Chapter 12, Chemical KineticsThis chapter is about:

1. numerical descriptions of how fast rxns. occur2. the intermediates that form during a rxn (re. mechanism)3. applying thermodynamics & the kinetic molecular

theory to go from the descriptive learning tounderstanding.

Our focus will be on gas and liquid phase (soln.)reactions.

Why care? Your health!

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I. Reaction Rates

A. Defining rate: change per unit time

rate = concentration change time change B. Example:

2 N2O5 (g) ÷ 4 NO2 (g) + O2 (g)

How could you measure the rate?

1. Measure [N2O5], [NO2], or [O2] at various times.2. Method: Pressure change, color change, etc.

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C. Define rate as increase in product conc. per unittime or decrease in reactant conc./unit time.

D. Let’s try O2 using data in Table 12.1, p. 444.

rate O2 formation = Ä O2 = [O2] at t2 ! [O2] at t1 Ä t t2 ! t1

Your text uses the 300-400 s interval, rate = 9 x 10!6 M/s.You should try a different time interval. (Pick one!)

Re. Fig. 12.1, does your answer seem reasonable relative tothe value at t = 300-400s?

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E. Is rate of NO2 formation equal to the rate of O2

formation (or the rate of N2O5 decomposition)?

1. Look at Fig. 12.1. (& Instructor animation: N2O5)

2. The balanced equation gives us the mole ratios.a) Stoichiometry: 1 O2, : 4 NO2 : !2 N2O5

b) Therefore equivalent rates (on M/s basis) are:

Ä [O2] = ¼ Ä [NO2] = !½ Ä [N2O5] Ä t Ä t Ä t

As Ä t approaches zero, Ä[X]'Ä t becomes the slope.

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So, if the O2 formation rate was 0.80 M/s, the NO2

formation rate should be , and the N2O5

decomposition rate should be .

(Remember, these rates are always positive numbers.)

F. Does the rxn. rate change with time? Would you have predicted this? Logic?

Try problem 12.2, p. 447.

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II. Rate Laws & Rxn. Order (Demolition derby?)

A. We are going to relate our numerical analysis ofthe rxn. rate to a mental picture (model) of whatsort of collisions are occurring during the rxn.

B. Consider the following general rxn.:

a A + b B ÷ Products

1. I can write a rate law for this reaction in the form:

Rate = k [A]m [B]n k: rate constant specific for this rxn., T, etc.

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2. Why does this equation work? (Collision-theory)a) Why does a k value have to be included?b) Why [A] times [B]? (Boltzmann?)c) Why is [A] raised to the m power? (Aside: Is a sometimes

equal to m [and b to n]?)

C. Reaction order

1. The overall reaction order = the sum of all theexponents in the rate law.

2. The reaction order “with respect to a specificreactant” = the exponent for that reactant. Do prob. 12.3, p. 449 to illustrate.

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D. If you have a balanced chemical rxn., you donot necessarily know the rate law.

The rate law must be determined experimentally!!!

How can we do that?

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III. Experimental Determination of a Rate Law One method is to measure initial rate at different [reactant]. Thensee what exponent (remember m and n?) best fits the data.

Rxn.: 2 NO (g) + O2 (g) ÿ 2 NO2 (g)

General Rate law: rate = k [NO]m [O2]n

Use Table 12.3 data, p. 450, to get values for variables.

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A. To get m, compare exp. #1 & #2, where [O2] isconstant, but [NO] changes by a factor of 2 (0.015M vs. 0.030 M). (Do math on board!)

If m = 1, the rate should double, if m = 2, the rate should quadruple, if m = 3, the rate should go up by factor of 8, etc.

Logic 21 = 2, 22 = 4, 23 = 8, etc.

From the data, we see that the rate quadrupled: 0.192 / 0.048 = 4, therefore, m = 2.

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B. Following same logic, we can determine valueof n.

What exps. should we compare in Table 12.3 toaccomplish this? Exp. # and exp. # .

Note: In this case we want [NO] to be constant.(Do math on board!)

