Basic Physical Chem Notes Full

7/29/2019 Basic Physical Chem Notes Full

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1. Fundamentals of Chemical Kin

1 Fundamentals of Ch

1.1 Concentrations

Chemical kinetics is the quantitchanging with time (Thermody




where γ is the stochiometric coetion. The + sign is used if Γ is a p

rA

Th l h i f




It has the rate law1

2

d[HBr

dt

where k 1 and k 2 are numbers (ca



2. Integration of simple rate laws

so that k 2[Br2] [HBr], the seconwill be negligible compared to 1 all order 3/2 (first order with resrespect to bromine). In the third second term in the denominator




both sides:

1

a

d[A

[A]

1

a

1

[A]d[A







Notice that the half life for a first concentration of A; this is a veryother reaction orders.

The time constant τ of this reae-folding time, or the 1/e time)




and in either case the value of k in the expression Q = α[A] + β.

Many textbooks will suggest

− ln




This rate law is second order oves−1. But if we run the reaction withen [B] will change very little dto B0 the whole time— and we ca

1




Look at the half-life now. By we find

t

Now, the half-life depends on t



3. Determination of rate laws

So, a plot of ln([B]/[A]) againThe closer A0 and B0 are to

poorly ∆0 is known; this methodmetric initial concentrations. It’under conditions where one reac




There are a couple of problem

1. The rate law might not hav

2. You might get tired beforeone of the ones you have a




is not linear, then the reaction h(the rate law is not the simple prdifficult to discover with other m

Note that in initial rates studwith the other reactants; you do




3.3.2 The single-run differenti

In this method, rates are taken frwhere other concentrations than(typically by flooding). At eachmeasure the slope of the concent



4. Elementary Reactions and Me

4 Elementary Reaction

If a chemical equation describes a

O + CH




0.6

0.8

1.0

i o n / [ A 0

] [B]

τ 1/(k + k




d[A]

dt= −k 1[A

Eq. (59) is now a separable diwith [A](t = 0) = A0, is



5. Exact and approximate analyt

5 Exact and approxima

rate equations

Any kinetic system composed ensteps has an exact analytic soluti




0.6

0.8

1.0

n / [ A 0

]

[C][A]




0.6

0.8

1.0

o n / [ A 0

]




gives separable equations for bosolutions:

[B] =k 1

k




“kinetically favored” product. Tnot mutually exclusive, and the

k 1 = 100

k 2 = 10−




5.2.2 Relative Rate Experimen

Consider elementary reaction of products P1 and P2.

A



6. Approximations

so

d[A1]

d[A2]=

k 1k 2

[A

[A

kd[A1]

= kd[



6. Approximations

d[A]

dt= −

d[B]

dt= k



6. Approximations

6.1 Validity of SSA

It is sufficient that the sum of alltermediate be much greater thanexample, this means

(k



6. Approximations

6.2 Other Simplifying App

Once again, consider the mechan

A



6. Approximations

0.6

0.8

1.0

a t i o n / A 0

exact[C]SSA

[C]REA[A]



6. Approximations

example) with a simple exponenfor this appearance is given in Ta

6.3 Rate determining steps



6. Approximations

6.4 Examples

6.4.1 Ligand substitution

The nucleophilic substitution rea



6. Approximations



6. Approximations

an intermediate. A page from Join Figure 7.

First-order kinetics suggestsstep. Try this:

k



6. Approximations

so

d[NO2]

dt

SSA≈ k 1[N2O5]

d[NO2]

d

SSA≈ 4k 1k 2k k

[N



6. Approximations

The first reaction is “fast”, sadded bromine as converted cocan either be complexed by Br− i140. The equilibrium gives us



6. Approximations

In fact, the complex can add the Br2N−

3 can react with azide cient than k 1. This additional compathway to products do not chathe reaction, but they make the e



7. Experimental Techniques

7 Experimental Techni

7.1 Elementary consideratio

Several questions must be answebe selected







in the composition of the mixturthe flow is stopped suddenly, thchange is monitored in time. See

Stopped flow is popular amoning enzyme reactions in the initia







and all components relax towardorder kinetics, τ = 1

k 1( Ae+Be)+k −1.

the equilibrium concentrations, bThe most common ways to d

P rapidly. To change T , the solut




0.2

0.4

0.6

0.8

1.0

- 55 C

0.1

0.2

0.3

0.4

0.5




is measured. The individual ratstant k 1/k −1 is also known.

