Thermodynamics
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Transcript of Thermodynamics
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Indicated by a degree symbol Gases at 1 atm Liquids are pure. Solids are pure. Solutions are at 1 M. Temperature is 25o C (298 K).
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Conservation of Energy—Energy can be neither created nor destroyed.
The amount of energy in the universe is constant.
Energy can change forms, but the amount does not change.
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Chemical (potential) energy is stored in the bonds of methane gas.
Combustion releases the energy stored in the bonds as heat.
Energy exchange between system & environment = enthalpy
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Spontaneous processes need no outside cause; they happen naturally.
Spontaneous = Fast (Thermodynamics is not
concerned with how a reaction happens—just whether it will happen.)
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Exothermicity?? (Tendency to reach lowest potential energy)
Partial explanation, but does not always hold true
Probability—how likely are arrangements of atoms (randomness or entropy)
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The amount of disorder or randomness in a system
Higher entropy is favored because random arrangements are more likely.
Ex.: Gas molecules are more likely to be evenly dispersed than clumped up in one place.
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1 mole of gas or 1 mole of liquid?
Pure water or a 0.5 M salt solution?
Gas at 25o C or gas at 100o C? 1 mole of gas or 5 moles of gas?
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In state changes--solid to liquid, liquid to gas or solid to gas
When solids dissolve When temperature increases When volume of a gas increases
(or pressure decreases) When a reaction occurs that
produces more moles of gases
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In any spontaneous process, there is an increase in entropy of the universe.
Think of the universe as being inclusive of a system and its surroundings—combined entropy must increase (Cell)
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Sign depends on direction of heat flow—heat added to surroundings increases entropy & heat absorbed from surroundings decreases entropy
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Magnitude—depends on temperature—greater impact of exothermic system if temperature is low
Analogy—giving $50 to a millionaire or to you—to whom does it mean more?
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Equal to –H/T negative—because the
surroundings are opposite from the system
H –change in heat in the system (from system’s perspective)
T—temperature in Kelvin
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ssurr is larger at lower temperatures because the denominator of the fraction is smaller .
The effect of heat flow for a process is more significant when temperature is lower.
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G Uses enthalpy, entropy and
temperature to predict the spontaneity of a reaction
If G is negative, then a process is spontaneous.
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G = H – TS Gis free energy is change in enthalpy Sis change in entropy All terms are from the point
of view of the system.
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Mathematical relationships: If H is negative and S is
postive—always spontaneous If H is positive and S is
negative—never spontaneous If both H and S have the
same sign—temperature dependent
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At what temperature is the following process spontaneous?
Br2 (l) Br2 (g)
H = 31.0 kJ/mol & S = 93.0 J/K mol
Calculate temperature when G is zero. (G = H – TS)
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At the exact boiling point, G is zero.
Set known values of H, S and/or T equal to zero and solve for unknowns.
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The enthalpy of vaporization of mercury is 58.51 kJ /mol and the entropy of vaporization is 92.92 J/K mol. What is mercury’s normal boiling point?
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The entropy of a perfect crystal at 0 K is zero.
Standard entropy values have been determined—Appendix 4
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To determine entropy change for a reaction, subtract the entropy values
S = npSoproducts – nrSo
reactants
n = number of moles (State functions depend on amount of substance present.)
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Calculate So at 25o C for the reaction:
2NiS(s) + 3O2 (g) 2SO2 (g) + 2NiO (s) So in J/K mol:
› SO2 248 J/K mol
› NiO 38 J/K mol
› O2 205 J/K mol
› NiS 53 J/K mol
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Calculate So for the reduction of aluminum oxide by hydrogen gas:
Al2O3(s) + 3H2(g) 2 Al(s) + 3 H2O(g)
Note: equal number of gas molAns: 179 /K (large & positive
because of water’s asymmetry)
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Cannot be directly measured Can be calculated using
enthalpy and entropy changes (G = H – TS)
Can be calculated from already determined standard free energy values: G = npGo
products – nrGoreactants
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Calculate H, S, and G using the following information for the reaction 2SO2 (g) + O2 (g) 2SO3(g).
SO2: H = -297 kJ/mol; So = 248 J/K mol
SO3: H = -396 kJ/mol; So = 257 J/K mol
O2: H = 0 kJ/mol; So = 205 J/K mol
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Using the following data, calculate G for the reaction Cdiamond (s) C graphite (s)
Cdiamond (s) + O2 (g) CO2 (g) G = -397 kJ
Cgraphite (s) + O2 (g) CO2 (g) G = -394 kJ
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Given the following free energies of formation, calculate the G for the reaction
2CH3OH (g) + 3O2 (g) 2 CO2 (g) + 4 H2O (g)
CH3OH (g): G = -163 kJ/mol
O2 (g): G = 0 kJ/mol
CO2 (g): G = -394 kJ/mol
H2O (g): G = -229 kJ/mol
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Entropy depends on pressure because changes in pressure affect volume.
G = Go + RTln(P) G = free energy at new pressure Go = free energy at 1 atm R = gas constant (8.3145 J/K*mol) T = Kelvin temperature P = new pressure
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From the previous equation, a relationship between G and Q or K can be derived.
(See p. 807) G = Go + RT ln (Q)
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CO (g) + 2H2 (g) CH3OH (l) Calculate G for this process
where CO is at 5.0 atm and H2 is at 3.0 atm.
Calculate Go using standard values for reactants & products.
Then use G = Go + RT ln (Q)
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At equilibrium, a system has used all of the free energy available.
G equals zero because forward & reverse reactions have the same free energy.
G = Go + RT ln (Q) = Go + RT ln (K) Go = - RT ln (K)
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For the synthesis of ammonia, Go = -33.33 kJ/mol of N2 consumed at 25o C.
Predict how each system below will shift in order to reach equilibrium:› NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2
atm
› NH3 = 1.00 atm; N2 = 1.00 atm; H2 = 1.00 atm
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NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2 atm
G = Go + RT ln (Q)
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G = Go + RT ln (Q)
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4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s) Use the data to calculate K for this
reaction at 25o C. Fe2O3: H = -826 kJ/mol; So = 90
J/K mol Fe: H = 0 kJ/mol; So = 27 J/K mol O2: H = 0 kJ/mol; So = 205 J/K mol
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