Zumdahl Ch 12 Kinetics powerpoint

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Zumdahl Ch. 12:

Kinetics

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Reaction Rate

 Change in concentration of a reactant or Change in concentration of a reactant or

product per unit time.

[A] means concentration of A in mol/L; A is the [A] means concentration of A in mol/L; A is the

reactant or product being considered. reactant or product being considered.

 

2 1

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The decomposition of NO

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The

decomposition

decomposition

of NO

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Instantaneous rate

 Value of the rate at a particular time.Value of the rate at a particular time.  Can be obtained by computing the Can be obtained by computing the

slope of a line tangent to the curve

at that point.

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Rate law

 Shows how the rate depends on the Shows how the rate depends on the

concentrations of reactants.

 Example: for the decomposition of Example: for the decomposition of

NO

NO₂₂:_:

2NO

2NO₂₂(₍g_g) ) →→ 2NO( 2NO(gg) + O) + O₂₂((gg))

Rate =

Rate = kk[NO[NO22]]nn

 k_k_{= rate constant}_{= rate constant}

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Rate law

Rate =

Rate = kk[NO[NO₂₂]]nn

 The concentrations of the products do not appear The concentrations of the products do not appear

in the rate law because the reaction rate is being in the rate law because the reaction rate is being

studied under conditions where the reverse studied under conditions where the reverse

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Rate law

Rate =

Rate = k_k[NO_[NO₂₂]_]nn

 The value of the exponent The value of the exponent nn must be must be

determined by experiment; it cannot

be determined from the balanced

equation.

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Types of rate laws

 Differential rate law (rate law) – Differential rate law (rate law) –

shows how reaction rate depends on

concentrations.

 Integrated rate law – shows how the Integrated rate law – shows how the

concentrations change over time.

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Rate laws: A summary

 Because we typically consider reactions Because we typically consider reactions

only under conditions where the reverse

reaction is unimportant, our rate laws

will involve only conc of reactants.

 Because the differential and integrated Because the differential and integrated

rate laws for a given reaction are related

in a well–defined way, the experimental

determination of either of the rate laws

is sufficient.

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Rate laws: A summary

 Experimental convenience dictates Experimental convenience dictates

which type of rate law is determined

experimentally.

 Knowing the rate law for a reaction is Knowing the rate law for a reaction is

important mainly because we can

usually infer the individual steps

involved in the reaction from the

specific form of the rate law.

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 We determine experimentally the We determine experimentally the

power to which each reactant

concentration must be raised in the

rate law.

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Method of initial rates

 The initial rate is determined for each The initial rate is determined for each

experiment at the same value of

experiment at the same value of t_t as _as

close to

close to t_t = 0 as possible._{= 0 as possible.}

 Several experiments are carried out using Several experiments are carried out using

different initial concentrations

different initial concentrations of each of _{of each of}

the reactants

the reactants, and the initial rate is _{, and the initial rate is}

determined for each run.

 The results are compared to see how the The results are compared to see how the

initial rate depends on the initial

concentrations

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Method of initial rates

for 2NO(g) + O

₂₂

(g)



2NO

₂₂

(g)

Expt.

Expt. [NO][NO]o o , , MM [O[O22]]o o , , MM Initial Initial rate,

rate, MM/s/s 1

1 0.0200.020 0.0100.010 0.0280.028

2

2 0.0200.020 0.0200.020 0.0570.057

3

3 0.0200.020 0.0400.040 0.1140.114

4

4 0.0400.040 0.0200.020 0.2270.227

5

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Method of initial rates

for 2NO(g) + O

for 2NO(g) + O₂₂ (g) (g)  2NO 2NO₂₂ (g) (g)

Expt.

Expt. [NO][NO]_o_o, , MM [O[O₂₂]]_o_o, , MM Initial Initial rate,

rate, MM/s/s 1

1 0.0200.020 0.0100.010 0.0280.028 2

2 0.0200.020 0.0200.020 0.0570.057 3

3 0.0200.020 0.0400.040 0.1140.114 4

4 0.0400.040 0.0200.020 0.2270.227 5

5 0.0100.010 0.0200.020 0.0140.014

Assume rate = [

Assume rate = [_NONO₂₂]/]/_tt

(a)

(a) Determine the rate law.Determine the rate law.

(b)

(b) Find the value of the rate constant Find the value of the rate constant

w/appropriate units. w/appropriate units.

Rate = k[NO]

Rate = k[NO]mm[O[O 2

2]]nn

Solve for m and n by looking closely at data. Solve for m and n by looking closely at data.