References on lineshape ana580 (1979), and H. Gunther, NMR1972.




7.5 Shock Tube Experiment

In a shock tube experiment, a loheated by the passage of a strondiaphragm that was holding bacperature can change by more tha



8. Construction of candidate mec

8 Construction of cand

8.1 Rate controlling steps in

The heuristic procedures I will dnism from a carefully determine




In Figure 13 we haveA −

The other steps will be fast compone with the largest free energy the section will be the RCS




rate

We “subtract out” the denoposition of TlHg, and a tran

1




so

[C]SSA≈ k 1[A][B

k −1 + k 3 +

Now




At low product concentrat

k [Fe2+][Tl3+]. At high prod

rate




the concentration of Tl2+ dproduct.

8.3 Application of “mechanample




Applying the SSA to Hg+, we fin

d[Hg+]

dt= k 1[V3+][Hg2+]−

[Hg+]SSA≈ k 1[V3+][Hg2



9. Kinetic-molecular theory of ga

9 Kinetic-molecular th

The “kinetic theory of gases” mak

1. Gases are composed of par




have

P

This gives the pressure of thethe molecules (their masses and




Notice in the last equation, I havethe molecular mass, and then in mass (which must be in kg, not often save annoying conversion

constant R = 8.314 J K−1 mol−1 r




probabilities:dN vxv yv z

N = f (v

This is a three-dimensional functequivalence of directions it can




Now we need to find b. We We can also calculate that averausing the usual formula for the a

v2x =

∞




that the Gaussian will be wider that the average speed along thfor the molecules to be moving e

evaluate b using v2x rather than

what value b has.)




expect to find a probability denspositive v. (Atkins (pp. 26–27) cmake a clearer distinction betweof velocity components, f (vx).)

Eq. (225) gave the probabilit



10. Testing the Maxwell distribut

What do these curves look l

pend on v; the v2 part is a parabtered at zero. So at low speeds th

as v2 increases the curve turns ovGaussian becomes small. Figure




In Doppler spectroscopy, themeasured with very high resoluways done with lasers since theyily. A molecule moving toward is higher than the frequency of th




10.1 Applications of the Ma

10.1.1 Average speed vWe use the usual approach to av




Lots of stuff divides out, leaving

0 =

2vm

0 2m




1.5

2.0x10-4

300K




spheres with diameter d. As our takes its center within a distancthe two will hit. (See Figure 17

“collision cylinder” of volume πare in that cylinder will collide w




for vrel is just the average speed, colliding pair:

vrel

h l /( +



11. Real gases

11 Real gases

11.1 p−V isotherms and th

If you put some gas into a pistonthe piston in holding the appara



11. Real gases

volume than it did before. The pfied the sample appears at slightuids expand with increasing temwhich you have liquid and gas to

As you raise the temperature



11. Real gases

have parameters that can be chan

11.2 The van der Waals equ

One of the first, and still the mosth d W l i d l



11. Real gases

low densities (that is, large molathat of an ideal gas. This is easyV m gets large, it dominates overmakes the second term become sgas law p = RT /V m.



11. Real gases

100

150



11. Real gases

rameters in the van der Waals eThe equations to estimate them f

a =



11. Real gases

been measured. At low to mediaccuracies better than one percen

At low temperature and low dthe molecules are more importanin a sample of gas is less than th



12. First Law of thermodynamics

12 First Law of thermo

We already calculated the averaggas: εtr = 3

2 kT . If the particles hrotate, for example), then we can




conservation of energy, says thasystem (ideal gas or otherwise) the heat flow into the system, q,

∆




We let the plunger rise slowly (side). We let it rise by 2 cm, thenwarms back up to room temperainto the gas?

