1

Physical chemistry 1

September 9

thermal equilibrium

The perfect gas law

1.2. The (perfect) gas law

Check Boyle's law pg. 34. with increasing pressure, temperature increases. decreasing

volume will increase pressure and temperature.

Combining all variables we have the perfect gas law:

a) The Perfect Gas Law is a combination of simpler gas laws

Boyle's law : pV = constant

=> pV = nRT, with these parameters:

- the pressure has to be about 1 atm

- assume no interactions between gas molecules

- assume perfectly elastic collisions

- molecules have no size (volume)

all these assumptions go into:

b)The Kinetic Molecular Theory:

pV = nRT = (1/3)nMc2 ,

where c = average speed, the root mean squared velocity, <v2>

1/2

To solve for c, we get:

c = (3RT/M)1/2

Now take into account that there are interactions and that molecules have size.

Molecules can have induced dipoles moment.

Real Gases

Deviations from behavior of perfect gases:

-molecular interactions

-molecules and atoms have a size (volume)

1.3. Molecular Interactions

-important at high pressure and at low temperature

(look at compression factor table in book; Z = pVm/RT

a) Compression factor: pV = nRT , V/n = Vm , pVm = RT ,

Z= (pVm)/(RT) = 1 , where Z is the compression factor and this is for

perfect gases. (Refer to graph).

b) Virial (force and energy) Coefficients:

pVm/RT = 1 + B/Vm + C/Vm2

2

y = 1 + BX + CX2...

pVm/RT = 1 ===> perfect gas behavior

B/Vm =======> two-body interactions

C/Vm2 ======> 3 and more body interactions

This holds for any gas and any curve.

1.4. The Van der Waals equations:

- molecules take up space, so Vreal = Vperfect + n*b , where b is looked up for a

specific gas

-molecules interact: Preal = Pperfect - a(n/Vreal)2

use Pperfect*Vperfect = nRT ===> [Preal + a(n/Vreal)2][Vreal-nb] = nRT , van der Waals gas

law. (Memorize this gas law)

a comes from molecular interactions, and b comes from molecules taking up space

Physical chemistry 1

September 12

CHAPTER 2: THE FIRST LAW OF THERMODYNAMICS

- deals with the conservation of energy

- can be used to calculate changes in the amount of relative energy changes associated

with physical and chemical processes. (temperature, size of the system has changed,

work)

Theromodynamics: the study of the transformation of energy. Energy is neither created

nor distorted, it is transformed.

Basic Concepts:

System: the part of world that we are interested in studying.

Surroundings: everything else, everything that is not the system.

Open System: molecules and energy can flow in and out. (energy + matter can be

exchanged with surroundings).

Closed System: energy can be transferred, matter cannot be transferred into surroundings.

Isolated System: cannot exchange energy or matter with the surroundings.

2.1 Work, Heat, and Energy

a) Definitions and general concepts:

-Energy - the capacity to do work

-Work is done against opposing force

-Heat transfer occurs as the result of temperature differences

work and heat are the two ways of transferring or converting energy

3

Endothermic and Exothermic processes. Diathermic boundary (thermally conducting),

endothermic will take in heat and exothermic gives off heat.

Adiabatic boundary (thermally insulating), no heat will exchange with the surroundings

in this boundary.

b) Molecular Interpretation of Heat and Work:

-is based on what happens in the surroundings

-heating - increased disorderly motion of molecules (thermal motion, random

movement)

-work - involes organized motion of molecules with respect to the surroundings

2.2 The Internal Energy, U:

- the internal energy, U, is the total (kinetic and potential) energy (of the molecules) in a

system.

- U is a state function, an extensive property (depends on how many molecules we have)

Δ U = Ufinal - Uinitial , where the units come out to joules.

molar internal energy : U/n = Um , where the unit is joules/mole ==> J/mol

a) molecular interoperation of U:

The equipartition theorem tells us that the average energy of each degree of freedom is

(1/2)KT

-monatomic gas ( has only translation, but no rotation)

Therefore Um(T) = Um(0) + 3/2RT, where Um is potential energy from electron

distribution and 3/2RT is the kinetic part.

-linear molecule: Um(T) = Um(0) + 5/2RT

-nonlinear molecule: Um(T) = Um(0) + 3RT

For a gas of nonlinear molecules:

Twice as much energy must be supplied to raise the temperature of the nonlinear gas, by

the same amount as a monatomic gas. (Different heat capacities).

b) Formulation of the First Law of Thermodynamics:

-Internal energy can be changed by heating or by doing work (equivalent ways of

changing the Internal Energy, U).

The internal energy of an isolated system is constant.

Δ U = q + w , for an isolated system ( no energy transfer), where q is energy and w is

work

which means q = 0, w = 0, and Δ U = 0 (definition of an isolated system)

4

P. Chem. September 14

equation for changes in internal energy of a system Δ U = q + W

an isolated will not let energy or matter transfer outside of the system.

2.3 Expansion Work

- work is done by a gas against an external pressure

- infinitesimal changes (ʃ and such)

dU = dq + dW (ʃ form)

a) work (some general aspects)

dW = -Fdz = -pAdz , where Adz is the dV and p is the external pressure.