Compare trials 2 and 4.

When [NO] doubles, rate

quadruples.

So rate  [NO]2. We

write m = 2.

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Now we find n, the power of O

Now we find n, the power of O₂₂ in the rate law. in the rate law.

Expt.

rate, MM/s/s 1

1 0.0200.020 0.0100.010 0.0280.028 2

2 0.0200.020 0.0200.020 0.0570.057 3

3 0.0200.020 0.0400.040 0.1140.114 4

4 0.0400.040 0.0200.020 0.2270.227 5

5 0.0100.010 0.0200.020 0.0140.014

Assume rate =

Assume rate = _NONO₂₂]/]/_tt

(a)

(a) Determine the rate law.Determine the rate law.

(b)

(b) Find the value of the rate constant Find the value of the rate constant

w/appropriate units. w/appropriate units.

Rate = k[NO]

Rate = k[NO]22[O[O 2

2]]nn

Compare trials 1 and 2.

When [O₂] doubles, rate

doubles.

So rate  [O₂]. We write n =

1.

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(b) Find the value of the rate (b) Find the value of the rate

constant w/appropriate units. constant w/appropriate units. Expt.

rate, MM/s/s 1

1 0.0200.020 0.0100.010 0.0280.028 2

2 0.0200.020 0.0200.020 0.0570.057 3

3 0.0200.020 0.0400.040 0.1140.114 4

4 0.0400.040 0.0200.020 0.2270.227 5

5 0.0100.010 0.0200.020 0.0140.014

Rate = k[NO]

Rate = k[NO]22[O[O 2

2]]11

k =

k = raterate

[NO][NO]22[O[O 2 2]]

k =

k = 0.028 mol L0.028 mol L-1-1 s s-1-1

(0.020 mol L(0.020 mol L-1-1))2 2 (0.010 mol L(0.010 mol L-1-1))

k = 7.0 x 10

k = 7.0 x 1033 L L22 mol mol-2-2 s s-1-1

Calculate the k value for each trial.

If slight

variations, take an average. If way

off, recalculate everything.

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Overall reaction order

 The sum of the exponents in the reaction The sum of the exponents in the reaction

rate equation.

Rate =

k

[A]

nn

[B]

mm

Overall reaction order =

Overall reaction order = n_n + ₊m_m

k

k = rate constant= rate constant

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End of lesson

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12.4 Integrated rate laws

 ObjectiveObjective: Students will be able to : Students will be able to

match integrated rate law data with

match integrated rate law data with

reaction order.

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Concept check

How do exponents in rate laws

compare to coefficients in balanced

equations?

Why?

Answer: No connection at all.

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 Rate = Rate = kk[A][A]

 Integrated rate law:Integrated rate law:

ln

ln[A] = –[A] = –ktkt + + ln_ln[A][A]_o_o

[A] = concentration of A at time [A] = concentration of A at time tt

k

k = rate constant = rate constant t

t = time = time [A]

[A]_o_o = initial concentration of A = initial concentration of A

First order reactions

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Plot of

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 Half–life:Half–life:

k

k = rate constant = rate constant

 HalfHalf––life is constant; it does not depend life is constant; it does not depend

on the concentration of reactants.

First order reactions

t_1/2 = 0.693

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Exercise

A first order reaction is 35%

complete at the end of 55 minutes.

What is the value of

What is the value of k_k?_?

k = 7.8 x 10

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 Rate = Rate = kk[A][A]22

 Integrated rate law:Integrated rate law:

[A] = concentration of A at time [A] = concentration of A at time t_t

k

t = time = time [A]

Second order reactions

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Plot of

ln

[C

₄₄

H

₆₆

] vs time and

plot of 1/[C

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k

k = rate constant = rate constant [A]

 Half–life gets longer as the reaction Half–life gets longer as the reaction

progresses and the concentration of

reactants decrease

reactants decrease..

 Each successive halfEach successive half–life is double the –life is double the

preceding one.

Second order reactions

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Exercise

For a reaction aA

For a reaction aA _ Products, Products,

[A]

[A]₀₀ = 5.0 _{= 5.0}M_M, and the first two half-lives are , and the first two half-lives are 25 and 50 minutes, respectively.

