PDF《EXAM 1》

1 EXAM

1 Physical constants/conversion factors

1 N =

1 kg m s-2

1 J =

1 N m =

1 kg m2 s-2

1 bar =

105 N m-2

1 atm =

101325 N m-2

1 bar = 750.

06 Torr

1 cal = 4.184 J

1 C =

1 A s,

1 V =

1 J C-1 ,

1 Ω =

1 V A-1 R = 8.3145 J K-1 mol-1 = 0.08315 (L atm) (mol K)-1 NA = 6.022136*1023 mol-1

1 L = 0.001 m3 k = 1.38*10-23 J/K, kN = Rn

10 L bar =

1 kJ Laws of thermodynamics 0th law C If A and B are in thermal equilibrium and B and C are in thermal equilibrium, then A and C are in thermal equilibrium. Defines Kelvin scale. 1st law C U is conserved. dw dq dU + = 2nd law C defines entropy and dir of time. rev dq S T Δ = ∫ 3rd law C absolute scale. S →

0 as T →

0 for pure crystal. Open: mass and energy can transfer Closed: only energy, not mass Isolated: neither energy nor mass Extensive: depend on size of system (n, m, V) Intensive: independent of size (T, p, Vbar) pV = nRT van der Waals:

2 ( )( ) a p V b RT V + ? = w F = ?A , also

2 2

1 ( )

2 f i f i w k d k = ? = ? ∫ A A A A A A Expansion work: ( ) ext ext w p A p = ? = ? Δ A V Surface work: ext dw dA γ = (where γext is surf tens in J/m2 ) Electrostatic work: dw Vde = If heat enters system, it is positive If system does work on surroundings, w <

0 If surroundings do work on system, w >

0 dw ∫ v may be, but is not neccesarily

0 path dq C dT = , Cp>

Cv

0 dw dq dU + = = ∫ v , so

2 2

1 1 U dU U U q Δ ∫ w Isothermal gas expansion 1. if pext = 0, , ΔU=q

0 ext w p V = ? Δ = 2. if pext=p2,

2 w p V = ? Δ 3. if pext=p (reversible),

2 1 V V nRT w dV V = ?∫

2 1

2 2

1 1 ln ln V V V p nRT dV nRT nRT V V ? = ? = ∫ p path ext V T U U dU C dT p dV dT dV T V ? ? ? ? ? ? = ? = + ? ? ? ? ? ? ? ? ? ? Constraints: reversible -- rev ext dU dq p dV = ? isolated -- dq

0 dw U = = Δ = dU dw adiabatic -- dq

0 = ? = constant V --

0 V V dw dU dq C dt = ? = = V V J dU C dT C dV η = ? for ideal gas, ΔU=0 for any isoT exp/comp also for isoT, ΔH = ΔU + nRΔT =

0 for ideal gas, ηJ=0 Joule free expansion q =

0 (adiabatic), w =

0 (pext = 0) isothermal, so ΔU, ΔH are

0 A constant internal energy process. Joule-Thomson expansion q =

0 (adiabatic), w = ΔU = - Δ(pV). ΔH = 0. A constant enthalpy process. Enthalpy H U pV ≡ + , p H q Δ = for reversible const P process p H Cp T ? ? ? = ? ? ? ? ? , dH p p JT C dT C dp = μ ? For an ideal gas p V C C R + v p C T p V Δ = Δ + Δ , C T = For an ideal gas V dU C dT = , p dH C dT = Reversible Adiabatic Expansion/Compression Monatomic IG:

3 2 V C R = ,

5 2 p C R = ,

5 3 p V C C γ ≡ =

1 2

1 1

2 T V T V γ ? ? ? ? ? = ? ? ? ? ? ? ? ? and

2 1

1 2 P V P V γ ? ? ? ? = ? ? ? ? ? ? ? ? Adiabatic expansion, gas cools Adiabatic compression, gas heats up

2 Irreversible Adiabatic Expansion/Compression

2 2

1 1 ( ) V V P T C R T C R P ? ? + = + ? ? ? ? T where pext=p2 (-wrev) >

(-wirrev) C less work recovered from irrev process Cycles ΔU, ΔH are state functions dq, dw are not state functions Thermochemistry (products) (reactants) rx i f i f i i H H H ν ν Δ = Δ ? Δ ∑ ∑ D D ΔHrx <

0, qp <

0, exothermic;

ΔHrx >

0, qp >

0, endothermic ΔHrx is the ΔH for isothermal reaction at constant p ΔHf is the ΔH for the creation of