W = -((ʃ from Vi to Vf)pex)dV

b) Free expansion =

when pex = 0 and W = 0

c) Expansion against a constant pex (external pressure, often air pressure for open

systems)

W = -pex ((ʃ from Vi to Vf))dV

= - pex(Vf - Vi)

= -pex(ΔV)

d) Reversible Expansion

- requires equilibrium all the time; the changes made have to be infinitesimal

W = -((ʃ from Vi to Vf)pex)dV

= - ((ʃ from Vi to Vf)p)dV

- in certain cases, we can know what to put in for p for p(V).

e) Isothermal Reversible Expansion

- for example: perfect gas

pV = nRT ==> p = nRT/V

W = -((ʃ from Vi to Vf)nRT/V)dV

= -nRT((ʃ from Vi to Vf)1/V)dV

= -nRT [lnV](from Vi to Vf)

= -nRT [lnVf - lnVi] = -nRTln(Vf / Vi)

ex. 2.1 pg. 53

5

Fe(s) + 2HCl(aq) ---> FeCl2(aq) + H2(g)

find Volume of H2(g) ʃ

open beaker pex is constant

W = -pex(ΔV) = -pex(ΔnRT/pex) = (Δ n)RT

= -........

2.4 Heat Transactions (exchanges)

- between bodies of different temperatures, T, energy is transferred as heat

- general: dU = dq + dWexp (+dWextra) , only concerned with dWexp --> pdV

- if there is no change in V, then there is no expansion work, Wexp.

===> dU = dq

For measurably large changes we use Δ U = qV , where the heat and

volume remain constant

September 16, 2011

Generally

dU = dq + dWexp (+dWextra)

if no change in V, there is no Wexp.

=> dU = dqv

Δ U = qv heat at constant V

"bomb" calorimeter, measures energy released as heat (at constant V)

a) calorimetry: the study of heat transfer

Δ T of the calorimeter is proportional to the energy absorbed by the calorimeter or

released as heat.

qv = C(Δ T) , where C is the calorimeter constant

Um(T) = Um(0) + ....RT

b) Heat Capacity

- U typically increases when T increases (and decreases when V increases)

- (δU/δT)v = Cv , heat capacity at constant V

- molar heat capacity Cv/n = Cv,m

-specific heat capacity Cv/m = Cv,s , for water Cv,s = 4.18 J/gK

-for small steps in T (small Δ T), Cv can often be considered constant

2.5 Enthalpy

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- enthalpy is heat transferred at constant p ( we allow the volume to change).

- if we have a gas, the volume increases as T increases, so the gas can do work

dU < dq

a) Enthalpy ( H)

H (enthalpy) = U (internal energy) + pV @ constant p

Δ H = Δ U + Δ (pV)

Δ U = Δ H - Δ pV

compare to: Δ U = q + W

H = qp

dH = dqp

fore measurably large changes:

Δ H = qp

b) Measurements of qp

- calorimetry at constant p can be done in a verssel opened to the atmosphere

- Δ H = Δ U + pex(Δ V) , for a gas

Δ H is approx. Δ U for liquids and solids

- for a perfect gas:

Δ H = Δ U + Δ nRT

or Δ U + nRΔ T

c) Variation of Enthalpy with T

-H increases as T increases

(δH/δT)p = Cp ( partial derivatives)

Cp/n = Cp,m molar heat cap. at constant p

- Cp,m = (δH/δT)p = a + bT + C/T2

ex: 2.4 on page 61

What is the molar change in the enthalpy of N2?

Δ Hm, N2

Ti = 25 degrees C

Tf = 100 degrees C

Δ H = (the ʃ from Ti to Tf of (a + bT + C/T2))dT

-most systems expand, i. e., do work on their surroundings when heated

==> Cp - Cv = nR for perfect gas

Cp(Δ T) - Cv(Δ T = nR(Δ T)

Δ H - Δ U = nR(Δ T)

7

September 19

isothermal reversible expansion

Δ U = 0

Δ U = q + W

2.6 Adiabatic Changes

-adiabatic expansion of a perfect gas is a decrease in T.

( no heat exchange => U decreases).

-Δ U = CV(Tf - Ti) = Wadiabatic (Wad)

Recall: Boyle's Law

Pressure changes with volume, while T is constant

- isothermal curve is higher than adiabatic expansion curve (in graph p. Vs. V.)

Reversible adiabatic expansion pressure falls a little faster than isotherm

piVigamma

= pfVfgamma

gamma = Cp,m/Cv,m > 1

q = 0 (no heat exchange)

Thermal Chemistry:

- the study of energy transfer as heat during chemical reactions

- exothermic reaction: Δ H < 0

- endothermic reaction: Δ H > 0

2.7 Standard Enthalpy Changes (Δ Hᶱ)

- standard state ( the state that pure compounds will be in at a 25 degrees Celsius and at 1

barr)

- phase state

a) Enthalpies of physical change

- Δvap Hᶱ or Δ Hvapᶱ

- Δfus Hᶱ

b) Enthalpies of Chemical Changes (chemical reactions)

- "thermochemical reaction"

A + B ---> C + D , Δ H = ...........

- all reactants and products should be in their standard states

- Hess's Law says that the reaction enthalpy, ΔrHᶱ, is the sum of the standard enthalpies of

a set of individual reaction equations.

