Chapter 19 Chemical Thermodynamics. No Review Quiz No Lab.
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Transcript of Chapter 19 Chemical Thermodynamics. No Review Quiz No Lab.
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Chapter 19
Chemical Thermodynamics
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Chemical Thermodynamics
• The study of the energy transformations that accompany chemical and physical changes.
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The Driving Forces
• All reactions (changes) in nature occur because of the interplay of two driving forces:
(1) The drive toward lower energy.
Decrease in enthalpy
(2) The drive toward increased disorder.
Increase in entropy
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Enthalpy
• Enthalpy is the internal energy of a system at constant pressure.
• We cannot measure enthalpy directly but we can measure changes in the enthalpy of a system.
• Changes in enthalpy normally are in the form of heat.
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Enthalpy (Energy) Change
• Exothermic reactions
–∆H
• Endothermic reactions
+∆H
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Energy Entropy
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Table 19.1 page 613
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P4O10 + 6H2O(l) → 4H3PO4
Use the information below to determine the ∆H.
4P + 5O2 → P4O10 ∆H = -2984 kJ/mol
H2 + ½ O2 → H2O(l) ∆H = -285.83 kJ/mol
3/2 H2 + P + 2O2 → H3PO4 ∆H = -1267kJ/mol
∆H = -369 kJ/mol
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2NH3 + 3O2 + 2CH4 → 2HCN + 6H2OUse the information below to determine the ∆H.
½ N2 + 3/2 H2 → NH3 ∆H = - 46 kJ/mol C + 2H2 → CH4 ∆H = -75 kJ/mol
½ H2 + C + ½ N2 → HCN ∆H = +135.1 kJ/mol
H2 + ½ O2 → H2O ∆H = -242 kJ/mol
∆H = -940 kJ/mol
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Another method to determine ∆H for a reaction
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∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants)
Appendix I in your notebook has ∆Hf values
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∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants)
Determine ∆H for the reaction using the above formula.Na(s) + O2(g) + CO2(g) → Na2CO3(s)∆Hf for Na2CO3(s) = -1130.8kJ/mol
∆H = -737.3 kJ/mol
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∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants)
Determine ∆H for the reaction using the above formula.
C2H5OH(g) + 3O2(g)→ 2CO2(g) + 3H2O(g)
∆H = -1277.4 kJ/mol
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Calculate the amount of energy released when 100.0g of C2H5OH is burned.
C2H5OH(g) + 3O2(g)→ 2CO2(g) + 3H2O(g)
∆H = -1277.4 kJ
-2772 kJ = 2772 kJ released
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Change in Entropy (∆S)
(+∆S) = increase in disorder (entropy)
(-∆S) = decrease in disorder (entropy)
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An Increase in Entropy (+∆S) Is a Force that Drives Physical and
Chemical Changes.
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Second Law of Thermodynamics• Every time a change occurs it
increases the entropy of the universe.
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Second Law of Thermodynamics• The second law implies that whenever
a change occurs, some of the energy will be wasted which will lead to an increase in the disorder of the universe.
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Entropy • Entropy is a complicated concept to truly
understand.• It may be best to think of entropy as the degree of
dispersion.– As matter or energy disperses (becomes more free to
move or becomes more spread out) entropy will increase.
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Entropy 1 Video ≈ 2:25
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Entropy increases when matter is dispersed.
• Production of liquid or gas from a solid “or” production of gas from a liquid results in the dispersal of matter.
• The individual particles become more free to move and generally occupy a larger volumewhich causesentropy toincrease.
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Increase in Entropy (+∆S)• Expansion of a gas.
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Increase in Entropy (+∆S)
• Formation of a mixture.
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Increase in Entropy (+∆S)• Dissolution of a crystalline solid in water.
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Increase in Entropy (+∆S)
• More particles are created.
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Increase in Entropy (+∆S)
• More particles are created.
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Decrease in Entropy (-∆S)• Is simply a reverse of the previous processes.
A gas dissolves in a liquid A precipitate forms
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Third Law of Thermodynamics
The entropy of any pure substance at 0 K is zero.
We can interpret this to mean that as temperature increases entropy increases.
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A more complex molecule has a higher entropy.
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Calculating Entropy Change
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∆S = ∑S(products)
─ ∑S(reactants)
Determine ∆S for the reaction using the above formula.
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
∆S = +95.64 J/mol K
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“Gibbs” Free Energy
• Allows use to determine whether a reaction is spontaneous or not spontaneous.
