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First Law of Thermodynamics
First law of thermodynamics: energy cannot Chapter 17 be created or destroyed.
The total energy of the universe cannot change Spontaneity, Entropy, and (though you can transfer it from one place to another). Free Energy ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings
Chapter 17 Chapter 17 2
First Law of Thermodynamics First Law of Thermodynamics
conservation of energy ΔE is a state function: internal energy change independent of how it was done exothermic rxn: releases heat to the surroundings (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 3 Chapter 17 4
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Thermodynamics and Spontaneity Comparing Potential Energy Thermodynamics predicts whether a process will proceed under the given conditions (spontaneous process). 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 The direction of spontaneity can be determined by before the rxn, the rxn is thermodynamically favorable. comparing the potential energy of the system at the start and the end. spontaneity ≠ fast or slow
Chapter 17 5 Chapter 17 6
Reversibility of Process Thermodynamics and Spontaneity 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 7 Chapter 17 8
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Thermodynamics and Spontaneity Factors Affecting Whether a Rxn is Spontaneous diamond vs. graphite 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 more stable compared to the products. conversion of diamond into graphite is spontaneous The entropy change, ΔS, is the difference in randomness of the reactants compared to the products.
Chapter 17 9 Chapter 17 10
Enthalpy (review) Entropy Entropy, S, is a thermodynamic function that increases as ΔH is generally measured in kJ/mol. the number of energetically equivalent ways of arranging - stronger bonds = more stable molecules the components (degree of freedom) increases.
ΔHrxn < 0: exothermic rxn; ΔHrxn > 0: endothermic rxn S is generally measured in J/mol.
ΔΗ = Σ (ΔH° ) − Σ (ΔH° ) rxn f products f reactants Random systems require less energy than ordered systems. (read section 17.1) endothermic rxn: the bonds in the products are weaker than the bonds in the reactant;
exothermic rxn: if the bonds in the products are stronger than the bonds in the reactants
Chapter 17 11 Chapter 17 12
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Changes in Entropy, ΔS Changes in Entropy, ΔS Entropy change is favorable when the result is a more Examples:
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 - solids dissociating into ions upon dissolving
Chapter 17 13 Chapter 17 14
Changes in Entropy, ΔS Entropy Change and State Changes When a material changes physical state, the number of ΔSsystem > 0 for a process in which the final condition is macrostates it can have changes as well. more random than the initial condition (favorable entropy) - the more degrees of freedom the molecules have, the more macrostates are possible;
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 15 Chapter 17 16
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Problem A Predict whether S is + or − for Entropy Change and State Changes Δ system each of the following: - solids have fewer macrostates than liquids, which have • Heating air in a balloon fewer macrostates than gases; • Water vapor condensing
• Separation of oil and vinegar salad dressing
• Dissolving sugar in tea
• 2 HgO(s) → 2 Hg(l) + O2 (g)
• 2 NH3 (g) → N2 (g) + 3 H2 (g)
+ − • Ag (aq) + Cl (aq) → AgCl(s) Chapter 17 17 Chapter 17 18
nd the 2 Law of Thermodynamics Temperature Dependence of ΔSsurroundings • The total entropy change of the universe must be When a system process is exothermic, it adds heat to the positive for a process to be spontaneous. surroundings, increasing the entropy of the surroundings. - for reversible process, ΔSuniv = 0. - for irreversible (spontaneous) process, ΔSuniv >0 When a system process is endothermic, it takes heat from ΔSuniverse = ΔSsystem + ΔSsurroundings the surroundings, decreasing the entropy of the surroundings. • If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount. The amount the entropy of the surroundings changes
- when ΔSsystem < 0 , ΔSsurroundings > 0 . depends on its initial temperature.
Chapter 17 19 Chapter 17 20
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Problem B Calculate the ΔS at 25ºC for the rxn: Temperature dependence of ΔSsurroundings surr. C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) The higher the original temperature, the less effect ΔHrxn = − 2044 kJ. addition or removal of heat has.
Chapter 17 21 Chapter 17 22
Problem C The rxn below has H = +66.4 kJ at 25°C. Problem C Δ rxn The rxn below has ΔHrxn = +66.4 kJ at 25°C.
2 O2 (g) + N2 (g) → 2 NO2 (g) 2 O2 (g) + N2 (g) → 2 NO2 (g) i. Determine ΔSsurroundings. ii. Determine the sign of ΔSsystem.
Chapter 17 23 Chapter 17 24
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Problem C The rxn below has ΔH = + 66.4 kJ at 25°C. rxn Gibbs Free Energy and Spontaneity iii. Determine whether the process is ΔSuniv = ΔSsys + ΔSsurr spontaneous -Τ ΔSuniv = ΔGsys
2 O2 (g) + N2 (g) → 2 NO2 (g) ΔGsys = ΔHsys− TΔSsys ΔH ΔS = ΔS - sys univ sys T -T ΔH -Τ ΔS = -T ΔS - sys univ sys T
-Τ ΔSuniv = -T ΔSsys + ΔHsys
-Τ ΔSuniv = ΔHsys - Τ ΔSsys
Chapter 17 25 Chapter 17 26
Gibbs Free Energy and Spontaneity Gibbs Free Energy, ΔG -Τ ΔSuniv = ΔGsys The Gibbs free energy is the maximum amount of work ΔSuniv is + when spontaneous, so ΔG is − energy that can be released to the surroundings by a system. And therefore, when ΔG < 0 process is spontaneous - valid at a constant temperature and pressure Possible ΔG values: ΔG = ΔH − T ΔS - often called the chemical potential energy because it is sys sys sys ΔG < 0: analogous to the storing of energy in a mechanical system ΔH < 0 and ΔS > 0 rxn exothermic & more random ΔG = ΔH − T ΔS sys sys sys ΔH < 0 and large and ΔS < 0 but small - since ΔSuniv determines whether a process is spontaneous, ΔG also determines spontaneity; ΔH > 0 but small and ΔS > 0 and large
-Τ ΔSuniv = ΔGsys or high temperature
Chapter 17 27 Chapter 17 28
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G = H − T S Gibbs Free Energy, ΔG Δ sys Δ sys Δ sys Free Energy Change and Spontaneity Possible ΔG values: ΔG > 0 : ΔH > 0 and ΔS < 0 never spontaneous at any temperature
ΔG = 0 : the reaction is at equilibrium
Chapter 17 29 Chapter 17 30
Problem D Problem D continued The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. ii. Determine if the reaction is spontaneous. i. Calculate ΔG for the reaction.