1 mole of compound Calorimetry

2 1

1 ( ) (Prod+Cal) T rx p T H T C dT Δ = ?∫ (for constant P)

2 1

1 1 ( ) (prod+cal) T rx V gas T H T C dT RT n Δ = ? + Δ ∫ (const V)

2 1

2 1 ( ) ( ) T rx rx p T H T H T C dT Δ = Δ + Δ ∫ rxn rxn rxn gas U H pV H n R Δ = Δ ? Δ = Δ ? Δ Entropy

0 rev dq T = ∫ v ,

0 irrev dq T <

∫ v rev dq dS T = ∫ ∫

2 2

1 1

1 1 q T q T ε = + = ? for isolated systems, ΔS >

0 C spontaneous, irreversible, ΔS = 0, reversible (or equilibrium)

2 2

1 1 rev dq S S S T Δ = ? = ∫ Entropy C Joule expansion of IG

2 1

2 1 ln V rev back V dq V dw RdV S T T V ? ? Δ ? ? ∫ ∫ ∫ R V Entropy C Reversible expansion of IG

2 1

1 2

1 ln V rev V dq p RTdV S pdV R T T V ? ? Δ ? ? ∫ ∫ ∫ p Entropy C Mixing of IGs at constant T, p [ ln ln rev mix A A B B dq S nR X X X T Δ ∫ where A A A tot tot n V X n V = = Entropy C Heating/cooling at constant V

2 1

2 1 ln T V V T C dT T S C T T Δ = = ∫ Entropy C Heating/cooling at constant p

2 1

2 1 ln T p p T C dT T S C T T Δ = = ∫ Entropy C Reversible phase change at constant T, p vap p v vap b b q H S T T Δ Δ = = ap Entropy C Irreversible phase change at constant T, p

1 1 ln fus T fus fus p p p T H T dT S C C C s T T T ?Δ Δ = + Δ = ? ∫ A Surroundings and Universe (if T, p surroundings constant) sys surr H S T Δ Δ = , universe sys surr S S S Δ = Δ + Δ Carnot Cycle if reversible, ΔSuniv=

0 Step 1→2

2 2

1 1

1 1

1 0 l V U q w pdV RT V Δ ∫ n 2→3 '

1 2 ( ) v U w C T T Δ = = ?

1 3→4

3 4

1 2

4 3

0 l V U w pdV RT V Δ ∫ n 4→1 '

2 1 ( ) v U w C T T Δ = = ?

2 ] X

2 2

4 3

1 1

2 1 ln( ) ln( ) q T V V q T V V = ,

1 1

1 4

3 2

1 2 T V V T V V γ ? γ ? ? ? ? ? = = ? ? ? ? ? ? ? ?

1 2

1 2

0 rev q q dq T T T = = ∫ v + Fundamental Equations dU TdS pdV = ? (valid for any closed system, rev/irrev) dH TdS Vdp = + Generally,

0 ext surr dU p dV T dS + ? <

Helmholtz free energy: A U TS = ? Under constant T=Tsurr, constant V, dA >

products endo Heterogeneous equilibria C equilibrium constant includes only gases, but ΔG° includes all products and reactants. Most stable compound at (T,p) has lowest (most negative) molar Gf Function What is kept constant? Greater than/less than dU S, V, ni ≤

0 dA T, V, ni ≤

0 dH S, P, ni ≤

0 dG T, P, ni ≤

0 Boltzmann equation ln S k = Ω where Ω is multiplicity (# poss. states)

1 ln t i i i S k p p = = ? ∑

1 2 ! i N n n n Ω = * * (general case;

ni = degeneracy) number of ways to mix up N things with degeneracies if no degeneracy, then just N! )

4 ! ( , ) !( )! N N k N k k N k ? ? Ω = = ? ? ? ? ? (binomial case, N=ntot) number of ways to pick k things from N objects Nk sequence of k items, each with N deg free maximizing Ω maximizes S A A n p N = , i i all p ε ε = ∑ Stirling'

s approximation ! n n n e ? ? ≈ ? ? ? ? , , ln ! ln n n n n = ? ln ln x n x = n Boltzmann distribution law *

1 exp j j E p Q kT ? ? ? = ? ? ? ? ? , exp t j j E Q kT ? ? ≡ ? ? ? ? ? ∑ relative populations: * * ( ) exp i j i j E E p p kT ? ? ? ? = ? ? ? ? As T→∞ or Ej→0, all states accessible As T→0 or Ej→∞, only ground state accessible More on the Partition Function sys i i Q q = ∏ For N independent, distinguishable particles, N Q q = For N independent, indistinguishable particles, ! N q Q N =

