Chapter 17Free Energy and Thermodynamics
Chemistry II
First Law of Thermodynamics
• First Law of Thermodynamics: Energy cannot be Created or Destroyed the total energy of the universe cannot change though you can transfer it from one place to another
ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings
First Law of Thermodynamics
• Conservation of Energy• For an exothermic reaction, “lost” heat from the system goes into the
surroundings
• two ways energy “lost” from a system, converted to heat, q used to do work, w
• Energy conservation requires that the energy change in the system equal the heat released + work done
Δ E = q + w
Energy Tax
• you can’t break even!• to recharge a battery with 100 kJ of useful
energy will require more than 100 kJ • every energy transition results in a “loss” of
energy conversion of energy to heat which is “lost” by
heating up the surroundings
e.g. if a light bulb is 5 % efficient 95% is lost as heat
Heat Tax
fewer steps generally results in a lower total heat
tax
Spontaneous and Nonspontaneous Processes
• e.g. in physics - will an object fall when dropped? in chemistry - will iron rust?
• thermodynamics predicts whether a process will proceed under the given conditions - spontaneous process
nonspontaneous processes require energy inputs
• In physics sponaneity is determined by looking at the potential energy change of a system
• In chemistry spontaneity is determined by comparing the free energy change of the system if the system after reaction has less free energy than before the
reaction, the reaction is thermodynamically favorable.
• spontaneity ≠ fast or slow
Comparing Potential Energy
The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.
High PE → Low PE
High ‘Free Energy’ → Low ‘Free Energy’
Reversibility of Process
• any spontaneous process is irreversible it will proceed in only one direction
• a reversible process will proceed back and forth between the two end conditions equilibrium results in no change in free energy
• if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction
Thermodynamics vs. Kinetics
Don’t confuse spontaneity (direction and extent of reaction) with speed
This reaction will require an energy input since ‘free energy’ of reactants < ‘free energy’ products
Diamond → Graphite
Graphite has less FE than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or
through many of your generations).
Factors Affecting Whether a Reaction Is Spontaneous
• The two factors that determine spontaneity (free energy) are the enthalpy and the entropy
• The enthalpy is a comparison of the bond energy of the reactants to the products. bond energy = amount needed to break a bond. Δ H
• The entropy factors relates to the randomness/orderliness of a system ΔS
• The enthalpy factor is generally more important than the entropy factor
Enthalpy, ΔH• ΔH related to the internal energy (kJ/mol)
• stronger bonds = more stable molecules
• if products more stable than reactants, energy released Exothermic (ΔH = negative)
• if reactants more stable than products, energy absorbed Endothermic (ΔH = positive)
Hess’ Law: Δ H°rxn = Σ(nΔ H°prod) - Σ(nΔ H°react)
Where n = stoichiometric coefficient
Changes in Entropy, ΔS• entropy change is favorable when the result is a more random system
ΔS is positive
• Some changes that increase the entropy are: reactions whose products are in a more disordered state.
(solid > liquid > gas) reactions which have larger numbers of product molecules than
reactant molecules increase in temperature solids dissociating into ions upon dissolving
Increases in EntropyMelting
Evaporating
Dissolution
Entropy• entropy is a thermodynamic function that increases as the number of
energetically equivalent ways of arranging the components increases, a measure of randomness, S S generally J/K
• S = k ln W k = Boltzmann Constant = 1.38 x 10-23 J/K W is the number of energetically equivalent ways, unitless
• A chemical system proceeds in a direction that increases the entropy of the universe
• Next we consider W,
W
Energetically Equivalent States for the Expansion of a Gas in a vacuum
The following are macro states
Macrostates → Microstates
This macrostate can be achieved throughseveral different arrangements of the particles
These microstates all have the same
macrostate
So there are 6 different particle
arrangements that result in the same
macrostate
Macrostates and ProbabilityThere is only one possible
arrangement that gives State A and one that gives State B
There are 6 possible arrangements that give State C
Therefore State C has higher entropy than either State A or
State B
The macrostate with the highest entropy also has the greatest
dispersal of energy, now imagine thousands of atoms!
