −394.4 H2O(g) −228.6 O3 (g) 163.2
Calculate ΔG° at 25 °C for the reaction CH4 (g) + 8 O2 (g) → CO2 (g) + 2 H2O(g) + 4 O3 (g)
Problem J
Chapter 17 51
Determine the free energy change ΔGo in the following rxn at 298 K: 2 H2O(g) + O2 (g) → 2 H2O2 (g)
Problem K
Chapter 17 53
Problem K continued
Chapter 17 54
Problem K continued
Chapter 17 55
- if a rxn can be expressed as a series of rxns, the sum of the ΔG values of the individual rxn is the ΔG of the total rxn (ΔG is a state function);
- if a rxn is reversed, the sign of its ΔG value reverses; - if the amount of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor; (the value of ΔG is an extensive property)
ΔGo Relationships
substance ΔG° (kJ/mol) H2 (g) + O2 (g) → H2O2 (g) −105.6
2 H2 (g) + O2 (g) → 2 H2O(g) −457.2
Determine ΔGo in the following rxn at 298 K: 2 H2O(g) + O2 (g) → 2 H2O2 (g)
Problem L
Chapter 17 57
Problem L continued
Chapter 17 58
Find ΔGºrxn for the following rxn using the given equations: 3 C(s) + 4 H2 (g) → C3H8 (g)
Problem M
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) ΔGº = −2074 kJ
C(s) + O2 (g) → CO2 (g) ΔGº = −394.4 kJ
2 H2 (g) + O2 (g) → 2 H2O(g) ΔGº = −457.1 kJ
Chapter 17 59
Problem M continued
Chapter 17 60
Problem M continued
Chapter 17 61
Problem M continued
Chapter 17 62
• The change in free energy is the theoretical limit as to the amount of work that can be done.
• If the reaction achieves its theoretical limit, it is a reversible reaction.
Free Energy and Reversible Reactions
Chapter 17 63
• In a real reaction, some of the free energy is “lost” as heat (if not most).
• Therefore, real reactions are irreversible.
Real Reactions
Chapter 17 64
Free Energy under Nonstandard Conditions ΔG = ΔG° only when the reactants and products are in their standard states;
- their normal state at that temperature - partial pressure of gas = 1 atm - concentration = 1 M
Under nonstandard conditions, ΔG = ΔG° + RT lnQ ( Q is the rxn quotient )
At equilibrium, ΔG = 0 , therefore ΔG° = −RTlnK
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Chapter 17 66
NO2 (g) 2.00
Calculate ΔG at 298 K for the rxn under the given conditions: 2 NO(g) + O2 (g) → 2 NO2 (g) ΔGº = −71.2 kJ
Problem N
NH3 (g) 2.0
Calculate ΔGrxn for the given rxn at 700 K under the given conditions: N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ
Problem P
substance ΔG°f (kJ/mol) N2O4 (g) +99.8 NO2 (g) +51.3
Problem R Calculate K at 298 K for the reaction N2O4 (g) 2 NO2 (g)
ΔGo from Appendix IIB
Chapter 17 74
Problem R continued
Chapter 17 75
Problem R continued
Chapter 17 76
Problem S Estimate the equilibrium constant for the given rxn at 700 K: N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ
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