• We will say a process is or is not “thermodynamically favored” instead of using the terms spontaneous or not spontaneous.
• This avoids common confusion with associating the term spontaneous with the idea that something must occur immediately or without cause.
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“Gibbs” Free Energy
• If a process is “thermodynamically favored” means the products are favored at equilibrium.
• The term “not thermodynamically favored” means the reactants are favored at equilibrium.
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Change in Free Energy (ΔG)
• It is the change in free energy (ΔG) that determines whether a reaction is thermodynamically favorable or not.
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∆G = ∑∆Gf(products) ─ ∑∆Gf(reactants)
Determine ∆G for the reaction using the above formula.
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
∆G = -1306 kJ/mol
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(-∆G) vs (+∆G) • If -∆G the process is “thermodynamically
favored” and the products are favored at equilibrium.
• If +∆G the process is “not thermodynamically favored” and the reactants are favored at equilibrium.
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Understanding ΔG
• Just because a process is spontaneous or “thermodynamically favored” (-∆G) does not mean that it will proceed at any measurable rate.
• A reaction that is thermodynamically favored, (spontaneous), may occur so slowly that in practice it does not occur at all.
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Understanding ΔG
• Thermodynamically favored reactions that do not occur at any measurable rate are said to be under “kinetic control”.
• These reactions often have a very high activation energy.
• The fact that a process does not proceed at a noticeable rate does not mean that the reaction is at equilibrium.
• We would conclude that such a reaction is simply under kinetic control.
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Example
The reaction of diamond (pure carbon) with oxygen in the air to
form carbon dioxide is thermodynamically
favored (spontaneous) but has an extremely
slow rate.
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Understanding ΔG
• A reaction with a with a +ΔG may be forced to occur with the application of energy or by coupling it to thermodynamically favorable reactions.
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+ΔG
• Energy can be used to cause a process to occur that is not thermodynamically favored.
• Using electricity to charge a battery.
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+ΔG
• Energy can be used to cause a process to occur that is not thermodynamically favored.
• Photoionization of an atom by light.
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+ΔG
• A thermodynamically unfavorable reaction may be made favorable by coupling it to a favorable reaction or series of favorable reactions.
• This process involves a series of reactions with common intermediates, such that the reactions add up to produce an overall reaction that is thermodynamically favorable (–ΔG).
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C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
∆H = -1277.3 kJ/mol
∆S = +95.64 J/mol K
∆G = -1306 kJ/mol
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∆G = 0
Reactants ↔ Products
-∆G = thermodynamically favored = products favored
+∆G = not thermodynamically favored = reactants favored
∆G is 0 = equilibrium
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∆G = ∆H – T∆S
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∆G = ∆H – T∆S
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)Determine ∆G given:
∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K
∆G = -1306 kJ/mol or -1,306,000J/mol
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∆G = ∑∆Gf(products) ─ ∑∆Gf(reactants)
Determine ∆G for the reaction using the above formula.
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
∆G = -1306 kJ/mol
∆G = ∆H – T∆S
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)Determine ∆G given:
∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K
∆G = -1306 kJ/mol
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Solve for ∆H: ∆G = ∆H – T∆S
∆H = ∆G + T∆S
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Solve for ∆S: ∆G = ∆H – T∆S
T
GHS
T
HGS
or
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Solve for T: ∆G = ∆H – T∆S
S
GHT
S
HGT
or
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CaO(s) + SO3(g) → CaSO4(s)
Calculate ∆S at 298K using the data given below.
∆H = -401.5 kJ/mol
∆G = -345.0 kJ/mol
∆S = -0.1896 kJ/mol K or -189.6 J/mol K
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∆G = ∆H – T∆S• Can be used to explain the driving forces
and thermodynamic favorability.
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C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
∆H = -1277.3 kJ/mol (exothermic)
∆S = +95.64 J/mol K (increased disorder)
∆G = ∆H – T∆S
∆G = (-) – [+(+)]
negative – positive = always negative
∆G = -1305.2 kJ/mol (thermodynamically favored)
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∆G = ∆H – T∆SWhat are the other possibilities?
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∆G = ∆H – T∆S+∆H = endothermic-∆S = decreased entropy
∆G = ∆H – T∆S∆G = (+) – [+(-)]
positive – negative = always positive
+∆G = not thermodynamically favored
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∆G = ∆H – T∆S+∆H = endothermic+∆S = increased entropy
∆G = ∆H – T∆S∆G = (+) – [+(+)]
positive – positive = ?