Entropy• Second law of thermodynamics:
For any spontaneous process, the entropy of the universe increases, ΔSuniv > 0
• Spontaneous processes increase the entropy of the universe
Δ S = Sfinal – Sinitial
• In our flask analogy, ΔS > 0
• Explains why heat travels from a substance at high temp. to a substance at low temp. but never the opposite
Entropy Change in State Change
• when materials change state, the number of macrostates it can have changes as well for entropy: solid < liquid < gasbecause the degrees of freedom of motion increases
solid → liquid → gas
Entropy Change and State Change
Heat Flow, Entropy, and the 2nd Law
Heat must flow from water to ice in order for the entropy of the universe to increase
Is Entropy Increases or Decreasing?
• Iodine sublimes
• A liquid boils
• A solid decomposes into two gaseous products
Heat Transfer and Changes in S of Surroundings
• 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 Δ Suniv > 0
ΔSuniverse = ΔSsystem + Δ Ssurroundings
• if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount when ΔSsystem is negative, ΔSsurroundings is positive e.g. when water freezes
• the increase in ΔSsurroundings often comes from the heat released in an exothermic reaction
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 the temperature it is at originally the higher the original temperature, the less effect addition or
removal of heat has
TH
S systemgssurroundin
Temperature Dependence of ΔSsurroundings
• Example:2N2 (g) + O2 (g) → 2N2O (g) ΔH = +163.2 kJ/mol
(i) Calculate the entropy change in the surroundings at 25 ºC
ΔSsurr. = -ΔHsys. / T = -163.2 kJ/298K = -0.548 kJ/K = -548 J/K
(ii) Determine the sign of the entropy change
3 mols on LHS, 2 mols on RHS so -ΔSsys
(iii) Determine the sign of Δ Suniverse , is reaction spontaneous?
Δ Suniverse = ΔSsystem + Δ Ssurroundings
-ve + -ve = -veNot spontaneous
Gibbs Free Energy, ΔG
• Enthalpy (ΔHsys) and entropy (ΔSsurr) are related by:
ΔSsurr. = -ΔHsys. / T
• Also: Δ Suniverse = ΔSsystem + Δ Ssurroundings
• Combining these: ΔSuniverse = ΔSsystem -ΔHsys. / T
• Now ΔSuniv can be calculated with the ‘system’ only
• Multiplying by –T: -TΔSuniverse = -TΔSsystem -ΔHsys.
• This is now called the Gibbs free Energy expression (remember we multiplied by a –ve)
Gibbs Free Energy, ΔG
• maximum amount of energy from the system available to do work on the surroundings
G = H – T∙S
ΔGsys = Δ Hsys – T Δ Ssys
The change in Gibbs free energy for a process at constant T and P is ~ -ΔSuniv
Δ Gsys = – T Δ Suniverse
• ΔG (Gibbs free energy) being –ve is a criterion for spontaneity• there is a decrease in free energy of the system that is released into the
surroundings; therefore a process will be spontaneous when ΔG is negative (i.e. ΔS +ve, since we mult. by - T)
Gibbs Free Energy, ΔG
• process will be spontaneous when ΔG is negative
ΔGsys = Δ Hsys – T Δ Ssys
Three possibilities:
(1) Δ G will be negative - spontaneous Δ H is negative and ΔS is positive
exothermic and more random
ΔH is negative and large and ΔS is negative but small
Δ H is positive but small and ΔS is positive high temperature favors spontaneity, nonspon. at low T
(2) ΔG will be +ve when ΔH is +ve and ΔS is −ve - nonspontaneous never spontaneous at any temperature
(3) when ΔG = 0 the reaction is at equilibrium
Which of the following is spontaneous?