September 21

2.8 Standard Enthalpy of Formation

Δf Hᶱ

8

- form compounds from their elements

- elements must be in their most stable state at pressure = 1 bar and Temperature = 25

degrees C

- Δf Hᶱ of elements (in their standard state) is 0

- Δr Hᶱ = Sum v*Δf Hᶱ - Sum v*Δf Hᶱ , where v is the coefficient

ex: 3H2O(g) + 3/2O2 (g) ---> 3H2O (l) Δr Hᶱ = ?

Δr Hᶱ = (3(-285.83kJ/mol)) - (3*0 + 3/2*0) = -857.49 kJ/mol

this works as long and its their standard state.

2.9 The Temperature Dependence of Δr Hᶱ

- we had: ΔH = Cp Δ T

dH = (Cp)dT

Δ H(Tf) - Δ H(Ti) = (the ʃ from Ti to Tf(Cp)dT

Δr Hᶱ (Tf) = Δr Hᶱ (Ti) + (The ʃ from Ti to Tf(Cp)dT , this is known as Kirchhoff's Law

Δ Cp = final - initial state

Δ Cp = Sum v*Cp,m - Sum v*Cp,m

2.11 Changes in Internal Energy, U ( more generally)

- Assume a closed system of constant composition (matter cannot be transferred in or out

of system)

- Internal energy, U, depends on p, V, T. ( really only need to know two to solve it)

a) General expression for dU, the change in internal energy

dU = ((δU/δV)T)dV + ((δU/δT) V)dT , where ((δU/δT) V) = CV and ((δU/δV) T) =

"internal pressure" = Pi T

therefore dU = (Pi T)dV + (CV)dT

c) Changes in U at constant pressure, p

we already have dU = Pi T dV + CVdT divide by dT throughout and require p to be

constant

(δU/δT)p = Pi T (δV/δT) p + CV , where (δV/δT) p = alpha V = expansion coefficient

alpha V = (dV/dT) p =====> alpha = (1/V)(dV/dT) p

2.12 The Joule-Thomson Effect

- general expression for dH

dH = ((δH/δp) T)dp + ((δH/δT) p)dT , where ((δH/δT) p) = C p

where ((δH/δp) T) = µ T = µ*C p

9

- µT is called the isothermal (constant T) Joule-Thomson coefficient (δH/δp) T

-µis the J-t coefficient (δT/δp) H

z = pVm/RT

sept 23

3. The Second Law of Thermodynamics

- thermodynamics can tell us in which direction a chemical reaction will happen (but not

the rate)

- the spontaneous direction of change is the one that doesn't require work

The 2nd law of thermodynamics (Kelvin)

"no process is possible in which the sole result is a complete conversion of heat in a

reservoir into work

The direction of spontaneous change

- what determines direction of spontaneous change?

- it is not the total energy of an isolated system (it is constant)

- not the energy decrease of a non-isolated system ( it has gone into the

surroundings)

** IT IS RELATED TO THE DISTRIBUTION OF ENERGY

3.1 The Dispersal of Energy

-dispersal of the total energy of an isolated system ("spreading out") gives us the

direction of spontaneous change.

3.2 Entropy

- is a measure of the energy dispersed in a process (physical or chemical)

the 2nd law of thermodynamics:

The entropy of an isolated system increases during a spontaneous change

====> Δ Stotal = Δ Ssystem + Δ Ssur > 0 , (Δ Ssystem is normally written as Δ S,

without the subscript).

a) the thermodynamic definition of entropy

- heat leads to random motion ( work does not)

- change in entropy ( of the system )

dS = dqreversible /T qrev = heat supplied reversibly

Δ S = ((ʃ from initial state to final)dq rev /T)

dU = dq + dW = 0

ex. self test 3.1, p. 97

10

calculate Δ S when p of a fixed amount of perfect gas is changed isothermally from pi to

pf.

Isothermal: Δ U = 0 ==> qrev = -Wrev , T is constant

Δ S = ((ʃ from pi to pf)-Wrev/T) ==> 1/T((ʃ from pi to pf)pdV)

V=nRT/p

dV=nRT(-1/p2)dp

therefore 1/T((ʃ from pi to pf)-pnRT/p2)dp

==> ((ʃ from pi to pf)-nR/p)dp

==> -nR[lnp] evaluated from pf to pi = -nR(lnpf - lnpi) = -nRln(pf/pi)

= nRln(pi/pf)

September 26

Boltzmann distribution (schematic)

- the change in entropy of the surroundings

Δ S = change in entropy in the system

Δ Ssur = change in entropy of the surroundings

Δ Stotal = Δ S + Δ Ssur

- Assume that the surroundings are at constant volume, V.

dUsur = dqsur

dSsur = dqsur, rev/Tsur = dqsur/T , we can take out the fact that it can be reversible.

the change in heat does not need to be reversible because we based it on dUsur = dqsur @

constant V.

measurable changes

Δ Ssur = qsur/Tsur

- Assume that the surroundings are at constant pressure, p.

Δ Ssur = Δ Hsur/Tsur

----------------------------------------------------------------------------------------------------------

-

ex. Self-Test 3.2, p. 98

Calculate the change in entropy in the surroundings, Δ Ssur, when 1 mol of N2O4 is

formed from 2.00 mol NO2 under standard conditions at 298K.