1 kT β = , U E N = , so

2 lnQ U kT T ? ? ? = ? ? ? ? ? ln U S k Q T = + , ln A kT = ? Q , ln T V Q kT N μ ? ? ? = ? ? ? ? ? ? , , ln T N Q P kT V ? ? ? = ? ? ? ? ? Absolute Entropy ln p S p T S T R p = ? D D At the given reference pressure, S°(T)=

0 (0 ) m b m b T T T fus vap p p T T m b C s dT H C dT H C g dT S K T T T T T Δ Δ ∫ ∫ ∫ D A p Integrate to whatever temperature desired (and adjust phase changes accordingly) For every chemically homogeneous substance in a pure crystalline state, as T→0 K, S→0. Phase Equilibria coexist dp S S S H dT V T V V V β α β α α β α → → ? ? ? Δ Δ = = = ? ? Δ Δ ? ? ? β fus m fus m H T p V T Δ Δ Δ ? ? Δ (for melting point increase) Clausius-Clapeyron:

2 2

1 1 ln

1 1 ln p d p H H dT RT p R T T ? ? Δ Δ ? ? ≈ ? = ? ? ? ? ? ? ? ? ?

2 sub vap fus H H H Δ = Δ ? Δ (watch signs!!)

1 1 fus s V ρ ρ Δ = ? A

5 EXAM

3 liquid mole fraction C xA gas mole fraction C yA F = C C P +

2 Raoult'

s Law (xA → 1) * * (1 ) A A A A B p p x p x = = ? * * * * ( ) C C C B C B x p y C p p p x = + ? * * * * ( ) C B C C B C y p x C p p p y = + ? * * * * * * ( ) B C C C C B C C C p p x p p p p p y y = = + ? Constructing T-x diagrams 1. Use Clausius-Clapeyron 2. p =

1 atm = 1.013 bar 3.

0 ln vap H p C p RT Δ = + 4. Use Raoult'

s Law If A is more volatile than B, then * * A B T T <

temp at which pure A has vapor pressure = p * * * B A A B p p x p p ? = ? , * B B B x p y p = For an ideal solution, * ln i i T p T p RT x μ μ = + A A i A In non-ideal solutions, calculate Δu = 2uAB C (uAA + uBB) if Δu >

0: positive deviation. like associates with like. p >

Raoult. Most common. if Δu <

0: negative deviation. mixing favored. p <

Raoult. If there is a minimum on coexistence curves, then dew line and bubble line touch C azeotrope. Henry'

s Law (xA → 0) A A p x K = where K is empirical constant for positive deviations, K >

p star for negative deviations, K <

p star Important terms: Mole fraction Molality C mB = (moles solute)/(kg solvent) = n B BB/(nAMA) Molarity C ? = nB/V B A Colligative Properties 1. Vapor pressure lowering: * * A A A B p p p x p Δ = ? = ? 2. Boiling point elevation: * b b b b T T T K m Δ = ? = B where *

2 ( ) A b b vap M R T K H = Δ 3. Freezing point depression: * f f f f T T T K m Δ = ? = ? B where *

2 ( ) A f f freezing M R T K H = Δ 4. Osmotic pressure: B RTc π = (and B V RTn π = ) also gh π ρ = Rate of reaction if aA + bB → cC + dD

1 1 [ ]

1 [ ]

1 [ ]

1 [ ] i N i i d A d B d C d D k C a dt b dt c dt d dt γ = t1/2 is the time at which [A]t = 0.5[A]0 Zero order reaction (A → products) [A]t = [A]0 C kt and t1/2 = [A]0/2k First order reaction (A → products)

0 [ ] [ ] kt t A A e? = ,

0 ln[ ] ln[ ] t A kt A = ? + , 1/

2 ln

2 t k = Second order reaction (2A → products)

0 1

1 (2) [ ] [ ] t kt A A = + , 1/2

0 1

2 [ ] t k A = Second order reaction (A + B → products)

0 0

0 0 [ ] [ ]

1 ln t t A B kt A B A B = ? special case i: [A]0=[B]0. Then

0 1

1 [ ] [ ] t kt A A = + , [A]t=[B]t special case ii: [B]0>

k1 C B is small but not zero) Assume [ ]

0 d B dt = , so

1 1

2 [ ] [ ]ss k A B k k ? = + '

0 [ ] [ ] k t t A A e? = where

1 2

1 2 '

k k k k k ? = + ( '

k A C ?? → ) 3) Pre-Equilibrium Approximation

1 2 k k A B C ←? → ?? → (valid when when k2 ........