ΔHº TΔSº
ΔHº
TΔSº
ΔHº
TΔSº
ΔHº TΔSº
ΔG, ΔH, and ΔS
Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has ΔH = +95.7 kJ and Δ S = +142.2 J/K at 25°C.
Calculate Δ G and determine if it is spontaneous.
Since ΔG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.
ΔH = +95.7 kJ, Δ S = 142.2 J/K, T = 298 K
ΔG, kJ
Answer:
Solution:
Concept Plan:
Relationships:
Given:
Find:
Δ GT, ΔH, ΔS
STHG
J 1033.5
2.142K 298J 1095.7
STHG
4KJ3
Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has ΔH = +95.7 kJ and ΔS = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
The temperature must be higher than 673K for the reaction to be spontaneous
ΔH = +95.7 kJ, ΔS = 142.2 J/K, ΔG < 0
T, K
Answer:
Solution:
Concept Plan:
Relationships:
Given:
Find:
TΔG, ΔH, ΔS
STHG
K
J3
KJ3
2.142TJ 1095.7
02.142TJ 1095.7
0 STHG
K 673T
T 2.142
J 1095.7
KJ
3
Entropy• entropy is a thermodynamic function that increases as the number of
energetically equivalent ways of arranging the components increases, a measure of randomness, S S generally J/K
• S = k ln W k = Boltzmann Constant = 1.38 x 10-23 J/K W is the number of energetically equivalent ways, unitless
• A chemical system proceeds in a direction that increases the entropy of the universe
• Next we consider W,
The 3rd Law of ThermodynamicsAbsolute Entropy
• the absolute entropy of a substance is the amount of energy it has due to 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 therefore, every substance that is not a
perfect crystal at absolute zero has some energy from entropy
therefore, the absolute entropy of substances is always +ve
Standard Entropies
• S° (J/mol.K)
• Extensive property (depends on amount)
• entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution
Substance S° J/mol-K
Substance S° J/mol-K
Al(s) 28.3 Al2O3(s) 51.00 Br2(l) 152.3 Br2(g) 245.3
C(diamond) 2.43 C(graphite) 5.69 CO(g) 197.9 CO2(g) 213.6 Ca(s) 41.4 CaO(s) 39.75 Cu(s) 33.30 CuO(s) 42.59 Fe(s) 27.15 Fe2O3(s) 89.96 H2(g) 130.58 H2O2(l) 109.6
H2O(g) 188.83 H2O(l) 69.91 HF(g) 173.51 HCl(g) 186.69
HBr(g) 198.49 HI(g) 206.3 I2(s) 116.73 I2(g) 260.57 N2(g) 191.50 NH3(g) 192.5 NO(g) 210.62 NO2(g) 240.45 Na(s) 51.45 O2(g) 205.0 S(s) 31.88 SO2(g) 248.5
Relative Standard EntropiesStates
• the gas state has a larger entropy than the liquid state at a particular temperature
• the liquid state has a larger entropy than the solid state at a particular temperature
Substance S°, (J/mol∙K)
H2O (l) 70.0
H2O (g) 188.8
Relative Standard EntropiesMolar Mass
• the larger the molar mass, the larger the entropy
Relative Standard EntropiesAllotropes
• the less constrained the structure of an allotrope is, the larger its entropy
Relative Standard EntropiesMolecular Complexity
• larger, more complex molecules have larger entropy than single atoms
• more available energy states (translational, rotational, vibrational) allowing more dispersal of energy through the states
SubstanceMolarMass
S°, (J/mol∙K)
Ar (g) 39.948 154.8
NO (g) 30.006 210.8
Relative Standard EntropiesDissolution
• dissolved solids generally have larger entropy
• distributing particles throughout the mixture
SubstanceS°,
(J/mol∙K)KClO3(s) 143.1
KClO3(aq) 265.7
Standard Entropy of Reaction, ΔSrxn
• Calculated similar to ΔHrxn
• Where n = stoichiometric coefficient
reactantsproducts SSS rp nn
Ex. 17.4 –Calculate ΔS for the reaction4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l)
ΔS is +, as you would expect for a reaction with more gas product molecules than reactant molecules
standard entropies from Appendix IIBΔS, J/K
Check:
Solution:
Concept Plan:
Relationships:
Given:Find:
ΔSSNH3, SO2, SNO, SH2O,
reactantsproducts SSS rp nn
KJ
KJ
KJ
KJ
KJ
)(O)(NH)O(H)NO(
reactantsproducts
8.178
)]2.205(5)8.192(4[)]8.188(6) 8.210(4[
)]S(5)S(4[)]S(6)S(4[
SSS
232
gggg
rp nn
Substance S, J/molKNH3(g) 192.8
O2(g) 205.2
NO(g) 210.8H2O(g) 188.8
Calculating ΔGrxn
• At 25 C and at temperatures other than 25 C: assuming the change in ΔHo
reaction and Δ Soreaction are small over a
limted temp. range
Δ Grxn = Δ Hrxn – T Δ Srxn
Ex. 17.6 – The reaction SO2(g) + ½ O2(g) SO3(g) has ΔHrxn = -98.9 kJ and ΔSrxn = -94.0 J/K at 25°C.