Chapter 17 31 Chapter 17 32
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Problem E Problem E continued The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. ii. Determine if the reaction is spontaneous. i. Calculate ΔG for the reaction.
Chapter 17 33 Chapter 17 34
Problem F Problem F continued 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.
Chapter 17 35 Chapter 17 36
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Problem G Problem G continued 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.
Chapter 17 37 Chapter 17 38
Problem G continued The Third Law of Thermodynamics
Absolute Entropy Read pages 656 - 659
Chapter 17 39 Chapter 17 40
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Problem H Calculate ΔS° for the rxn Changes in Entropy 4 NH3 (g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g) ΔSºreaction = ( ∑ np Sºproducts) − ( ∑ nr Sºreactants) Substance S°, J/mol⋅K NH 192.8 the standard entropy change is the difference in absolute 3 (g) O 205.2 entropy between the reactants and products under 2 (g) standard conditions; NO(g) 210.8 H2O(g) 188.8
Chapter 17 41 Chapter 17 42
Problem H continued Problem H continued
Chapter 17 43 Chapter 17 44
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Problem I Calculate ΔS° for the rxn Problem I continued
2 H2 (g) + O2 (g) → 2 H2O(g)
Substance S°, J/mol⋅K
H2 (g) 130.7
O2 (g) 205.2
H2O(g) 188.8
Chapter 17 45 Chapter 17 46
Calculating ΔGo Problem J Calculate ΔG° at 25 °C for the reaction CH4 (g) + 8 O2 (g) → CO2 (g) + 2 H2O(g) + 4 O3 (g) o o o 1. ΔG rxn = Σ nΔG f (prod.) − Σ mΔG f (react.) Substance ΔG°f, at 25 °C kJ/mol
at temperatures other than 25 °C: CH4 (g) −50.5 o O2 (g) 0.0 assuming the change in ΔH reaction and CO2 (g) −394.4 o ΔS reaction negligible H2O(g) −228.6 O3 (g) 163.2 Another way: 2. ΔG°rxn = ΔH°rxn − T ΔS°rxn or,
3. ΔG°total = ΔG°rxn 1 + ΔG°rxn 2 + ...
Chapter 17 47 Chapter 17 48
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Problem J continued Problem J continued
Is the rxn spontaneous?
Chapter 17 49 Chapter 17 50
Problem K Determine the free energy change ΔGo in the Problem K continued following rxn at 298 K:
2 H2O(g) + O2 (g) → 2 H2O2 (g)
Chapter 17 51 Chapter 17 52
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Problem K continued Problem K continued
Chapter 17 53 Chapter 17 54
ΔGo Relationships Problem L Determine ΔGo in the following rxn at 298 K: 2 H2O(g) + O2 (g) → 2 H2O2 (g) - 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 substance ΔG° (kJ/mol) rxn (ΔG is a state function); H2 (g) + O2 (g) → H2O2 (g) −105.6
2 H2 (g) + O2 (g) → 2 H2O(g) −457.2 - 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)
Chapter 17 55 Chapter 17 56
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Problem L continued Problem M Find ΔGºrxn for the following rxn using the
given equations: 3 C(s) + 4 H2 (g) → C3H8 (g)
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 57 Chapter 17 58
Problem M continued Problem M continued
Chapter 17 59 Chapter 17 60
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Problem M continued Free Energy and Reversible Reactions
• 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.
Chapter 17 61 Chapter 17 62
Real Reactions Free Energy under Nonstandard Conditions
• In a real reaction, some of the free energy is “lost” as ΔG = ΔG° only when the reactants and products are in heat (if not most). their standard states; • Therefore, real reactions are irreversible. - 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
Chapter 17 63 Chapter 17 64
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Free Energy under Nonstandard Conditions Problem N Calculate ΔG at 298 K for the rxn under the given conditions:
2 NO(g) + O2 (g) → 2 NO2 (g) ΔGº = −71.2 kJ
substance p (atm)
NO(g) 0.100 O2 (g) 0.100
NO2 (g) 2.00
Chapter 17 65 Chapter 17 66
Problem N continued Problem N continued
Chapter 17 67 Chapter 17 68
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Problem P Calculate ΔGrxn for the given rxn at 700 K Problem P continued under the given conditions:
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ
substance p (atm)
N2 (g) 33
H2 (g) 99
NH3 (g) 2.0
Chapter 17 69 Chapter 17 70
Problem P continued
Chapter 17 71 Chapter 17 72
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Problem R Calculate K at 298 K for the reaction Problem R continued
N2O4 (g) 2 NO2 (g) substance ΔG°f (kJ/mol) o N2O4 (g) +99.8 ΔG from Appendix IIB NO2 (g) +51.3
Chapter 17 73 Chapter 17 74
Problem R continued 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
Chapter 17 75 Chapter 17 76
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Problem S continued
End of Chapter 17
Spontaneity, Entropy, and Free Energy
Chapter 17 77 Chapter 17
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