NO2 is reactant

N2O4 is product

we need to figure out the reaction enthalpy, Δr Hᶱ, at standard conditions for

2 NO2----->N2O4

Find values for Δf Hᶱ from Table 2.6

need to form NO2 Δf Hᶱ/(kJ/mol)

11

1) 1/2N2 + O2 -----> NO2+33.18

2) N2 + 2O2 -------> N2O4 +9.16

reverse (1) and x 2, keep (2) as is

2 NO2------> N2 + 2O2 -33.18 x 2

N2 + 2O2 -----> N2O4 +9.16

------------------------------------- ---------------

2 NO2-----> N2O4 -57.20 kJ

this is Δ H of the system, but were asked of the surroundings.

Δ Ssur = qsur/Tsur = Δ Hsur/Tsur = -Δr H/Tsur = +57.20kJ/298K = =192J/K , the

entropy of the surroundings increased

-----------------------------------------------------------------------------------------------------------

b) The Statistic View of Entropy

- Energy is quantized, atoms and molecules can have only specific values of energy

(those values are different energy levels)

- Boltzmann distribution at a given Temperature, T

- Boltzmann formula for entropy:

S = klnW , where W is the # of microstates in a system, where S is the Statistical

Entropy. The more microstate mean that the entropy is larger, more disorder.

c) Entropy as a state function

- should be independent of path ( we don't need to know how we got there)

(pressure and volume diagram: not available)

(description of pressure and volume diagram: basically it goes in a circle to same

point. at same point there is no change in entropy)

(isotherm graph: not available)

(description of isotherm graph: it is pressure vs. volume with two curves, both

going from -infinity to zero)

(adiabatic graph: not available)

(descriptiong of adiabatic graph: p vs. V with two curves where the start off at one

point and curve downward away from it. isotherm curve is above adiabat curve)

(carnot cycle graph: not available)

(description: similar to adiabatic, but a point comes off the end of a one curve

graph instead of beginning and with two curves)

September 28

12

c) Entropy As A State Function

- entropy should be independent of path.

(graph drawn, NA.)

- (path ʃ of dqrev/T) = 0

The Carnot Cycle. (figure 3.6 in the book, study it )

isotherm = constant T

adiabat = no heat exchange

path AB is isothermal reversible expansion at Th

Δ S = qh/Th , qh > 0

path BC is adiabatic , q = 0

Δ = 0

Temp. falls from Th to Tc

path CD is isothermal reversible compression at Tc

Δ S = qc/Tc , qc < 0

qc is heat flowing out of the system

Path DA: adiabatic q = 0

T increase from Tc to Th

Sum up the four terms:

((the path ʃ)dS) = qh/Th + qc/Tc

calculate qh and qc for a perfect gas

AB: isothermal reversible expansion

Δ U = 0 ==> q = - Wh

qh = -Wh = -(-(ʃ from VA to VB of)pgas*dV)

= nRTh*ln(VB/VA) > 0

similarly, for path CD:

qc = nRTc*ln(VD/Vc) < 0

Use the adiabatic expansion to find the relationship between VA, VB, Vc, VD and Th and

Tc

VA*V T c +VD*TD

c

c = Cv,m/R

VA*Thc = VD*Tc

c

and VB*Thc = Vc*Tc

c

13

=======>>>> VA/VD = VB/Vc ==> qh/qc =

= [nRTh*ln(VB/VA)]/[nRTc*ln(VD/Vc)]

= -Th / Tc

(the path ʃ of dS) = qh / Th + qc/Tc = 0

qh/Th = -qc/Tc

Ta Da.

- we need to show that the path ʃ = 0 for any material, not just for a perfect gas.

Efficiency of a heat engine

η = ada? = greek letter = looks like n = the work performed/heat absorbed =

abs(W)/abs(qh)

-the efficiency is always between 0 < η < 1

- all reversible engines have the same efficiency, ada?, regardless of their construction.

September 30

Matt is a dashing ginger.

efficiency = η = abs(W)/abs(qh)

(read up on c, d in book)

e) The Clausius Inequality

- more work is done in a reversible process than in an irreversible one

abs[dWrev] >= abs[dW]

dqrev > or = dq , comes from dU = dq + dW = dqrev d dWrev

==> entropy:

dS = dqrev/T (no subscript on dS means it is the system)

dS > or = dq/T ==> Clasusius inequality

3.3 Entropy Changes Accompanying Specific Processes

a) Expansion

- for a reversible process

Δ S = qrev/T = nRln(Vf/Vi)

we also know that:

Δ Stotal = Δ S + Δ Ssur = 0 <=== reversible system (one positive, one negative)

- expansion against constant external pressure:

{Read this part in book}

- expansion against vacuum

14

W = 0 ===> q = 0 , when letting a gas expand in a vacuum

Δ Ssur = 0

Δ S = nRln(Vf/Vi)

= - Δ Stotal = nRln(Vf/Vi) > 0 , for expansion

b) Phase Transitions

- Changes in ordering

- Phase transitions are considered equilibrium situations, so they are reversible

Δ S = Δtransition H/ T transition Δ H < 0 ==> exothermic, then the system

becomes more ordered, the surroundings become more disordered.

