chemical kinetics rates associated with chemical reactions –how much of a goes away in a given...

Chemical Kinetics

• Rates associated with chemical reactions– How much of A goes away in a given time?– How much of C appears in a given time?– Units usually M/s (or Ms-1) (M = “molar” = mol/L)

• Sometimes mol/L s [I don’t care for this]

– General form: ([A] means “concentration of A”)

• How does chemical change actually take place?

– “mechanism” of a reaction

. ., mol

i eL s

Rearrangement / re-“partnering” (of atoms)

[A]

t

Some Early Goals

• Understand concept of reaction rate• Define various rates

– Of loss, of formation– Average vs. instantaneous

• Be able to relate “rate of A” to “rate of B (C, D, etc.)” for a given reaction– Related by stoichiometry (coefficients)

• Calculate rates of loss or formation from plots of [ ] vs. time

Copyright © Houghton Mifflin Company. All rights reserved. 12–2


Figure 12.1 (in Zumdahl! Old text; analogous with Fig. 13.2 in Tro)

Concentration vs. time plot for a given reaction.

Can you figure out the balanced equation for the reaction that is occurring?

Balanced Equation?

Pick a given time interval to focus on (see board, ICF)

• Twice as many moles of NO appear as O2 (in that given time interval)

rate of formation of NO is twice that of O2

ratio of those coefficients must be 2 : 1


• The same amount of NO2 is lost as the amount of NO that is produced (same rates)

coefficients must be same (i.e., 1:1; 2:2)• Balanced equation is thus:

2 NO2 (g) → 2 NO(g) + O2 (g)


Figure 12.2 (Zumdahl) Representation of a Reaction Represented By:

2 NO2 (g) → 2 NO(g) + O2 (g)


Example(s) 1

A + 2 B → 3 C + D

• If the average rate of decomposition of A is 5.4 x 10-2 M/s over a given time interval, what is the rate of formation of C over that same time interval?

• Write an equation showing the relationship between the rate of formation of C and the rate of formation of D.


Example(s) 2

• If the rate of formation of B at some time equals 3.4 M/s and the rate of loss of A equals 1.7 M/s during that same time interval, which could be the balanced chemical equation for the reaction?

a) A → B

b) 2 A → B

c) A → 2 B

d) None of the above

Backtrack a bit: specifics on defining “rates”

From handout:• Consider a reaction represented by:

aA + bB cC + dD (A, B, C, & D are (aq) or (g))


Negative sign makes the value positive

(Also from handout)



Figure 12.1 (revisit)

Is the rate of loss of NO2 constant as time goes by?

Look at different time intervals (first just “eyeball” them; no calcs yet):

1st 50 s? 3rd 50 s? 5th 50 s?

Since rate is constantly changing, must distinguish “average” rate from

“instantaneous” rate

• The rate of reaction at t = 0 s is not the same as the rate at t = 50 s!– The following calculation is an “average”

rate for the interval t = 0 – 50 s:


22

-5

[ ]Rate of loss of NO =

0.0079 M - 0.0100 M 0.0021 M4.2 x 10 M/s

50. s - 0 s 50. s

NO

t


Table 12.2 Average Rate (in M/s) of Decomposition of Nitrogen Dioxide as a

Function of Time

To calculate an “instantaneous” rate, draw a tangent line (and find its slope)

• See board (Review of “slope” concept)• A line is characterized by a single slope

y = mx + b; m is slope or “steepness”

• If y = [A] and x = t, then slope “rate”!

On a plot of [X] vs. t, |slope| = rate

• Curves don’t have a single slope—but:– each point on a curve can be said to have a

slope equal to the slope of the line tangent to the curve at that point.


See Next Two Slides for illustration


Tangent line to the curve at t = 100 s

Summary: to calculate an “instantaneous rate” (at a given time) from a plot of [X] vs. t:


1. Draw a tangent line to the curve at the point associated with the desired time

2. Pick any two points on that tangent line (better if they are somewhat far apart from one another)

3. Calculate [X]/t (slope) using those two points. • The absolute value of this slope is the

rate of interest (i.e., loss or formation)


Figure 12.3 (and Table 12.3)


Example(s) 3

• See Handout Sheet with plot

Next Step: What things do you think should affect the rate of a reaction?