25 and 50 minutes, respectively.

a)

a) Write the rate law for this reaction.Write the rate law for this reaction. rate =

rate = kk[A]_[A]22

b)

b) Calculate _Calculatekk._.

k

k = 8.0 x 10 = 8.0 x 10-3-3 _M_M–1–1minmin–1–1

c)

c) Calculate [A] at Calculate [A] at tt = 525 minutes. = 525 minutes. [A] = 0.23

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12.1-3 Quiz

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 Rate = Rate = kk[A][A]00 = = kk

 Integrated:Integrated:

[A] = –

[A] = –kt_kt + [A]_{+ [A]}_o_o

[A] = concentration of A at time [A] = concentration of A at time t_t

k

t = time = time [A]

Zero order reactions

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Plot of [A] vs time

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k

k = rate constant_{= rate constant}

[A]

[A]_o_o = initial concentration of A_{= initial concentration of A}

 Half–life gets shorter as the reaction Half–life gets shorter as the reaction

progresses and the concentration of

reactants decrease.

Zero order reactions

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Concept check

How can you tell the difference among 0

How can you tell the difference among 0thth, 1, 1stst, ,

and 2

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Summary of the rate laws/graphing methods

Summary of the rate laws/graphing methods

O r d e r

Rate law Integrated rate

law plot slope

0 Rate = k[R]0 [R]

0 - [R]t = kt [R]t vs t -k

1 Rate = k[R]1

Most important!

ln([R]_t/[R]₀) = -kt ln[R]_t vs t -k

2 Rate = k[R]2 1/[R]

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Summary of the rate laws

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Examples

1.

1. NN₂₂OO_4(g)_4(g) 2NO 2NO_{2 (g)}_{2 (g)} Rate = k[NRate = k[N₂₂OO₄₄]]

k =3.20 x 10

k =3.20 x 10-3 -3 minmin-1 -1 at 25Cat 25C

Starting with 4.50 x 10

Starting with 4.50 x 10-2 -2 MM, what is the concentration , what is the concentration

remaining after 20.0 minutes? remaining after 20.0 minutes?

2. How long would it take for 90% of the N

2. How long would it take for 90% of the N₂₂OO₄₄to to decompose?

decompose?

3. What is the t

3. What is the t_1/2_1/2for this reaction?for this reaction?

0.0422 M

720 min

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More examples

4. Ozone decomposes: 2O

4. Ozone decomposes: 2O_3(g)_3(g) 3O 3O_{2 (g)}_{2 (g)} in a second in a second

order reaction; rate = 550. L/mol

order reaction; rate = 550. L/molmin at 400.K How min at 400.K How

long will it take for the concentration to drop from 0.100 long will it take for the concentration to drop from 0.100

M

M to 0.0300 to 0.0300 MM? ?

5. How long will it take to decompose 90.0 % of a 5. How long will it take to decompose 90.0 % of a

compound that reacts in a zero order reaction with k = compound that reacts in a zero order reaction with k =

2.05 x 10

2.05 x 10-3 -3 mol/Lmol/L__s? s?

0.0418 min?

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4. Ozone decomposes: 2O

4. Ozone decomposes: 2O_3(g)_3(g) 3O 3O_{2 (g)}_{2 (g)} in a second order reaction; in a second order reaction; rate = 550. L/mol

rate = 550. L/molmin at 400.K How long will it take for the min at 400.K How long will it take for the concentration to drop from 0.100

concentration to drop from 0.100 MM to 0.0300 to 0.0300 MM??

1

1 - - 11 = kt = kt [R]

[R]_t_t [R] [R]_o_o

1 - 1 = (550 L/mol1 - 1 = (550 L/mol••min) tmin) t 0.0300

0.0300 M_M 0.100 _0.100M_M

33.3 L/mol – 10.0 L/mol = (550 L/mol

33.3 L/mol – 10.0 L/mol = (550 L/mol••min) tmin) t

t = 33.3 – 10.0 L/mol = 23 min

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5. How long will it take to decompose 90.0 % of a compound that

reacts in a zero order reaction with k = 2.05 x 10

reacts in a zero order reaction with k = 2.05 x 10-3 -3 mol/Lmol/L__s?s? [A]

[A]_t_t = -kt + [A] = -kt + [A]_o_o [A]

[A]_t_t – [A]_{– [A]}_o_o = -kt_{= -kt}

If we assume an initial concentration of 1.00

If we assume an initial concentration of 1.00 MM, , then we can solve the problem. Otherwise, not then we can solve the problem. Otherwise, not

enough information. enough information.

0.10

0.10 MM – 1.00 – 1.00 MM = -kt = -kt -0.90

-0.90 M_M = -(2.05 x 10_{= -(2.05 x 10}-3-3 mol/L mol/L••s) ts) t

t = 0.90 mol/L t = 0.90 mol/L

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End of lesson

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Exercise

Consider the reaction aA

Consider the reaction aA  products. _products.

