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CHEMICAL THERMODYNAMICS AP Chemistry Chapter 19 Notes

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REVIEW 1st Law of Thermodynamics Energy in the universe is conserved

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REVIEW Path Function value depends on how process takes place (i.e. q, w)

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heat q = mC p T

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REVIEW State Function value independent of path - is a defined reference or zero point (i.e. H)

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Enthalpy - H zero point H f 0 of elements in natural form at 25 0 C H f 0 of elements in natural form at 25 0 C

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Enthalpy - H types: H rxn = n H f 0 (prod) - n H f 0 (react)

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Enthalpy - H types: H fus, H vap

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ENTROPY S a measure of randomness or disorder of a system

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Entropy is NOT conserved. The universe seeks maximum disorder.

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2nd Law of Thermodynamics for a spontaneous process, entropy increases

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Entropy - state function zero reference: S = 0 for a perfect crystal at absolute zero

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S solid < S liquid << S gas

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S 0 = standard entropy of elements & cmpds at P = 1 atm; T = 25 0 C units = J/K mol [Appendix L, text]

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Calculate S using Hess Law S(rxn) = nS 0 (prod) - nS 0 (react)

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Example: Ca(s) + C(gr) + 3/2 O 2 CaCO 3 (s)

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S univ = S sys + S surr if S univ > 0 process spontaneous

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S univ = S sys + S surr if S univ < 0 process ?

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S univ = S sys + S surr if S univ = 0 process ?

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to relate S and H consider

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H 2 O(s) H 2 O(l) where the water is the system & everything else is the surroundings

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Temperature Dependence S = J/mol (1/T) S surr = - H/T

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S sys S surr S univ spon + +

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S sys S surr S univ spon Y - -

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S sys S surr S univ spon Y N + -

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S sys S surr S univ spon Y N + - ? D - +

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S sys S surr S univ spon Y N + - ? D - + ? D

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For a phase change:

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GIBBS FREE ENERGY G

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most abstract of thermodynamic state functions

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w 1 w = reversible work w 2 P V

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Definition: G = w + PV w: reversible work PV: pressure/volume work isothermal, reversible path

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G = w + P V + V P at constant P P = 0 so V P = 0 G = w + P V

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G = w + P V + V P at constant P G = w + P V at constant V V = 0 so P V = 0 G = w = useful work

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G cannot be measured must measure G over a process

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ZERO REFERENCE G = 0 for elements in stable form under Standard Thermodynamic Conditions T = 25 o C ; P = 1 atm

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G f 0 = standard Free Energy of Formation from the elements Appendix L, text

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G follows Hess Law: G 0 (rxn) = n G f 0 (p) - n G f 0 (r)

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Summary of Laws of Thermodynamics Zeroth Law: Heat Gain = Heat Loss

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Summary of Laws of Thermodynamics First Law: Law of Conservation of Energy

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Summary of Laws of Thermodynamics Second Law: Defines Entropy

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Summary of Laws of Thermodynamics Third Law: Defines Absolute Zero

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GIBBS HELMHOLTZ EQUATION

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G H T combine

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G = -aT + H G, H are state functions, thus a must be a state function G = - S(T) + H

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Gibbs Helmholtz Equation G = H - T S

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Units on the State Functions

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thus, a process is spontaneous if and only if G is negative

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Spontaneity controlled by enthalpy (minimum energy)

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Sponaneity controlled by enthalpy entropy (maximum disorder)

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Sponaneity controlled by enthalpy entropy both

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Predict Spontaneity IF H(-) and S(+) G = - H - T(+ S) G spontaneous

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Predict Spontaneity IF H(+) and S(-) G = + H - T(- S) G > 0, => NOT spontan

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Summary of Spontaneity H S G Spont yes no or - ? or - ?

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Uses of the Gibbs Helmholtz Equation

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1.Find the molar entropy of formation for ammonia.

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2.Elemental boron, in thin fibers, can be made from a boron halide: BCl 3 (g) + 3/2 H 2 (g) ->B(s) + 3HCl(g)

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Calculate: H 0, S 0 and G 0. Spontaneous? Driving force?

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3. Using thermodynamic information, determine the boiling point of bromine.