• (on board, first)


• [ ]’s of reactants (Why? Invokes theory)

• T (Why? Invokes theory. Details later. Takes energy to break bonds?)

• “Activation Energy”, Ea also key (part of theory)

• Presence of a catalyst (later)


(Differential) Rate Laws (and how to determine them)

• Definition and symbols (see board)• How do you find orders and k?

– Find orders using “Method of Initial Rates”• Do different trials with different initial

concentrations• See how the initial rate varies

– Use “short method” (if numbers “easy”)– Use “brute force” method (if necessary)

– Find k using substitution (only AFTER the orders have been determined

Practice I with Rate Laws

Write the rate law (format only; no values for orders and k) for the following:

1)2 HI(g) H2(g) + I2(g)

Rate Law:

2) 2 C2H4(g) + O3(g) 2 CH2O(g) + ½ O2(g)Rate Law:

3) 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)Rate Law:


R = k[HI]m

R = k[C2H4]m[O3]n

R = k[NO]m[H2]n

Meaning of “orders”(see boardwork)

The order indicates how sensitive the rate is to changes in concentration of a given reactant:

R = k[A]m

1)Zeroth Order—Rate is not dependent on [A]

2) First Order—Rate is proportional to [A]

3) Second Order—Rate is proportional to [A] squared (i.e., [A]2)

4) Third Order—Rate is proportional to [A] cubed



Example Problem on Handout

a) Doubling the concentration of both A and Bb) Tripling the concentration of both A and Bc) Tripling the concentration of A and doubling Bd) Tripling the concentration of B and doubling Ae) Halving the concentration of bothf) Halving the concentration of A and doubling B


Initial Rates Data

NO2-

1) Find the orders of NH4+ and NO2

-

2) Calculate k (on board)

Back up a bit…[Actually, already done last class!]

Definition of “Reaction Rate”

• To get one unambiguous rate,– Divide “rate of loss” or “rate of formation”

by coefficient in the balanced equation– See handout (and board)



Example Problem 2 on Handout

Determine the rate law (in terms of the rate of

formation of CH2O) and the value of k.


Table 12.5 Another Set of Initial Rates Data

“Brute force” method

• Calculate (from given data)

• Create an equation by putting the above on the left side of the equal sign, and then substituting into the numerator and denominator on the right side using the rate law: R (trial z) = k [A](trial z)

m [B](trial z)n

• Do on board (if not yet done)


(trial x)

(trial y)

R

R

Calculation of k (if not yet done)

• Earlier found that n = 1 and m = 1 Now find k:


R = k[C2H4]m[O3]n

1-1-37-8-

-12

sM 10 x 2.0M) 10 x M)(0.50 10 x (1.0

M/s 10 x 1.0 kk

• Pick any trial you want, and substitute values into rate law:

• (Trial 1): 1.0 x 10-12 M/s = k(1.0 x 10-8 M)1(0.50 x 10-7 M)1

Careful with units!

Calculation of k (if not yet done)

• Remember, k has units!– The units of k are not “fundamental”—they

change depending on the overall order of the reaction of interest. Use algebra:

1st order overall: R = k[A] k = R/[A]


MsM

11s

s 2nd order overall: M-1s-1

3rd order overall: M-2s-1

Etc.


• logba = x bx = a; b is “base”; a is “argument” • If no “b” present, it is an implied 10 (i.e., base 10 log)• loge = ln (i.e., “natural log”); e is a special number (like )

• lnx is the inverse of ex; “undo” one another: ln(ex) = x !


Integrated Rate Laws (How does [ ] vary with time?)

• If -[A]/t (rate of loss of A) depends on [A], then [A] must also depend on t– Precise relationships (for each situation: 0th, 1st,

and 2nd order) are given by calculus (which is not required for this course)

– However, much can be conceptually rationalized without calculus, and that will be my focus.

– At least one equation will need to be memorized; I will stress/show how to derive others from it.

• Board Work (0th , 1st , and concept of “half life”)


• A first order reaction is 35% complete at the end of 55 minutes. What is the value of k?

React 5

Half Life “Quick Quiz”


What is the half life for this reaction (trial)?