- since it is reversible it should hold that Δ Stotal = 0

c) Heating

- can calculate Δ S from heat capacity:

Δ S = ((the ʃ from Ti to Tf)dqrev/T)dT = ((ʃ from Ti to Tf)C p dT/T Cp,m = C p/n

Sm (T) = Sm(0) + ʃ from 0 to Tf (Cp,m (S,T)dT/T + Δfus H/Tf + ʃ from Tf to TB (Cp,m

(l,T)dT/T + ΔVap H/TB + ʃ from TB to T(Cp,m (g,T)dT/T

3.4 The Third Law of Thermodynamics

- there is no thermal motion at T = 0 , So the entropy, S = klnW ,(W is not work, it is

the # of microstates)

a) The Nernst Heat Theorem

- for any physical or chemical transformation, the change in entropy, Δ S ---> 0 and T ---

-> 0, provided that all the substances are perfectly ordered.

The Third Law of Thermodynamics

- the entropy of all perfect crystalline substances is zero at T = 0.

oct. 2

The Third Law of Themodynamics

The entropy of all perfect crystalline substances is zero at T = 0K

- compounds can have residual entropy , which means they cannot obtain that perfect

crystalline state.

S > 0 and T > 0 , S is always > 0

b) Third-Law Entropies

- standard ( third-law ) entropy Sᶱ(T)

15

a substance in its standard state at Temperature, T, with respect to S(0) = 0

- standard reaction entropy

Δr Sᶱ = ∑products(v)Smᶱ - ∑reactants(v)Smᶱ

Smᶱ molar standard/entropy ====> none of these values are zero at T not = to

zero

CONCENTRATING ON THE SYSTEM

- it is sometimes easier to find the Δ Ssur , we will make use of this to construct equations

that directly give us Δ Stotal

3.5 The Helmholtz and Gibbs Energies

- dS = dqrev/T dS > or = dq/T

- use Causius inequality

dS > or = dq/T <----------- heat transferred irreversibly

a) Criteria for Spontaneity:

- at constant V

Δ U = qV

dqV = dU

==> dS > or = dU/T <====> TdS > or dU

dA = dU - TdS < or = 0 A = U - TS , where A is Helmholtz energy

, where dU = -TdSsur

dA = -TdSsur - TdS < or = 0

= -T(dSsur +dS) < or = 0 , whenever > or = to zero it is spontaneous

- at constant p

- assume only expansion work ( no "extra work")

dqp = dH dS > or = dqp/T

dG = dH - TdS < or = 0 (called Gibbs energy G = H - TS)

all parameters dG, dH, dS and T are for the system

c) Maxiµm work

dA = dU - TdS < or = for spont. process

dU = q + W ===> W > or = dU - q (max. value W can take is dA, the Helmholtz

energy) Wmax = dA

3.6 Standard molar Gibbs Energies

Δr Gᶱ = Δr Hᶱ - T(Δr)Sᶱ

standard Gibbs energies of formation

Δr Gᶱ = ∑products(v)*Δf*Gmᶱ - ∑react(v)Gmᶱ

oct. 4

16

COMBINING THE FIRST AND SECOND LAWS

3.7 The Fundamental Equation ( dU, the change in internal energy)

First law

dU = q + W = qrev + Wrev

(pressure & volume graph)

Second law

dS = (dqrev)/T

dU = TdS - pdV ---------> fundamental change in energy

3.8 Properties of the Internal Energy

dU = (δU/δS) VdS + (δU/ δV)SdV

, where (δU/ δV)SdV = -p

, and where (δU/δS) VdS = T

a) The Maxwell Relations

- assume a function of f(x,y) , the differential of f is:

df = (δf/δx)ydx + (δf/y)xdy

(δ/δy(δf/δy)y)x =

= (δ/δx(δf/δy)x)y

This should apply to the differential dU as well (δ/δV(T))S =

= (δ/δS(-p) V

3.9 Properties of the Gibbs Energy

- G = H - TS ( useful for discussion of chemical reactions and phase transtions)

a) General Considerations

dG = dH - d(TS) = dH - dTS - TdS

dH = dU + d(pV) = dU + Vdp + pdV

dU = d(TS) - pdV

===> dG = Vdp - SdT

b) The Variation of Gibbs Energy with Temperature

- use (δG/δT) p = -S = [G - H ]/T , (G = H - TS)

for practical purposes ( for calculating eq constants ch. 6) it is more useful to know how

G/T varies with T

(Graph of G/T vs T depicted.)

(δ(G/T)/δT ) p (use chain rule) = 1/T(δG/δT) p + G(-1/T2)

1/T[(G - H)/T] - G/T2= -H/T

2

[δ(G/T)/δT] p = -H/T2

Gibbs-Helmholtz equation

17

[δ(Δ G/T)δT] p = -Δ H/T2

( chapter 3 Road Map thingy)

oct. 7

4.1 Physical Transformations of Pure Substances

- phase changes occur without chemical change in chemical compositioin.