C. Now that we know m = 2 and n = , do wehave enough information to determine k?

Solve for k in the space below:(Do math on board!)

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Should we do Worked Conceptual Ex. 12.4 (p.453) in class? Comment on Key Concept problems!!!

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IV. Integrated Rate Law: 1st-order Reaction

A. For rxn, A ÿ products, that is 1st order overall:rate = !ÄA / Ät = k [A] (m = 1)

Recall from your calculus:!ÄA / Ät = k [A] is ~same as:!dA / dt = k [A]

which we can rearrange to get: dA / [A] = !k dt

which we can integrate to get: ln [A]t /[A]0 = !k t

This is the integrated 1st order rate equation.

If you prefer base 10: log [A]t /[A]0 = !k t /2.30314

B. Because [A]0 means [A] at time = 0, it is aconstant, we can simplify to:

log [A]t = !k t /2.303 + log [A]0

1. This should remind you of the equation for a straightline: y = m x + b (See Fig. 12.6, p. 455, next page.)

Not the same m as the exponent in the general rate law.

2. Can you do a linear regression on your calculator tosolve for m & b? (You do Probs. 12.7 & 12.8.)

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V. Half-life of a First-order Reaction

A. A number of interesting rxns follow 1st orderdecay kinetics. Example: decay of 3H (tritium, t½

= 12.3 yr), re. SRS (Savanna River Site).

B. Because of the nature of this type of rxn. thehalf-life is a constant. (What is half-life?) (1st

order CD animation)

C. Rxn. half-life: time required for amount ofreactant to decrease to ½ of its original value.(Fig. 12.7, p. 459)

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D. Mathematically: log [A]t /[A]0 = !k t /2.303 log (½) = !k t½ /2.303

!0.3010 = !k t½ /2.303 So, t½ = (!0.3010 × !2.303)/ k = 0.693/ k

E. This means that: t½

Surprising? (Do Prob.12.9, p. 460 and Conceptual

Prob 12.10)

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VI. Radioactive Decay Rates

A. Applications:1. Medicine, particularly internal exposure.2. Archeology, etc. 14C dating3. Geology: How old are rocks, age of the earth?

B. Try an application with 14C dating: prob. 12.13,p. 463.

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VII. Second-order ReactionsMany reactions do not fit 1st order kinetics. Of these, many do

fit 2nd order kinetics.

A. For rxn that is 2nd order in A: ÄA / Ät = k [A]2

Example: 2 NO2 (g) ÿ 2 NO (g) + O2 (g)

B. The integrated rate law is: 1/[A]t = k t + 1/ [A]0

C. Does this look like y = m x + b? Look at Figure12.8 & the figures on p. 466 in Example 12.12.

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Worked Example 12.12 (Graphical determination of rate law)

Left-hand graph should give a straight line if rxn is 1st order.

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Right-hand graph should give a straight line if rxn is 2nd order.

What is the order of this rxn with respect to NO2?

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A copy of Table 12.4 will be provided w/Exam I.

1. Fig. 12.8: plot of [A] vs. t for a 2nd order rxn.

2. Which of the figures on p. 466 is a better fit to astraight line? What does this tell you about thekinetics of NO2 decomposition?

D. Try prob. 12.15 outside of class.

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VIII. Zeroth- Order Reactions (See Fig. 12.9, p. 467)

Rare. Text example notes Pt catalyzed decompositionof NH3. Collisions appear to be unimportant. This iszeroth order (with respect to ?) under the conditions given:

1. Relatively large amount of NH3

2. Relatively small Pt surface

See CD animation for Surface Reaction-hydrogenation

Can you imagine conditions where [NH3] & collisionswould become important re. rate?

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3. Enzymes can sometimes be viewed as surfacecatalysts. Assays to determine the amount of agiven enzyme (ex.: the enzyme PAH for PKUdiagnosis) are often set up to be zeroeth order withrespect to the reactants.

Would the slope change if we changed [A]0?

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IX. (Now 12.11) Reaction Mechanisms (diff. approach)Time to see if there is more to the analysis that we have been doing than

meets the eye. A rxn mechanism describes the specific changes that occurduring a chemical rxn. This stuff is massively useful in drug development!