Solving this problem require




12.2 General pV work

In the piston-and-cylinder apparsure pext is

dw = − p




is still true even if the external plong as we know what it is; that

w = −




occur nearly at equilibrium.) Sec ble to calculate changes in a systeasily if we assume the process not depend on paths, we can suband get the same answer with l




12.3 Enthalpy

For any process at constant voluwork, then w = 0, so that

∆U =




conditions. It is conventional to the heat capacity when the volucapacity at constant pressure. Th

CV =




For an ideal gas, U and H are bon V or p (since the product pV

CV =

∂U

∂T




12.4.2 Reversible adiabatic exp

We now have the tools to analyzWe did the reversible isothermalis a little harder because the temsion goes on




We know for an ideal gas thatute for nR in Eq. (308) to get

CV ln

T 2T 1

=




so that the final temperature is 2an apparatus that uses a mediuma vacuum chamber perhaps a ftant applications is in spectrosctemperature their spectra are hop




298.15 K is the heat absorbed by

3C(s) + 3H

is carried out at 1 bar pressure tit i ∆ H−◦ Th ∆ i di t




in a series of steps, calculate the up to get the overall enthalpy chthe enthalpy change for the isom350 K and 1 bar pressure, the foll

1 C l l f 35




Table 4 Heat capacity data for cChemistry WebBook (webbook.n

T/K C p (propene )/ J/mo

298.15 63.79




Writing Cc p = a + bT , Cp

p = cthat

∆ H = ∆r H −◦298.15 +

350

298.15∆C p 350




it is nearly impossible to carry Therefore, most enthalpies of fo been determined by measuringand products and using Hess’s L



13. Mathematical interlude

13 Mathematical interl

13.1 Properties of partial de

For many kinds of thermodynamanipulate partial derivatives e




The van der Waals equation is mapply the chain rule:

= −V




We can find a relation between th

µ =

∂

∂

= −



14. Prelude to the Second Law: th

then the differential dz is exact if∂ f (x, y)

∂ y

H ’ t i i l l C



14. Prelude to the Second Law: th

For this to be exact, we would ha∂ p

∂T

F id l

∂p

nR A



15. The Second Law

15 The Second Law

15.1 Statements of the Seco

We now come to one of the mostscientific “laws” the Second Law



15. The Second Law

It is easy to come up with prhotter body. For example, we canto thermal equilibrium with an opiston out, so that the gas expaextract heat from the object. Now



15. The Second Law

1, that ∆U = 0 for the whole cyclThe path 1 → 2 is a reversibl

path 2 → 3 is a reversible isothermq2→3 > 0. Finally, the path 3 →Therefore, overall, q > 0, and si



15. The Second Law

Eq. (363) we have

[Θ1(T , V 1)

Since this is true no matter whatth t



15. The Second Law

We know from the uniqueness ofunction, so dS must be an exactus

∂

∂V (Θ(T )CV (T ))



16. Examples of entropy calculat

16 Examples of entrop

16.1 Reversible isothermal

In an isothermal change, T is coconstant so dq = dw = pdV ;




We must calculate the entropthe cooling of sample B, and add

For the heating process, we h

dS =




Similarly, if the surroundings are

∆Ss

So under those conditions it surroundings This idea is extrem




Under what conditions will t

1

2+

1

4



17. Gibbs and Helmholtz energie

17 Gibbs and Helmhol

In any isolated system, for any sp0. It is useful to regard the entirean isolated system. This outlook



17. Gibbs and Helmholtz energie

These are called the Gibbs and With them the spontaneity criter



18. Alphabet soup: Maxwell relastate

Atkins uses Eq. (414) to sho114–117) that ∆ A for a process gcan be extracted from the procemum amount of non-expansioncess. Along the way, in those ar




Eq. (420) is one of four Maxwell reone is interested in (like, how thand things one can measure (likwith pressure, and so on.) The oway, by applying the Euler crite