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Thermodynamic Definition of Equilibrium G eq = 0

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by definition G = H - TS & H = E + PV

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thus, G = E + PV - TS take derivative of both sides dG = dE + PdV + VdP - TdS - SdT

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for a reversible process TdS = q

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derivative used for state function while partial derivative used for path function

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if the only work is PV work of expansion PdV = w

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First Law of Thermodynamics E = q - w dE = q - w = 0

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thus q = w or TdS = PdV

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by substitution dG = 0 + PdV + VdP - PdV - SdT or dG = VdP - SdT

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lets assume we have a gaseous system at equilibrium, therefore, examine K p at that constant temperature

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at constant T SdT = 0 thus dG = VdP

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but

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if condition 2 is at standard thermodynamic conditions, then G 2 = G 0 and P 2 = 1 atm

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thus

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determine G for aA + bB cC + dD where all are gases

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G = G prod - G react = cG C + dG D - aG A - bG B

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but and likewise for the others

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and

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but

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Thus, G = G 0 + (RT) ln Q

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But aA + bB cC + dD G = 0

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thus or in general

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THERMODYNAMICS&EQUILIBRIUM

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3. Using thermodynamic information, determine the boiling point of bromine.

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FACT: Product-favored systems have K eq > 1. FACT: Product-favored systems have K eq > 1. Thermodynamics and K eq

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Therefore, both G˚ rxn and K eq are related to reaction favorability. Therefore, both G˚ rxn and K eq are related to reaction favorability. Thermodynamics and K eq

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K eq is related to reaction favorability and thus to G o rxn. The larger the value of K the more negative the value of G o rxn Thermodynamics and K eq

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G o rxn = - RT lnK where R = J/Kmol Thermodynamics and K eq

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Calculate K for the reaction N 2 O 4 2 NO 2 G o rxn = +4.8 kJ K = 0.14 When G 0 rxn > 0, then K 0, then K < 1 G o rxn = - RT lnK Thermodynamics and K eq

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G, G˚, and K eq G is change in free energy at non-standard conditions. G is change in free energy at non-standard conditions. G is related to G˚ G is related to G˚ G = G˚ + RT ln Q where Q = reaction quotient G = G˚ + RT ln Q where Q = reaction quotient

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G, G˚, and K eq When Q K, reaction is spontaneous. When Q K, reaction is spontaneous. When Q = K reaction is at equilibrium When Q = K reaction is at equilibrium When G = 0 reaction is at equilibrium When G = 0 reaction is at equilibrium Therefore, G˚ = - RT ln K Therefore, G˚ = - RT ln K

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Product favored reaction –G o and K > 1 In this case G rxn is < G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion. G rxn is < G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion. G, G˚, and K eq

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Product-favored reaction. 2 NO 2 N 2 O 4 G o rxn = – 4.8 kJ Here G rxn is less than G o rxn, so the state with both reactants and products present is more stable than complete conversion. G, G˚, and K eq

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G o rxn is the change in free energy when reactants convert COMPLETELY to products. G o rxn is the change in free energy when reactants convert COMPLETELY to products. Thermodynamics and K eq Overview

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K eq is related to reaction favorability. When G o rxn < 0, reaction moves energetically downhill

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4. For the following reaction, calculate the temperature at which the reactants are favored.

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THERMODYNAMICSOFCHEMICALREACTIONS

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5. How much useful work can be obtained from an engine fueled with 75.0 L of hydrogen at 10 C at 25 atm?

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6. The reaction to split water into hydrogen and oxygen can be promoted by first reacting silver with water. 2 Ag(s) + H 2 O(g) Ag 2 O(s) + H 2 (g) Ag 2 O(s) 2 Ag(s) + 1/2 O 2 (g)

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Calculate H 0, S 0 and G 0 for each reaction.

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Combine the reactions and calculate H 0 and G 0 for the combination. Is the combination spontaneous?

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At what temperature does the second reaction become spontaneous?

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7. The conversion of coal into hydrogen for fuel is: C(s) + H 2 O(g) CO(g) + H 2 (g)

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Calculate G 0 and K p at 25 0 C. Is the reaction spontaneous?

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At what temperature does the reaction become spontaneous? Calculate the temperature at which K = 1.0 x

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8. The generation of nitric acid in the upper atmosphere might destroy the ozone layer by: NO(g) + O 3 (g) NO 2 (g) + O 2 (g)

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Calculate G 0 (reaction) and K at 25 0 C.

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