A. Terms

1. A single step in mechanism is called an elementaryreaction or an elementary step.

2. The molecularity of a rxn refers to the # of separatemolecules/atoms on reactant side of elementary rxn.a) unimolecular = oneb) bimolecular = twoc) termolecular = three (These are uncommon.)

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B. Let’s look at a mechanism for:

NO2 (g) + CO (g) ÿ NO (g) + CO2 (g)

1. Experiments have suggested a two step process:

1st step: NO2(g) + NO2(g) ÿ NO(g) + NO3(g)

2nd step: NO3(g) + CO(g) ÿ NO2(g) + CO2(g)

See Fig. 12.14, p. 475

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(Think about molecules colliding and/or breaking apart!)

The molecularity of the 1st step is .The molecularity of the 2nd step is .

2.While a number of the components in the balancedmolecular equation are unstable, one component ofelementary rxns. is very unstable. That is NO3. Acomponent like NO3 that exists transiently (made &then used up) in the mechanism is a reactionintermediate.

3. Notice that the sum of the individual elementary stepsmust add up to the overall, balanced molecular

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equation:

1st step: NO2(g) + NO2(g) ÿ NO(g) + NO3(g)

2nd step: NO3(g) + CO(g) ÿ NO2(g) + CO2(g)

NO2(g) +NO2(g) + NO3(g) + CO(g) ÿ NO(g) + NO3(g) + NO2(g) + CO2(g)

Net: NO2(g) + CO(g) ÿNO(g) + CO2(g)

The terms that occur on both reactant and productssides of this summed equation are shown in bold. When these terms are dropped, you must regeneratethe original, balanced overall rxn. for the mechanismto have a chance of being correct.

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Do Prob. 12.18, p. 477. a) Equation for overall rxn?b) Molecularity of each step?

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X. Rate Laws for Elementary Reactions (simple)

A. 1st order example, decay of O3: O3 ÿ O2 + O

If this is 1st order in O3, what is the rate law?

B. Second order example, rxn. of CH3Br w/ OH!:See SN2 Molecular Movies animation. Also on 4th floor computers.

CH3Br + OH! ÿ CH3OH + Br!

If this is 1st order in both CH3Br & OH! what isthe rate law?

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Important: When the overall rxn. occurs in asingle step, then the rate law is determined bythe molecular equation.

Because this rxn. has a single step mechanism:

CH3Br(aq) + OH!(aq) ÿ CH3OH(aq) + Br!(aq)

rate = !Ä [CH3Br] / Ät = k [CH3Br] [OH!]

Do prob. 12.19, p. 479.

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XI. Rate Laws for Overall ReactionsBack to the crossword puzzle analogy.The balanced molecular equation for a rxn. only gives

information about stoichiometry, not necessarily about themechanism. However, the rate law for the rate limiting step in amechanism does define the rate law for the overall rxn.

A. Introductory comments:

1. Find an every day analogy of a multistep process with a clearrate limiting step. The rate limiting step is the slowest step. (Shopping?) Ever worked on or seen an assembly line?

When a rxn. occurs with more than one elementary step, usuallyone of the elementary steps is much slower than the others. This isthe rate limiting step.

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B. Multi step Reactions with an Initial Slow Step

1. Let’s go back and look at the rxn. of NO2 with CO:

Balanced rxn.: NO2 (g) + CO (g) ÿ NO (g) + CO2 (g)

Proposed mechanism: k1

1st step: NO2(g) + NO2(g) ÿ NO(g) + NO3(g)

k22nd step: NO3(g) + CO(g) ÿ NO2(g) + CO2(g)

The 1st step is the “slow” step in this sequence. By slow,I mean slow relative to the 2nd step. That is, k1 << k2.

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Rate of the overall process can’t be faster than rate ofstep 1. Can we write a rate law for the 1st elementary rxn.?

Rate = !Ä [NO2] / Ät = k1 [NO2]2

The experimentally determined rate law for this rxn.does turn out to be:

Rate = !Ä [NO2] / Ät = k [NO2]2

Therefore, our mechanism predicted the correct rate law.