+ or - ∆G = may be thermodynamically favored
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∆G = ∆H – T∆S-∆H = exothermic-∆S = decreased entropy
∆G = ∆H – T∆S∆G = (-) – [+(-)]
negative – negative = ?
+ or - ∆G = may be thermodynamically favored
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What is the determining factor for thermodynamic favorability when the
factors ∆H and T∆S contradict?
∆G = ∆H – T∆S
∆G = (+) – [+(+)]
Temperature
∆G = ∆H – T∆S
∆G = (-) – [+(-)]
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A process may be thermodynamically favored at one temperature and not at another.
• Is it thermodynamically favored for ice to melt or water to freeze?
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CaO(s) + SO3(g) → CaSO4(s)The data given below are for 298K. Calculate ∆G using the ∆H and ∆S
values below at 2463K.
∆H = -401.5 kJ/mol∆G = -345.0 kJ/mol∆S = -189.6 J/mol K
∆G = +65.5 kJ/mol at 2463K
Conclusion: ∆S becomes a larger factor as temperature increases
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CaO(s) + SO3(g) → CaSO4(s)Estimate the temperature at which the reaction
becomes spontaneous.∆H = -401.5 kJ/mol∆S = -189.6 J/mol K
T = 2118K = 1845°C
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CaO(s) + SO3(g) → CaSO4(s)Estimate the temperature at which the reaction
becomes spontaneous.∆H = -401.5 kJ/mol∆S = -189.6 J/mol K
T = 2118K = 1845°C
Why is this temperature only an estimate of the temperature?
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CaO(s) + SO3(g) ↔ CaSO4(s)Calculate ∆S at 298K using the data given below.
∆H = -401.5 kJ/mol∆G = -345.0 kJ/mol
∆S = -0.1896 kJ/mol K
• As the temperature increases above ≈ 2118K the reaction becomes nonspontaneous as ∆G becomes positive. Why?
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Boiling & Equilibrium• Boiling is a reversible reaction:
H2O(l) ↔ H2O(g)
• What is the value of ∆G for the
boiling water?
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Estimate the boiling point of ethanol (C2H5OH) .
BP = 350K = 77°C
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Estimate the boiling point of ethanol (C2H5OH) .
Estimated BP = 350K = 77°C
Formula Names Boiling Point
C2H5OH Ethanol
Ethyl alcohol
Hydroxyethane
Grain alcohol
78.4°C
Why is there a discrepancy in the boiling point?
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Thermodynamic Favorability (∆G) and
Equilibrium (K)
∆G K Reaction
Negative >1 Thermodynamically favored (products favored)
0 =1 At equilibrium
Positive <1 Not thermodynamically favored (reactants favored)
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∆G° = -R T lnK
• R = 8.314 J/K
• Used to determine:(1) K when given ∆G°
(2) ∆ G° when a reaction is at equilibrium.
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∆G° = -R T lnK
• Calculate ∆G° for a reaction at equilibrium where K = 1.
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∆G° = -R T lnK
• If K ≠ 1 then ∆G° ≠ 0.
• This means the reactions is moving to the right towards products or moving to the left towards reactants and is not at equilibrium.
• LeChatlier’s Principle should help you to understand this.
• What does changing the temperature do to an equilibrium system.
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What does changing the temperature do to an equilibrium system?
The reaction shifts right or left favoring the products or reactants
and raising or lowering K.
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Every reaction has a temperature at which K = 1 and the reaction is neither
favoring products nor reactants
Reaction Kelvin
TemperatureK
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The equilibrium constant for a reaction at 0°C is 1.657 x 10-5. What is ∆G°?
∆G° = 24980J/mol = 24.98 kJ/mol
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Calculate the equilibrium constant for a reaction at 298K if ∆Gº = -228.59 kJ.
K = 1.17 x 1040
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K vs ∆G
• Compare K and ∆G from the last two problems.
• ln K = -11.0079 and K = 1.657 x 10-5 and ∆G° = 24.98 kJ/mol
• ln K = 92.264 and K = 1.17 x 1040 and ∆G° = -228.59 kJ/mol
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5 Formulas
1. ∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants)
2. ∆S = ∑S(products)
─ ∑S(reactants)
3. ∆G = ∑∆G(products)
─ ∑∆G(reactants)
4. ∆G = ∆H – T∆S
5. ∆G° = -RTlnK
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Entropy 2 Video ≈ 8:21