Physical Chemistry Example Problems III

Hello and welcome back to www.educator.com, welcome back to Physical Chemistry.0000

Today, we are going to continue with our example problems for molecular spectroscopy.0005

Let us jump right on in.0011

Under a correction for anharmonicity, the vibrational term is this thing right here.0014

G of R equals ν sub E × R + ½ - X sub E ν sub E × R + ½².0022

This is the correction for anharmonicity for just the vibrational term.0028

Deriving expression for δ G of R and recall from the image of part B,0033

recall from the image of the true potential energy curve, this thing down here, that δ G goes to 0 as R increases.0040

In other words, δ G is the difference between the energy levels.0047

As you go up, the energy levels, the difference between them decreases until you are at 0.0050

Show that the maximum vibrational quantum number is given by this expression.0058

In other words, our max is the number of vibration states before the molecule actually flies apart.0063

Let us get started.0072

Part A, this is going to be just an algebra problem.0076

G of R, we have G of R is equal to ν sub E × R + ½ - X sub E ν sub E × R + ½².0080

Δ G of R is equal to G of R + 1 – G of R, the upper state - the lower state.0100

That is going to equal ν sub E × R + 1 + ½.0117

There is a reason I cannot do a ½ symbol, that saved my life.0128

- X sub E ν sub E × R + 1 - ½ - ν sub E, + ½ not – ½.0133

Crazy, + and -.0162

This one is – X sub E ν sub E × R + ½².0165

This is going to equal ν sub E.0178

Actually you know what, I wonder if I should do it that way?0183

Ν sub E × R + 3/2.0198

This is just a big long algebra problem.0201

We are looking for δ G, we are looking for the frequency of absorption or emission0205

when you go from 1 vibrational level to another vibration level.0210

- X sub E ν sub E R + 3/2² – ν sub E × R + ½.0215

- and -, this is going to be + X sub E ν sub E × R + ½².0227

Let me go ahead and expand that.0237

We get ν sub E × R + 3/2 ν sub E – X sub E ν sub ER² - 3 X sub E ν sub E.0239

I hope you got my algebra correctly, please confirm it for me.0260

-3X sub E ν sub E × R -9/4 X sub E ν sub E - ν sub E × R - ½ ν sub E + X sub E ν sub ER²0264

+ X sub E ν sub ER + ¼ X sub E ν sub E.0289

This goes with that, this goes with that.0302

When I put all the terms together, we are going to end up with ν sub E - 2 X sub E ν sub E × R -2 X sub E ν sub E.0310

Δ G is equal to ν sub E - 2 X sub E ν sub E × R + 1.0325

Or I could take out a ν sub E and I can write it is 1 – 2X × R + 1.0341

Either one of those is just fine.0354

That is the expression that we are looking for.0359

This gives me the frequency of absorption or emission for that transition.0362

Part B, as R increases δ G goes to 0.0371

Let us just set δ G in this equation equal to 0 and solve the equation.0388

Let us set δ G equals 0 and solve for R max.0400

We have ν sub E - 2 × X sub E ν sub E × R max + 1.0415

Use this equation first one, this version of the equation, either one is fine, equal to δ G which is 0.0435

We get 2X sub E ν sub E × R max + 1 is equal to ν sub E, what we solve the ν sub E cancels.0446

We are going to end up with R sub max equals ν sub E / 2X sub E ν sub E – 1.0464

Or if you want to cancel the ν sub E, you may or may not actually prefer not to cancel because ν sub E0479

and X ν sub E are actually individual spectroscopic parameters and I like to see them.0485

Or you can write it as R sub max equals 1/2 X sub E – 1.0490

Again, either one is fine.0500

This one happens to be my preference, not a big deal.0502

Example 2, using the equations we derived in example 1, derive an expression for D sub E,0510

the dissociation energy of a molecule.0516

Recall that the dissociation energy of the molecules taken from a minimum0519

of the potential energy curve not a 0 point energy of the oscillator.0522

D sub E is from the minimum to the energy of dissociation.0527

The D sub 0 that is from the 0 point to the dissociation.0531

It is that little difference.0535

The difference is the 0 point energy of the harmonica oscillator.0537

Part B, looking at the equation for δ G very carefully, is there a way to find this ν sub E and X sub E ν sub E from a graph of some sort.0542

Let us go ahead and do part A.0556

Let u see, the dissociation energy is equal to the energy of R max.0562

It is just equal to ν sub E × 1 / 2X sub E - 1 + ½ - X sub E ν sub E × 1/ 2 X sub E - 1 + ½².0578

This is equal to ν sub E × 1/ 2 X sub E – ½ - X sub E ν sub E × 1/ 2 X sub E – ½².0604

When I expand this I get, and multiply it out, I get ν sub E / 2X sub E - ν sub E / 20626

– X sub E ν sub E / 4X sub E + X sub E ν sub E or 2 X sub E² 2X sub E – X sub E ν sub E all / 4.0645