[A]

[A]₀₀ = 5.0 _{= 5.0}M_M and _andk_k = 1.0 x 10_{= 1.0 x 10}–2 –2 (assume the (assume the

units are appropriate for each case). units are appropriate for each case).

Calculate [A] after 30.0 seconds have Calculate [A] after 30.0 seconds have

passed, assuming the reaction is: passed, assuming the reaction is:

a)

a) Zero order_{Zero order} b)

b) First order_{First order} c)

c) Second orderSecond order

4.7

4.7 M_M

3.7

3.7 M_M

2.0

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12.5-6 Reaction mechanisms

 Objective: Students are able to

use kinetics data to choose

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 Most chemical reactions occur by a Most chemical reactions occur by a

series of elementary steps. series of elementary steps.

 An An elementary stepelementary step is a reaction is a reaction

whose rate law can be written from whose rate law can be written from

its stoichiometry. its stoichiometry.

 An An intermediateintermediate is formed in one is formed in one

step and used up in a subsequent step and used up in a subsequent

step, thus is never seen in the step, thus is never seen in the

overall balanced equation. overall balanced equation.

Reaction mechanism

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For the reaction:

For the reaction:

NO

NO₂₂(₍g_g) + CO() + CO(gg) ) →→ NO( NO(gg) + CO) + CO₂₂((gg)) The kinetics data show Rate = k[NO

The kinetics data show Rate = k[NO₂₂]]22

Why?

Because the reaction appears to take

place through a two-step mechanism:

Step 1. (slow)

Step 1. (slow)

NO

NO_2(g)_2(g) + NO + NO_2(g)_2(g)  NO NO_3(g)_3(g) + NO + NO_(g)_(g)

Step 2. (fast)

NO

NO_3(g)_3(g) + CO + CO_(g)_(g)  NO NO_2(g)_2(g) + CO + CO_2(g)_2(g)

Note: The rate law reflects the collision between two NO₂ molecules in the slow

step…

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 The step 2 you just saw was a

bimolecular step (involves two colliding particles).

 A bimolecular step always has a rate law like:

 Rate = k[A]2 where A collides with A, or

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 The sum of the elementary steps The sum of the elementary steps

must give the overall balanced must give the overall balanced

equation. equation.

 The mechanism must agree with The mechanism must agree with

the experimentally determined rate the experimentally determined rate

law. law.

A reaction mechanism must

meet these requirements:

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So does the proposed mechanism work?

So does the proposed mechanism work?

NO

NO₂₂((gg) + CO() + CO(gg) ) →→ NO( NO(gg) + CO) + CO₂₂((gg)) Rate = k[NO

Rate = k[NO₂₂]]22 Step 1. (slow)

Step 1. (slow)

NO

NO_2(g)_2(g) + NO + NO_2(g)_2(g) _ NO NO_3(g)_3(g) + NO + NO_(g)_(g)

Step 2. (fast)

NO

NO_3(g)_3(g) + CO_{+ CO}_(g)_(g) _ NO NO_2(g)_2(g) + CO + CO_2(g)_2(g) Yes, because:

Yes, because: 1)

1) The steps add up to the overall equation;The steps add up to the overall equation;

2)

2) The slow step is always the rate-The slow step is always the

rate-determining step.

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Picture story time…

Picture story time…

NO

NO₂₂(₍g_g) + CO() + CO(gg) ) →→ NO( NO(gg) + CO) + CO₂₂((gg))

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 Unimolecular – step involving one Unimolecular – step involving one

molecule; first order.

 Bimolecular – step involving the collision Bimolecular – step involving the collision

of two species; second order.

 Termolecular – step involving the collision Termolecular – step involving the collision

of three species; third order [rare].

Elementary steps (molecularity)

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 A reaction is only as fast as its A reaction is only as fast as its

slowest step.

 The rate-determining step (slowest The rate-determining step (slowest

step) determines the rate law and the

molecularity of the overall reaction.

Rate-determining step

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More difficult example:

 Sometimes the mechanism involves

a fast step that reaches equilibrium, followed by a slow, rate-determining step.