Calculate ΔG at 125C and determine if it is spontaneous.
Since ΔG is -, the reaction is spontaneous at this temperature, though less so than at 25C
ΔH = -98.9 kJ, ΔS = -94.0 J/K, T = 398 K
ΔG, kJ
Answer:
Solution:
Concept Plan:
Relationships:
Given:
Find:
ΔGT, ΔH, ΔS STHG
kJ 5.61J 105.61
0.94K 398J 1098.9
STHG
3KJ3
Calculating ΔGrxn with ΔG f
• at 25C:ΔGo
rxn = ΣnpΔGof(products) - ΣnrΔGo
f(reactants)
• Like ΔHf°, ΔGf° = 0 for an element in its standard state:
• Example: ΔGf° (O2(g)) = 0
52
Substance G°f kJ/mol
Substance G°f kJ/mol
Al(s) 0 Al2O3 -1576.5 Br2(l) 0 Br2(g) +3.14
C(diamond) +2.84 C(graphite) 0 CO(g) -137.2 CO2(g) -394.4 Ca(s) 0 CaO(s) -604.17 Cu(s) 0 CuO(s) -128.3 Fe(s) 0 Fe2O3(s) -740.98 H2(g) 0 H2O2(l) -120.4
H2O(g) -228.57 H2O(l) -237.13 HF(g) -270.70 HCl(g) -95.27
HBr(g) -53.22 HI(g) +1.30 I2(s) 0 I2(g) +19.37 N2(g) 0 NH3(g) -16.66 NO(g) +86.71 NO2(g) +51.84 Na(s) 0 O2(g) 0 S(s) 0 SO2(g) -300.4
Ex. 17.7 –Calculate ΔG at 25C for the reactionCH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g)
standard free energies of formation from Appendix IIBΔG, kJ
Solution:
Concept Plan:
Relationships:
Given:Find:
ΔGΔGf of prod & react
reactantsproducts GGG frfp nn
kJ 3.148)]kJ 0.0(8)kJ 5.50[(]kJ) 2.163()kJ 6.228(2)kJ 4.394[(
)]G(8)G[()]G()G(2)G[(
GGG
24322 OCHOOHCO
reactantsproducts
fffff
frfp nn
Substance Gf, kJ/molCH4(g) -50.5O2(g) 0.0
CO2(g) -394.4H2O(g) -228.6O3(g) 163.2
ΔG Relationships
• if a reaction can be expressed as a series of reactions, the sum of the ΔG values of the individual reaction is the ΔG of the total reaction ΔG is a state function
• if a reaction is reversed, the sign of its ΔG value reverses
• if the amounts of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor the value of ΔG of a reaction is extensive
ΔG°rxn ≠ Reaction Rate
• Although ΔG°rxn can be used to predict if a reaction will be spontaneous as written, it does not tell us how fast a reaction will proceed.