That gives me ν sub E / 2X sub E - ν sub E / 2 - ν sub E / 4 X sub E + ν sub E / 2 - X sub E ν sub E/ 4.0671

These go away and I'm left with, this is dissociation energy, my final answer,0702

my expression for dissociation energy equals ν sub E / 4 X sub E – X sub E ν sub E/ 4 or ν sub E / 4 × X sub E × 1 - X sub E².0715

This is the expression that I was looking for.0750

Notice if you make dissociation energy in terms of ν sub E and X sub E, this is extraordinary.0754

Now, this X sub E spectroscopic parameter.0760

This X sub E is a lot less than 1, it is very small.0765

If you look and if you remember some of the previous example problems that we did, it is very tiny.0770

It is much less than 1.0776

The second term here or this one right here is 0.0782

The second term is close to 0.0791

We usually just take the first term and we rewrite the dissociation energy as approximately equal to ν sub E / 4 X sub E.0799

Or if you want because X sub E ν sub E tends to be a single spectroscopic parameter.0824

At least that has listed a lot of data, you can just do ν sub E² / 4X ν sub E X sub E, that is fine too.0828

There we go, that takes care of that.0836

Part B, we had δ G is equal to ν sub E - 2X sub E ν sub E × R + 1.0841

Let us rearrange this.0856

Let us write δ G as a function of R is equal to - 2X sub E ν sub E × R + 1 + ν sub E.0866

Y equals MX + B, a linear equation.0883

We can plot δ G.0900

And again, I have values for δ G.0904

I can pick out a δ G values.0906

Δ G vs. R + 1.0908

0 + 1, 1 + 1, 2 + 1, 3 + 1, I have all these values, I can tabulate them.0914

In this particular case, the ν sub E is the one you intercept when I get it, when I plot it.0920

And -2 X sub E ν sub E, left of the slope.0928

I can go ahead and divide by -2 to make it the X sub E ν sub E.0935

This type of graph is called a rich mono plot, you should look it up on www.google.com.0940

Now as R increases the plot starts to deviate from linear behavior, deviate from linearity.0964

Interesting is when you look at a plot, you will see the plot.0984

From your perspective looking at a plot, it is going to be like this and it is going to start curving down.0987

The plot deviates from linearity as R increases.0992

If you get values of R, I will start to get 8, 9, 10, 11, and so on.0995

It is going to start to deviate from linearity because this energy for the vibrational term,1000

the vibrational energy which we have which is ν × R + ½ - X sub E ν sub E × R + ½².1010

It is actually part of an infinite series.1023

When we solve the Schrödinger equation for the anharmonic oscillator,1026

what we get is we get an infinite series for our energy.1029

We only took the first two terms because really all that we need.1032

It is really all that we need because we are not often going to find ourselves in the 15 or 20 vibrational state.1036

It is extremely high temperatures for something like that.1043

We are just going to take that.1048

As R increases, these other terms of the series start to become important.1050

It starts to deviate from linearity.1054

What you end up actually getting is because R increases the plot deviates from linearity because of this.1057

Yes, we only take the first two terms of the series solution to the energy of the anharmonic oscillator.1065

The graph itself will look something like this.1081

It is going to start pretty linear.1084

It looks nice but then you know it starts to deviate.1086

This is δ G, this is the R + 1, this axis.1092

How many vibrational states are there for BR2 before the molecule dissociates.1100

In other words, it breaks apart.1104

For BR2, the ν sub E is that and X sub E ν sub E equals that.1106

We have an expression.1111

We have that R max is equal to ν sub E / 2X sub E ν sub E – 1.1113

We are just going to put it in and find R max.1124

The number of vibrational states before the molecule falls apart.1127

R max, I’m just going to put the values in.1133

The ν sub E 325.321 divided by 2 × X sub E ν sub E is 1.0774 inverse cm -1.1140

When I do that, I get R sub max is equal to 149 vibrational states.1161

In other words, I can excite this molecule BR2 through the 149 vibrational states.1171

It is not going to hit the 150th.1178

At that point, it is just going to fly apart.1180

Let us calculate the dissociation energy for this too, we have an expression for that.1189

Approximately equal to ν sub E / 4 X sub E ν sub E.1196

We have ν sub E / 4X sub E which we said was equal to ν sub E²/ 4 X sub E ν sub E.1209

We get 325.321 inverse cm² divided by 4 × 1.0774 inverse cm.1220

We are left with the dissociation energy equal to 24,557.067 inverse cm.1238

Of course, if you want to convert that to Joules, you can go ahead and do that, not a problem.1251

If the R equals 0 to R equals 0 vibronic transition.1259

This is not vibrational, this is vibronic.1264

This is the transition that takes place between vibrational states from 1 electronic state to another electronic state.1267

If the R = 0 to R equals 0 vibronic transition for carbon monoxide equals 64748.5 inverse cm, find the value of T sub E.1277

This is the difference between the two minima of the potential energy curves below.1288