In our table, we find simple expr∂U

∂V

S

∂U



19. The chemical potential

19 The chemical poten

Our development so far has beention. To make thermodynamics uaccount for variable composition




in a solution, for example) as

µi ≡

h h l l ll h




If we consider an isothermal cha

∆G =



20. Conditions for material equil

19.2 Multiple phases

In many problems more than opreparation of a Grignard reagecally present in solution, in conboth solid and liquid water are



20. Conditions for material equil

20.1 Phase equilibrium

Consider the simplest sort of phapure substance between two phain a closed flask). Eq. (443) says t



21. Reaction equilibrium in ideal

Since the chemical potential of sudrop the superscript on µi and m

0 =




The equilibrium condition, E

k

∑ i=1

νi

µ−◦i (T

Collecting the standard state




where all the species pressures anotation,

K −◦ p

O ilib i diti i




First, an Erratum: beginning minus sign that should be in frEq. (464) should read

∆rG−◦T




Substituting that result into Eq. (

d ln K −◦ p (T

dT




21.2 ExamplesI want to give several explicit extions. I will use the reaction

N2




Constant∆ H In the crudest appthalpy of reaction, ∆r H −◦ , is constThis is not a bad approximation iapproximation, the van’t Hoff eqchange in the equilibrium consta




Constant ∆C p If the temperatuto assume that ∆r H −◦ is constant approximation, we can assume itants and products are constant. Cp m = 7

2 R quite accurately from




where the “per mole” in the unhave

ln K −◦ p (375) =

=




That gives a quadratic equation me, than using the quadratic forpressure is 10 bar, I have x1 = 0.7

In the constant volume case, Let z be the extent of reaction at




and since this is a 2-component s

x1 =

(Of course we could also calculaThis ICE table approach wr




will come out either positive or function value of that sign with tat 4.5, so my new limits are 4.5halving the width of the interval

Bisection gets you successive




Figure 24 shows how this worklution set to this week’s homewtemperature, so you can see how



22. Equilibrium in real gases and

22 Equilibrium in real22.1 Definition of activity

For ideal gases we found that




Fugacity coefficients for pure

lnφ( p) =

f




1. In an ideal solution, the intesame as those between A a

2. In an ideally dilute solution, only A–A and A–B interac nsid d




the vapor pressures of the two psince that would mean that if yoa benzene sample its vapor prewould expect that the vapor preswould be nearly that of benzene




and for fewer substances in otherWebBook has Henry’s Law consFigure 7.15 in Atkins shows h

real solution; Raoult’s Law workLaw for the minority component




22.4 Activity conventions fo22.4.1 Solvent activities

Eq. (533) was

µ µ




The activity coefficient is lessthe solution and appear in the vaattractive interactions between adeviations” from Raoult’s Law.

Activities calculated in the w




the mole fraction scale. This chvalues of the tabulated numbersor ∆ H that you would calculate.

µB =



23. Electrochemistry

23 ElectrochemistryIn an electrochemical cell, it is pthat take place during a redox ridized species travel through w




23.2 Electrochemical cellsIn an electrochemical cell, we hadifferent metals) immersed in elin one solution, or they can be ibridge which allows ions (positiv




solution, connect them with a saa copper rod in the second, andthe two electrodes, you will readcopper rod being positive with how to predict those things from




state Gibbs energy change, we ha

∆rG =

where Q is the usual ratio of prostoichiometric powers. Dividing




The standard cell potential is therium constant is

ln K



24. Temperature Dependence of

24 Temperature Depen

Most reactions go faster with incused to describe the T dependen

k




elementary. A sharper interpretaTolman’s theorem: the activation eage energy of molecules in the prall the molecules. There is no reathe activation energy should be i




235002000

T/




24.1 Thermodynamic funct

There is a large and relatively sustate theory” or “activated compsolute prediction of rate coefficiecases but the theory has found w



25. Statistical mechanics

25 Statistical mechanic

Statistical mechanics, the last mathat connects the molecular propmodynamic properties of the ma




The actual calculation of parof interacting particles (for examnumber of possible quantum stamous. Practical calculations requproaches are known, but let’s mo




With B = 1.9225 cm−1, at 300 Kevaluate the partition function I plenty high to converge the sum

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