There is something of a mutually reinforcing relationshipbetween the mechanism and the rate law: The rate lawhelps to confirm the mechanism and the mechanism givesus insight into the rate law.

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What is the order of the NO2-CO rxn.: Overall? With respect to NO2? With respect to CO?

Does this seem logical?

In other words: does it bother you that the rate law forthe rxn.:

NO2(g) + CO(g) ÿ NO(g) + CO2(g)

has no [CO] term in it?

Try Prob. 12.20, p. 483

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C. Multi step Reactions with Initial Fast Step New!What if step 1 in the mechanism isn’t the rate limiting step?

1. Example: 2 NO(g) + 2 H2(g) ÿ N2(g) + 2 H2O(g)

2. Proposed mechanism: k1

Step 1: 2 NO W N2O2 fast, reversible k!1

k2Step 2: N2O2 + H2 ÿ N2O + H2O slow, rate limiting

k3Step 3: N2O + H2 ÿ N2 + H2O fast

Do these steps sum to give the overall balanced equation?

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3. Based on our previous logic, the rate law should be:

rate = k2 [N2O2] [H2] This isn’t practical or useful.Bummer!

a) Not practical because we can’t measure [N2O2] easily.b) Not useful because we can’t control [N2O2] easily.

4. Is there a way to cope?a) Because the 1st rxn is fast, it is essentially at equilibrium.

(By definition: at equilibrium, rateforward = ratereverse)

b) Therefore ratef = k1[NO]2 and rater = k!1[N2O2]

c) Since ratef = rater then k1[NO]2 = k!1[N2O2]

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d) Rearrange k1[NO]2 = k!1[N2O2] to solve for [N2O2]:

k1 [N2O2] = )) [NO]2 & substitute this back into:

k!1

rate = k2 [N2O2] [H2] to get:

k1 rate = )) k2 [NO]2[H2] Practical and useful!

k!1

We usually combine all of the constant terms into a newconstant k.

Do Prob. 12.21, p. 483 on your own.

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XII. Rxn Rates & Temp: Arrhenius Equation(Now 12.9 & 10) This equation will help us relate the rate constant, k,to other chemical ideas (collision theory, steric hindrance, etc.). Ithelps to unify our thoughts about kinetics.

A. Reaction rates tend to increase with increasingtemperature. Why?

1. Consider the rxn.: O(g) + HCl(g) ÿ OH(g) + Cl(g)In this rxn. the H!Cl bond must be broken. Thinkabout overlap of electron clouds. ÷ ²O H!Cl ÿ OþHþCl ÿ O!H Cl

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2. Do electrons normally repel each other? (Y or N?)

3. The intermediates in a rxn. are usually very unstable& high in potential energy. (See Fig 12.12) Important terms:

a) activation energy (Ea)b) transition state (synonym: activated complex)

Try to relate the process of going from reactants to thetransition state to some human activity. (Mountainclimbing?)

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B. Some of the thoughts behind this part of kineticsstarted with analysis of collision rates in the gas phase.

1. Air, at normal T & P: each gas molecule has about109 collisions/sec. Imagine a reaction occurring witheach collision. While this does seem consistent withthe rates of chemical rxns. that cause explosions, itclearly is not consistent with the rates of the majorityof chemical rxns.

Therefore, for most rxns., only a small percentage of thecollisions lead to a reaction occurring. These productivecollisions occur with appropriate energy and orientation.

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2. The fraction of collisions with enough energy (seeFig. 12.13) to reach Ea is:

f = e!Ea/RT (Units for T??)

At T= 298K, a rxn. with Ea = 75 kJ/mol has f = 7 x 10!14.

With a collision rate of 109 /s, how long (ave)would youhave to wait to get a collision with sufficient energy?

What would f be for this reaction at 373 K?

Can you see why T has such a large effect on rxn rate?

This takes care of part of the energy considerationmentioned before. What about orientation?