T sub E is the difference between this level and this level.1293

That is P sub E, the difference between the potential energy minima.1302

The data below gives us, for the ground state our ν sub E is this and our X sub E ν sub E is this.1309

For excited electronic state, the excited state ground state 1518.2, 19.4.1316

Let us see what we have got here.1329

Let us do this one very carefully.1331

Let me go ahead and go back to black here.1334

For the 0 to the 0 vibronic transition, we have the electronic energy, the rotational energy,1338

and the electronic vibration rotational energy.1351

For the 0 to 0 vibronic transition that means R equal 0 in both the upper and lower states, lower electronic states.1354

The equation for the total energy of a molecule, we have the total energy1388

is equal to the electronic energy + the vibrational energy + the rotational energy which is F of J.1398

For electronic transitions, the rotational energies are small compared to the vibration and electronic but we are going to ignore it.1419

We are going to drop that.1426

Our expression for total energy is just going to be our electronic energy of a molecule in a given state + the vibrational energy.1428

We have an equation for the observed, the calculated frequency of absorption,1447

that is going to equal the energy of the total energy of the upper state – the,1455

we will use a single prime for the total energy of the lower state I will use a double prime for that.1463

And we have expression for that, it is going to be the electronic energy prime + ν sub E prime × R prime + ½ - X sub E prime ν sub E prime × R + ½² - the total energy for the lower which consists of the electronic energy of lower state + ν sub E double prime × R double prime + ½1475

- X sub E double prime ν sub E double prime R double prime + ½².1493

When we expanded and solve this, we get it in a previous lesson we got ν equal to the upper electronic energy - the lower electronic energy + ½ ν sub E upper -1/4 X sub E ν sub E upper - ½ ν sub E lower -1/4 X sub E ν sub E lower1538

+ ν sub E upper R upper – X sub E upper ν sub E upper R upper × R upper + 1.1604

We end up with this equation right here.1618

This gives us the frequency of the observed spectral line for the transition between vibrational states when lower to upper state.1622

Now, because R single prime equals 0, in other words the lower and the upper are both 0,1635

these terms right here, let me go to blue now.1645

These terms, they dropped out.1648

We get just this term, this term, this term, and this term.1653

I’m going to call this term 1, I’m going to call this term 2.1663

I’m going to call this term 3, I’m going to call this term 4.1668

We get that the ν observed is equal to 1 - 2 + 3 – 4.1673

The difference in the electronic energies, the electronic energy of the upper –1689

the electronic energy of the lower that is our TE.1696

TE equals the upper electronic - the lower electronic.1701

It equals that.1711

What we end up with is and because it is a 0 to 0 transition vibronic, it is actually the ν 00 transition frequency1715

that equals to T sub E + ½ ν sub E upper - 1/4 X sub E upper ν sub E upper - ½ ν sub E lower -1/4 X sub E lower ν sub E lower.1730

We have all of these numbers already.1759

We know what ν 00 is.1762

It actually gave it to us, the problem gave it to us.1765

64,748.5 is equal to T sub E + 1/2 1518.2.1769

1518.2 – ¼ × 19.4, this is the data that was in there.1785

- ½ 2169.81 – ¼ × 13.29, I get 64748.5 is equal to T sub E + 754.25 - 1081.58.1797

When I do the math, I get my T is equal to 65,075.33 inverse cm.1829

This is the difference in energy between the two electronic states.1841

Example problem 5, the following data were obtained from an analysis of the rotational spectrum of CLO.1852

Use this data to find B sub E and Α sub E.1859

We have R values 0, 1, 2, 3, and we have B sub R values.1864

You do not need all the data, we are just going to use R equals 0 and R equals 1.1870

I can choose any that I want.1874

I can do 0 2, 0 3, 1 2, 1 3, or 2 3, it does not really matter.1876

Using the data for R equals 0 and R equals 1, we will take those two.1880

Our equation is B sub R, this is the vibration rotation interaction correction.1894

It is equal to B sub E - Α sub E × R + ½.1900

We are looking for α sub E and B sub E.1907

B sub 0 that is 0.62054.1912

It is equal to B sub E - R is 0.1921

It is - ½ α sub E.1928

For B sub 1, I have 0.61474 that is equal to B sub E - 3/2 Α sub E 1.1933

When I put 1 into here, 1 + 1/2 is 3/2.1947

When I subtract one from the other, I end up with Α sub E is equal to 0.0058 inverse cm1953

and I go ahead and put this into equation 1.1967

And I end up with B sub E is equal to 0.62344 inverse cm.1972

There you go.1985

Now, if I want to I can take 0 1, 1 2, 2 3, 0 2, I can take any combination.1986

I can take 0 1, 1 2, 2 3, I can get three different values and I can take the average.1995

It is not really necessary, you can just take two data points and go ahead and find that value.1999

Thank you so much for joining us at www.educator.com.2006

We will see you next time, take care, bye.2009