- the system minimizes it's Gibbs energy

dG = ......surroundings + ......system

- phase transitions can be classified based on how thermodynamic functions change at the

transition

PHASE DIAGRAMS

for examples (Graph of p vs. T)

- they are maps of the physical states of a substance

- helpful when discussing mixtures

4.1 The Stability of Phases

- it is related to Gibbs energy (another way of saying it is related to entropy)

a) The number of phases

- a phase is uniform in chemical composition and physical state

- # of pahses in a system, P (capital P, not to be confused with p for pressure)

ex.

a gas or a gas mix P = 1

NaCl dissolved in H2O P = 1

Ice + water P = 2

dispersion P > 2

(Phase transitions diagram in book)

b) Phase Transitions

- spontaneous conversion, occurs at given p and T

- at Ttrs the Gibbs energy is minimized at a given p. (trs = transition)

c) Thermodynamic Criteria of Phase Stability

- use the molar Gibbs energy Gm = G/n and this is called CHEMICAL POTENTIAL

(symbol used is µ)

- basically take the Gibbs energy and divide by moles and call it µ

-The Second Law of Thermodynamics:

At equilibrium µ is the same throughout the sample regardless of how many

phases are present

18

Previously we had:

dG = Vdp - SdT - (µ)dn

ex.

-------------------

- -

- - µ 1

- -

-------------------

- -

- - µ 2

- -

-------------------

Transfer of amount of substnace (dn) from one location to the other

dG = -µ1dn + µ2dn = (µ2 - µ1)dn

b) The Phase Rule (gibbs Phase Rule)

- F = C - P + 2

(F = degrees of freedom, C = # of components, P = # of phases, 2 is the amount of

variables, which are p and T)

oct. 12

Homework 3.7b and 3.22 was graded.

THERMODYNAMIC ASPECTS OF PHASE TRANSITIONS

- criterion for phase equilibrium:

both phases have the same µ.

4.4 The Dependence of (Phase) Stability on the Condition

- µ changes with p and T dµ = Vmdp - SmdT

(δµ/δT) p = - Sm

(δµ/δp)T = Vm

vapor pressure of a liquid subjected to an external pressure

dµ(l) = dµ(g)

µ(l) = µ(g)

Vm,ldpext = Vm,gdp <---- vapor pressure

Vm,ldpext = (RT/p)dp

dµ = Vmdp - SmdT

19

4.5 The Location of Phase Boundaries

- we can find phase boundaries by using

µ(α : p,T) = µ(β : p,T)

a) The Slope of Phase Boundaries

Vm,αdp - Sm,αdT = Vm,βdp - Sm,βdT

rewrite "final - initial"

(Sm,β Sm,αdT = (Vm,β - Vm, αdp

-----> ΔtrsSm = ΔtrsVm

dp/dT = [ΔtrsSm]/[ΔtrsVm] = [ΔtrsH]/[T*ΔtrsV] (Known as Clapeyron equation)

melting Δ H > 0

Δ V > 0 but small

dp/dT >> 0

boiling (vaporization)

Δ H > 0 and similar to Δ H for melting

Δ V >> 0

dp/dT >0 , not as large as the other

- furthermore, for the liquid to gase phase transition:

Δvap V is approximately V(m,gas) = RT/P

dp/dT = [ΔvapH]/[T(RT/P)]

=> dlnp = dp/P = [ΔvapH/R] x dT/T2 = d (Clausius-Clapeyron Equation)

20

5. Simple Mixtures

- relationship between chemical potential and composition of the mixture (composition in

mole fraction, goes from 0 to 1)

- the chemical potential tells us how other thermodynamic quantities change with

composition

- in this chapter: mostly binary mixtures

XA + XB = 1

THE THERMODYNAMIC DESCRIPTION OF MIXTURES

5.1 Partial Molar Quantities

a) Partial molar volume

[side note: partial molar volume graph was on last year's exam. Figure 5.1 pg. 157, saved

in photos.]

[Figure 5.2 pg. 158, saved in photos]

- if we add a volume of compound A to a mixture of A and B, the total volume does not

always increase by exactly the added volume A.

- the partial molar volume VA is the change in total volume per mole of A added

VA = (δV/δnA)p,T,n'

book uses

VJ = (δV/δnJ)p,T,n'

- the total colume changes as:

dV = (δV/δnA)p,T,n'dnA + (δV/ δnB)p,T,n'dnB

= VAdnA + VBdnB

Matt = cumslutA, Jason = cumslutB

cumslutA + cumslutB = Partial clothes coming off

- molar volumes are always > 0

- partial molar volumes can be > 0, = 0, < 0

b) Parital Molar Gibbs Energy

- for each compound in a mixture

= chemical potential, µ, of a compound in a mixture

µJ = (δG/δnJ)p,T,n'

[Figure 5.4 pg. 159, saved in photos]

- previously we had dG = Vdp - SdT

- now we have::

[dG = Vdp - SdT + µAdnA + µBdnB +....]

- at constant p and T

dG = myAdnA+ µBdnB

- wider significance of µ

21

- we had dU = TdS - pdV

now dU = TdS - pdV + µAdnA + µBdnB + .....

µJ = (δU/δnJ)S,V,n' also for dH and dA

d) The Gibbs-Duhen Equation:

- general starting point differential of G

dG = µAdnA + µBdnB + nAdµA + nBdµB

{nAdµA + nBdµB = 0}

==> ∑nJdµJ = 0

for example: dµA = - (nB/nA)dµB

ex: we can do the same thing for partial molar volumes:

dVA = -(nB/nA)dVB integrate, start from having no amount of B

∫ from V*A to VA dVA = - ∫ from 0 to VB (nB/nA)dVB

oct. 17

5.2 The Thermodynamics of Mixing

- Gibbs energy <--- state function ==> we can use (final - initial) states to find out about

the spontaneity of a process.