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C. Back to our previous example w/gas phase rxn.:

O(g) + HCl(g) ÿ OH(g) + Cl(g)

1. What if O collides with the Cl side of the HClmolecule, instead of the H side? ÷ ²

H!Cl O ÿ No H to O bonding can occur.

2. Define a steric factor, p: fraction of collisions w/appropriate orientation for the rxn. to occur. If twoatoms of a diatomic reactant are about the same size,p . 0.5. (Electrostatics?)

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3. For the example above with HCl reacting w/ O, wouldp be < 0.5 or > 0.5? Review molecular modeling:

3 representations of HCl cylinder

ball and stick

All to same scale. Align atom centers.

space filling (geometrically more accurate than the 2above)

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º »

a) Recall the O (red) has to hit the H (white)

b) Is the surface area of the H greater than the surface area of

the Cl (green)?

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D. We can now try to set up an equation predicting

rxn. rates w/ the collision theory described here:

1. collision rate = Z [A] [B] (This was 109 above.)2. rxn rate = p × f × collision rate = p × f × Z [A][B]

Remind you of our general rate law? rate = k [A][B]

Clearly, p × f × Z = k. Therefore: k = p Z e!Ea'RT

We now can relate structure, velocity, energy, etc. tounderstanding & predicting rxn rates. Thanks Arrhenius!

p Z is sometimes called A, the frequency factor.

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XIII. (Now 12.10) Using the Arrhenius equation

A. Rearrange k = A e!Ea/RT to: ln k = ! Ea/RT + lnA

Can you see y = mx + b ?

By measuring k a range of different T values we can get a goodestimate of Ea. (Do prob. 12.17 on your own.)

B. Why would anyone want to do this?

Knowledge is power!

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1. If you understand a rxn., you can modify it to changethe rate, improve efficiency, and so on.

2. Enzyme inhibitors based on transition-state analogs.

By Baldwin et al., 1995

Green & blue represent theA & B subunits of HIVprotease. The red molecule(KNI-272) is a transition stateanalog that is an effectiveinhibitor of the enzyme. FromPDBSum accession code 1hpx.

http://www.ebi.ac.uk/pdbsum/

2-dimensional map of inhibitor-enzyme interactions:50

(From PDBSum code 1hpx)

Covalent bonds of theinhibitor are shown inpurple. Everything else ispart of the enzyme. (Ex.: Val82(A) is the valine residue atposition 82 of the A subunit.)

Green dotted lines areionic interactions orHydrogen Bonds.

Red eyelashes representHydrophobic Interactions.

3-D view of inhibitor-enzyme interactions:51

(From PDBSum code 1hpx)

The thicker tube structurerepresents the inhibitor (KNI-272).

The thinner tubes represent thoseparts of the enzyme that interactdirectly with the inhibitor.

Standard CPK color-coding forelements.

grey: carbon red: oxygen

blue: nitrogen yellow: sulfur

What element appears to be strangelyabsent?

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The active site of the enzyme is a different environmentthan bulk solvent. Hence, different energy pathway.

3. Catalytic antibodies = made-to-order catalysts.

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XIV. Catalysis (PKU & PAH? HIV protease?)

A . Catalysts work by decreasing activation energy. (SeeFig. 12.18, p. 485)

B. Enzymes (and catalysts in general) do this by creatinga different environment (different from what?), whichallows for a different (lower energy) reaction pathway. For some enzymes, the environment and pathway isvery well understood. See HIV protease.

C. Do catalysts speed up the rate of the reverse rxn. too?

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Do Conceptual Prob. 12.22, p. 486-7

rxn: 2 A + C2 ÿ 2 AC A = red B = blue C = yellow

Relatives rates: 1 1 2 3

a) What is the order of the rxn in A, B, C?

b) Write the rate law.

c) Write a mechanism that agrees with the rate law.

d) Identify all catalysts & intermediates in mechanism.

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XIII. Homogeneous & HeterogeneousCatalysis (Read on your own.) Fig. 12.20, p. 489.

The Interlude on pp. 490-491 gives you a nice, briefintroduction into an important part of biochemistry.

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