Answer: About 2.5 s


Figure 12.7 A Plot of [A] versus t for a Zero-Order Reaction

Note: k is always a positive qty. The negative sign makes it + (b/c slope is negative)

0[A][A]R k k

t


[ ] vs. t Plot for2 N2O5 → 4 NO2 + O2

0

0.02

0.04

0.06

0.08

0.1

0.12

0 100 200 300 400 500

Time (s)

[N2O

5]

(M)

Is the reaction 0th order in N2O5? 1st order?

How do you know? (See next slide as well)


Figure 12.4 A Plot of In[N2O5] versus Time


Table 12.6 Summary of Kinetics Info on 0th, 1st, and 2nd Order Reactions

(A bit later)

0

[A]

[A]ktt e

Fig. 13.11, Tro

0[A] [A] ktt e

2nd Order Integrated Rate Law

• Back to board

• Use idea that in 2nd order: “rate is more sensitive to changes in [A] than

in 0th or 1st order cases”

Gets “increasingly” slower as time goes by! Not “exponential” decay; decays “more slowly”

Half life gets longer and longer (takes “forever” to get to zero—see next slide)


2nd Order


What is the second half life for this reaction (trial)?

The 3rd half life?

)1(2/1t )2(2/1t )3(2/1t

For 2nd order processes, half life increases with time (really, increases as [A]o decreases)

Fig. 13.5, Tro


Figure 12.6 (a) A Plot of In[C4H6] versus t (b) A Plot of 1/[C4H6] versus T


0

[A]

[A]ktt e


Mechanisms

• See Handout (in part)

Mechanism for the Iodination of Acetone (Exp 20)


(fast, equilibrium)

(slow)

(fast)

CH3CCH3

O

H3O+

CH3CCH3

OH+

CH3C

OH

C

H

H

H

+

H2O CH3C

OH

C

H

HH3O+

Step 1

Step 2

Step 3 CH3C

OH

C

H

H

I I

CH3C

O

C

H

H

I

CH3C

OH

C

H

H

I

I-

CH3C

OH

C

H

H

I

Step 4H2O

H3O+(fast)

+

+

+

+

k1

k2

k3

k4

+

+

+

+

Mechanism Ideas Discussed Earlier

• For elementary reactions (steps) only, the rate law is “knowable” from the balanced equation of the elementary step.

• The rate of an overall reaction that occurs in more than one step can only be as fast as the slowest step:

Roverall Rslow (Rslow is also called Rrls or Rrds)


Mechanism Ideas Discussed Earlier (Cont’d)

• A mechanism dictates (predicts) an overall rate law for the reaction

• If the predicted rate law does NOT match the experimental rate law (the “actual” rate law), the proposed mechanism is “wrong” (rxn does not occur by that mechanism)

• If the predicted rate law DOES match experiment, the mechanism is “possibly” correct, but not necessarily correct.– More than one mech can predict the same rate law!


The Rate Law for an Elementary Step Has Orders Equal to Coefficients

Elementary occurs in one collision Every collision “matters”

[recall the iodination reaction—collisions after the slow step did not “matter” because the rate was limited by the slow step]

Twice as many collisions means twice the reaction rate. 2x the [ ] means 2x the collisions!

Order is 1 for each species involved in the collision. NOTE: If a species is involved twice in the collision

[i.e., it collides with itself], its order will be 2.



Table 13.3 (Tro) Examples of Elementary Steps and Their Rate Laws

NOTE: You can “know” the rate laws for elementary steps only (using collision theory—higher [ ] more collisions)

Examples (from Handout)

• What are the rate laws predicted by:– Mechanism 1?

– Mechanism 2?

– Mechanism 3?


R = k[NO2][F2] (Rslow)

R = k[NO2]2[F2] (Rslow)

R = k[NO2][F2] (Rslow)

If the actual rate law is R = k[NO2][F2], what can you conclude? #2 is not the mechanism!

Temperature Dependence of k—Ea and the Arrhenius Equation

• Board Work (PE curves, etc.)



Figure 12.10 (Zumdahl) A Plot Showing the Exponential Dependence of the Rate Constant on Absolute Temperature

RT

Ea

Ae

k

Arrhenius Equation (“Law”):


Figure 12.11 a & b (Zum.) (a) The Change in Potential as a Function of Reaction Progress (b)

A Molecular Representation of the Reaction

How can one determine Ea experimentally?