- first example: mixing of perfect gases (refer to book for depiction)

(we know that this is a spontaneous process)

a) Δmix G of (two) perfect gases

- assume constant T

dGm = Vmdp -SmdT

perfect gas Vm =RT/p

∫ from Gmᶱ to Gm of dGm = ∫ from pᶱ to p RTdp/p =

= RTln(p/pᶱ)

- each of the pure gases has chemical potential, µ.

µ = Gm = Gmᶱ + RTln(p/pᶱ) , pᶱ <--1 bar

write as:

µ = Gmᶱ + RTlnp (short hand method since divide by 1 bar anyways

(standard pressure)) (Gmᶱ for pure gas at 1 bar) (RTlnp for pure gas at pressure p)

- compare final and initial state:

Before mixing:

Gin. = nA*µA + NB*µB

= nA (µAᶱ + RTlnp) + nB (µBᶱ +RTlnp)

After mixing:

Gfin = nA (µAᶱ + RTlnpA) + nB (µBᶱ + RTlnpB)

Difference:

Δmix G = nARTln(pA/p) + nBRTln(pB/p)

22

Both terms are negative, since pA + pB = p

- rewrite using nJ = XJ*n , where n = total # of moles, X = mole fraction = chi

and XJ = pJ/p , where p = total pressure

Δmix G = nRT(XAlnXA + XBlnXB) < 0 , spontaneous process.

b) Other Thermodynamic Mixing Function

- we know that (δG/δT) p = -S [From dG = Vdp - SdT, Vdp is crossed out because it is

held constant]

S = -nR(XAlnXA + XBlnXB) > 0 ===> Δmix H = 0 , for a perfect gas, not for a

system in general.

5.3 Chemical Potential of Liquids

- chemical potential of a liquid varies with composition.

- a liquid and a vapor in equilibrium µst have the same chemical potential.

- µ (chemical potential) of a liquid can be related to its vapor pressure.

a) ideal solutions

- similar molecules

- use an asterisk (*) for pure substances

Liquid - vapor equilibrium:

for pure substance:

µA*(l) for the liquid

µA*(g) = µAᶱ(g) + RTpA* for gas

pA* = pA/pᶱ

==> µA*(l) = µAᶱ(g) + RTlnpA* (1)

If another substance B is present in the liquid, what happens to the chemical potential?

µA (l) = µAᶱ(g) + RTlnpA (2)

pA < pA*

combine equations 1 and 2

==> µA (l) - µA*(l) = RTln(pA/pA*)

µA (l) - µA*(l) is for the liquid, ln(pA/pA*) is for the gas, vapor

oct. 21

µA (l) (liquid A in a mixture) - µA *(l) (pure liquid A)= RTln(PA/PA*)

[µ= chemical potential]

-approximation

pA = XA (pA*)....................Rault's Law

pA = partial pressure of A in a gas mixture

23

pA* = Vapor pressure of pure A

XA = mole fraction of A in the liquid

[Refer to Textbook for Mole Fraction Diagram]

- for an "ideal solution" this law holds at all concentrations

=> µA (l) - µA *(l) = RTlnXA

µA (l) = XA*(l) + RTlnXA

b) Ideal-Dilute Solutions

- Henry's Law

pB = XB (KB), where B is the solute of the component we do not have a lot of

(must be at a low concentration)

KB is an experimental constant with units of pressure, but may have other units as

well, pressure/concentration.

- the Ideal-dilute solutions:

Obey Raoult's Law for the solvent and Hentry's Law for the solute

THE PROPERTIES OF SOLUTION

-The mixing of liquids:

similar to the mixing of gases (?)

a) Ideal Solutions

-as for gases

Gi = nA (µA *) + nB (µB*)

Gf = nA (µA) + nB (µB)

= nA (µA * +RTlnXA) + nB (µB* + RTlnXB)

final - initial:

=> ∆mix G = nARTlnXA + nBRTlnXB

[XA = nA/(nA + nB)]

= ntotalRT(XAlnXA + XBlnXB)

5.5 Colligative Properties

- depend on number of molecules (concentration), not on their identity.

- lowering of vapor pressure leads to:

=> elevation of boiling point, TB

=> depression of freezing point Tfus, Tm

=> osmotic pressure of solvent in the presene of a solute

-assume dilute solution and non-volatile solute (no solute in the vapor)

[Refer to graph of mole fraction A, XA]

a)

24

oct. 26

colligative properties:

µA (l) = µA *(l) + RTlnXA

b) The Elevation of Boiling Point, Tb (lowercase b for boiling point)

solvent A, Solute B

Find T where:

µA,l(T) = µA,g*(T)

=> µA,g*(T) = µA,l*(T) + RTlnXA

=> lnXA = [µA,g*(T) - µA,l*(T)]/RT = ∆vapGm/RT

dlnXA/dT = d(∆mixGm/T)/dT x 1/R = 1/R(-∆mix Hm/T2)

ʃ from lnXA, where XA = 1 to lnXA of dlnXA = - ∆mix Hm/b ʃ from T* to T of

1/T2dT

lnXA = - ∆mix Hm/R(1/T* - 1/T)

use sum of mole fraction must = 1

XA + XB = 1 => XA = 1 - XB, where XB is small

T is approximately T*

ln(1 - XB) = - ∆mix H/R[(T-T*)/(T* x T)]

=> +XB is approximately + ∆mix H/R(∆T/T*2)

=> ∆ T = XB x K (sometimes known as Kb)

c) Depression of Freezing Point, Tfusion (Tf), Tmelting (Tm)

d) Change in Solubility with Change in T.