• Example:
C(s, diamond) + O2(g) CO2(g) ΔG°rxn= -397 kJ<<0
But diamonds are forever…. ΔG°rxn≠ rate
Why Free Energy is ‘Free’
• If ΔG –ve: change in Gibbs free energy = max amount of available ‘free’ energy to do work, should be called ‘Gibbs available energy’
• For many reactions ΔG < ΔH e.g.
C(s) + 2H2(g) →CH4(g)
ΔHrxn = -74.6 kJ, ΔSºrxn = -80.8 J/K, ΔGºrxn = -50.5 kJ
• Exothermic reaction but only half as much available to do work
• Entropy is –ve, so some of emitted heat must inc. entropy of the universe in order to make ΔGºrxn -ve
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 theoretically a reversible reaction
Real Reactions
• in a real reaction, some of the free energy is “lost” as heat if not most
• therefore, real reactions are irreversible
Energy is lost – when the battery does work on the motor some energy is lost in the wires as heat (resistance to e- flow)
ΔG under Nonstandard Conditions Δ G = Δ G only when the reactants and products are in their standard
states there normal state at that temperaturepartial pressure of gas = 1 atmconcentration = 1 M
under nonstandard conditions, ΔG = ΔG + RTlnQQ is the reaction quotient
at equilibrium ΔG = 0ΔG = ─RTlnK
ΔG under Nonstandard Conditions Spilled water spontaneously evaporates even though ΔG◦rxn for the
vaporization of water is +ve..why?
We do not live at std. conditions
Q = PH2O
Under std. conditions water does not vaporize
ΔG rxn = ΔG◦rxn
= +ve
Spontaneous in reverse direction
Water condenses
ΔG◦rxn = +8.56 kJ/mol
ΔG rxn = -ve
Non-std.Conditions
Example – ΔG nonstandard• Calculate ΔG at 427°C for the reaction below if the PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0
atmN2(g) + 3 H2(g) → 2 NH3(g)
Q = PNH32
PN21 x PH2
3
(2.0 atm)2
(33.0 atm)1 (99.0)3= = 1.2 x 10-7
ΔG = ΔG° + RTlnQ
ΔG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
ΔG = -46300 J = -46 kJ
ΔH°rxm = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
ΔS°rxn = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
ΔG° = -92380 J - (700 K)(-198.2 J/K)
ΔG° = +46400 J
Free Energy and Equilibrium:Relating ΔG°rxn to K
• Standard free energy change of a reaction and the equilibrium constant are related
• K becomes larger as ΔG° becomes more –ve
• If ΔG° is large and –ve then reaction has a large equil. constant, if ΔG° is large and +ve reaction has a small equil. Constant, favors reactants at equilibrium
• ΔG = ΔG + RTlnQ and at equilibrium Q = K, ΔG = 0
ΔG°rxn = -RT lnK
Free Energy and Equilibrium:Relating ΔG°rxn to K
• ΔGrxn = ΔGrxn + RT lnQ and at equilibrium Q = K, ΔGrxn = 0
ΔG°rxn = -RT lnK
• When K < 1, lnK is –ve and ΔG°rxn is +ve When Q = 1, reaction is spontaneous in reverse direction
• When K > 1, lnK is +ve and ΔG°rxn is -ve When Q = 1, reaction is spontaneous in forward direction
• When K = 1, lnK is 0 and ΔG°rxn is 0 Reaction is at equilibrium under std. conditions
Example - K
• Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C
N2(g) + 3 H2(g) 2 NH⇌ 3(g)
ΔG° = -RT lnK
+46400 J = -(8.314 J/K)(700 K) lnK
lnK = -7.97
K = e-7.97 = 3.45 x 10-4
since K is << 1, the position of equilibrium favors reactants
ΔH° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
ΔS° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
ΔG° = -92380 J - (700 K)(-198.2 J/K)
ΔG° = +46400 J