• Take ln of both sides of the Arrhenius equation

• Swap lnA term with the other term on right• Get the following (see board [and posted file]):

12–58

ATR

Ek a ln

1ln

bxmy

Plot lnk vs. 1/T ; Find slope, set equal to –Ea/R !

12–59

ln k (no units)

8.123

10.79

12.79

14.35

15.59

16.61

17.46

18.18

18.79

19.32

19.79

20.20

20.57

20.90

1/T (K-1)

0.001667

0.001429

0.00125

0.001111

0.001000

0.000909

0.000833

0.000769

0.000714

0.000667

0.000625

0.000588

0.000556

0.000526

Example 13.7 in Tro

12–60

ln k

8.123

10.79

12.79

14.35

15.59

16.61

17.46

18.18

18.79

19.32

19.79

20.20

20.57

20.90

1/T (K-1)

0.001667

0.001429

0.00125

0.001111

0.001000

0.000909

0.000833

0.000769

0.000714

0.000667

0.000625

0.000588

0.000556

0.000526

)molJ )(8.314 10 x 1.12( 1-1-4 KKmRER

Em a

a

kJ/mol 93.1 J/mol 16.8193

Collision Theory Explains Arrhenius Equation Behavior

• Exponential factor: tells fraction of collisions that have a KE great enough for reaction to occur (≥ Ea ) Value goes from 0 – 1; depends on T! Greater T, greater exponential factor (b/c avg KE

T greater T, greater KE


• Steric (orientation) factor (p): tells fraction of collisions that have the proper orientation to make products Value typically goes from 0 – 1 (text notes exception)

RT

E

RT

E aa

e z pAe

k

Graphical / Physical Interpretation of the Exponential Factor

• (see next slide)



Figure 13.14 (Tro): Plot Showing the Fraction of Collisions with a Particular

Energy at T1 and T2, where T2>T1

aE

RTe

The fraction spoken about here (represented by the colored

areas under the curve) is equal to the value of the exponential factor:

whose value goes from:0 (at T = 0) to 1 (as T )N

um

ber

Kinetic*NOTE: This fraction depends on two things: Ea and T.


Actual shapes of KE distribution curves for a sample at two temperatures

Note that the areas under the curve are the same, but the

peak for the lower T curve is at a much smaller KE

KE (arbitrary units)

# of

par

ticle

s (w

ith a

giv

en

KE

)

KEcollision ≥ Ea

KEcollision < Ea

An intrinsic property of the reaction (mechanism) [does not change with T]

Average KE increases with T, so more collisions have a higher KE.

Graphical / Physical Interpretation of the Steric (Orientation) Factor (p)

• (see board, then next slide)


What kinds of orientations at the point of collision would lead to reaction?

Hint: Which bonds are made and broken?

p probably < 0.2

What is a catalyst and how does it “work”?

• Typical definition: A species that speeds up a reaction without being consumed. – Okay, but really only part of the story

• how does it speed up the reaction?• How can it not be consumed? (Some even say it is “not

involved in the reaction” How silly! Is it “magic”?!)


Catalysts (cont.)

• Some say that a catalyst “lowers the activation energy” for a reaction.– That’s roughly true, but not precisely true.– It cannot change the activation energy of “the

exact process”, because the activation energy is determined by the collisions in that process…


• A catalyst changes the mechanism of a reaction! Creates a new pathway that avoids the original reaction’s “slow step”!– The new pathway’s activation energy is generally smaller

than the original one’s. So there is truth to the idea noted above. But see next slide for an oversimplification…


Figure 12.15 (Zumdahl): A catalyst creates a new mechanism (pathway) that has a lower overall Ea

In actuality, the catalyzed pathway must have at least one inter-mediate (it can’t be one step!)


Figure 12.16 (Zumdahl): Addition of a Catalyst Increases the Number of Collisions That Have the Energy Needed to React (i.e., to

“Get Over” the Activation Energy Barrier) without raising the temperature

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chemical kinetics rates associated with chemical reactions –how much of a goes away in a given...

Documents

given time interval

rate of loss

rate of b c

time plot

balanced equation

average rate of decomposition

different time intervals

balanced chemical equation