[Refer to figure of Colligative properties: osmotic pressure]

e) Osmotic Pressure

-osmosis: transfer of solvent from a dilute solution to a concentrated one

At Equilibrium:

pure solvent Solution applied with π (greek Pi)

µA*(p) = µA (p + π)

= µA *(p) + ʃ from p to p+π of Vmdp + RTlnXA

0 = ʃ from p to p+π of Vmdp + RTlnXA =

= -RTlnXA = ʃ from p to p+π of Vm,Adp = πVm,A

-RT(-XB) = πVm,A

RTXB = πVm,A [ Van't Hoff's Equation ]

25

PHASE DIAGRAMS OF BINARY SYSTEMS

5.6 Vapor Pressure Diagrams

[p vs. T diagram showing solid, liquid, gas Refer to Photos 10/26]

oct. 28

5.6 Vapor Pressure Diagrams ( at a given T)

- start from p-T diagrams

=> pressure-composition diagrams

-composition in the liquid

XA = pA/pA* XB = pB/pB*

XA + XB = 1

a) Composition of the Vapor

YA = pA/ptotal YB = pB/p (it is understood that p is ptotal)

pA+ pB = p

YA + YB = 1

-ideal mixture: (Raoult's Law holds)

YA = pA/p = XApA*/(XApA* + (1-XA)pB*) , where 1-XA = XB

= XApA*/(XA (pA* - pB*) + pB*), if > 0, then A is more volatile than B

[Refer to "mole fraction of A in the vapor" diagram] [figure 5.32 in the book]

Watching Matt watching me watching him, while watching me.

b) Interpretation of pressure-composition diagrams:

-construct figure 5.32 from 5.29 to 5.31

-in 5.32:

point a: total vapor pressure of liquid mixture

point b: composition of vapor in contact with the liquid mixture (at that same T)

horizontal axis: overall composition

XA , YA

- the line connecting points a and b is called a tie line, (or conode)

[Refer to diagram of "different pressure: different ratios of liquid and vapor"]

c) The Lever Rule

- can measure relative amount from the "lengths" of the parts of the tie line.

5.7 Temperature-composition diagram (at a given p)

26

-helpful when doing distillations

[Refer to "Temperature-composition phase diagram"]

a) distillation of mixtures

[Refer to azeotropes diagram]

b) azeotropes

[Refer to "liquid-liquid phase diagram" and "liquid-solid phase diagram"]

27

nov 2

[Refer to handouts or emailed notes on oct. 31]

Chapter 6 Chemical Equilibrium

- use chemical potential, µ, to find the equilibrium composition of chemical reactions

- relationship between equilibrium constant and change in Gibbs energy in reactions,

∆rGᶱ

ex. of equilibrium constant in acid base reaction:

HCl(aq) <====> H+(aq) + Cl

-(aq)

K = [H+][Cl

-]/[HCl]

- chemical reactions ----> dynamic equilibrium

reaction can be almost complete or still have a lot of reactants left

- many chemical reactions involve transfer of electrons, which leads to electrochemistry

SPONTANEOUS CHEMICAL REACTIONS

- the spontaneous direction of a chemical reaction is the one that lowers the Gibbs free

energy at constant p, T

6.1 Gibbs Energy Minimum (equilibrium point)

a) Reaction Gibbs energy, ∆rG

-assuming A <===> B (can be isomer)

dnA = -d(ξ) , dnB = d(ξ) , ξ is Greek letter, xi, for the extent of the reaction

- Then dG = µA dnA + µBdnB

= - µAdξ + µBdξ

= (µB - µA)dξ , (as time changes dξ and dG may also change).

∆rG = (δG/δξ)p,T = µB - µA

µB - µA .... product - reactant (not yet final - initial)

- µA and µB are at the composition of the reaction mixture. since they change with

composition (nA, nB), then the ∆rG will be changing as the reaction proceeds.

- equilibrium at ∆rG = 0, when µB = µA for a given composition nB, nA

b) Exergonic and endergonic reaction

∆rG < 0 => forward reaction A ----> B is the spontaneous one (exergonic)

∆rG > 0 => reverse reaction B ----> A is spontaneous (endergonic)

6.2 The Description of Chemical Equilibrium

a) Perfect gas equilibria

- assume A and B are perfect gases

28

µJ (J being either A or B) = µJᶱ + RTlnp , where p is p/pᶱ

∆rG = µB - µA = µBᶱ - µAᶱ + RTlnpB - RTlnpA ,

where µBᶱ - µAᶱ = ∆rGᶱ, then:

= ∆rGᶱ + RTln(pB/pA) ,

where ∆rGᶱ is given from a table and pB/pA is called the reaction quotient