 Example:

 2NO + 2H₂  N₂ + 2H₂O

has an experimentally determined rate law of Rate = k[NO]2[H

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 2NO + 2H₂  N₂ + 2H₂O

Rate = k[NO]2[H 2]

The proposed mechanism is:  Step 1 2NO  N₂O₂

(fast, equilibrium step)

 Step 2 N₂O₂ + H₂  N₂O + H₂O (slow)

 Step 3 N₂O + H₂  N₂ + H₂O (fast)

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Rate = k[NO]2[H 2]

 Step 1 2NO  N₂O₂ (fast, equilibrium step)

 Step 2 N₂O₂ + H₂  N₂O + H₂O (slow)

 Step 3 N₂O + H₂  N₂ + H₂O (fast)

Overall 2NO + 2H₂  N₂ + 2H₂O

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Rate = k[NO]2[H 2]

 Step 1 2NO  N₂O₂

(fast, equilibrium step)

 Step 2 N₂O₂ + H₂  N₂O + H₂O

(slow)

 Step 3 N₂O + H₂  N₂ + H₂O

(fast)

Overall 2NO + 2H₂  N₂ + 2H₂O

Answer: Because the N₂O₂ is an

intermediate; the rate law never includes intermediates.

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For the equilibrium step 2NO  N2O2

We can write an equilibrium expression… K_eq = [N₂O₂]/[NO]2

(products over reactants; see Ch 13) We rearrange this to give:

[N₂O₂] = K_eq [NO]2

Now if we substitute that into the Step 2 rate law, rate = k [N₂O₂][H₂], we get:

Rate = k K_eq [NO]2 [H 2]

Combining the two constants gives us: Rate = k [NO]2 [H

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The final result is that the Step 2 rate law gives us the observed rate law for the entire reaction …

… exactly as expected for reactions involving a fast step that reaches

equilibrium followed by a slow, rate-determining step!

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Concept check

The reaction A + 2B

The reaction A + 2B _ C has the _{C has the} following proposed mechanism: following proposed mechanism:

A + B D

A + B D (fast equilibrium)(fast equilibrium) D + B

D + B  C C (slow)(slow)

Write the rate law for this mechanism. Write the rate law for this mechanism.

rate = k[A][B]

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End of lesson

 Read the rest of the handout.Read the rest of the handout.

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12.7 Microscopic view of

reaction rates

 Objective: Students will be able to use Objective: Students will be able to use

collision theory to explain kinetics

collision theory to explain kinetics

phenomena.

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 In order to react, molecules must In order to react, molecules must

1)

1) collide, collide,

2)

2) with sufficient energy, andwith sufficient energy, and

3)

3) in the proper orientation.in the proper orientation.

Collision model

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Collision model

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 = minimum energy needed to = minimum energy needed to

produce a chemical reaction. produce a chemical reaction.

Activation energy,

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Effect of temperature on the

number of reaction-producing

collisions

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Analogy: Effect of studying on

getting into UC Berkeley

Less studying More

studying

(66)

Effect of temperature on the

number of reaction-producing

collisions

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Change in potential energy

Ea

H

(68)

Change in potential energy

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 Collision must have enough energy to Collision must have enough energy to

produce the reaction (must equal or

exceed the activation energy).

 Orientation of the reactants must Orientation of the reactants must

allow formation of any new bonds

needed to produce products.

For reactants to form products

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A

A =₌ frequency factor_{frequency factor} E

E_a_a == activation energy (J/mol)activation energy (J/mol)

R

R =₌ gas constant (8.3145 _{gas constant (8.3145}

J/K

J/K·mol)·mol)

T

T =₌ temperature (K)_{temperature (K)}

Arrhenius equation

/

=

Ea RT

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Linear form of Arrhenius

equation

 

a

E

1

ln( ) =

ln

R T

 



 

 

k

+

A

y = m

(72)

Linear form

of Arrhenius

equation

slope =

-E

_a_a

/R

E

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…

or use:

ln

k

₂₂

=

-E

_a_a

1

–

1

k

₁₁

R T

₂₂

T

₁₁

to find k value at a given

temp.

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Try p. 557

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Exercise

Chemists commonly use a rule of thumb

that an increase of 10 K in temperature

doubles the rate of a reaction. What

must the

must the activation energyactivation energy be for this be for this statement to be true for a temperature

statement to be true for a temperature

increase from 25

increase from 25 ooC to C to 3535 ooC? C?

E

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 A substance that speeds up a reaction A substance that speeds up a reaction

without being consumed itself.

 Provides a new pathway for the Provides a new pathway for the

reaction with a lower activation

energy.

Catalyst

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Catalyzed vs uncatalyzed pathway

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Catalyst reduces E

_a_a

; increases

number of effective collisions

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 Most often involves gaseous Most often involves gaseous

reactants being adsorbed on the reactants being adsorbed on the

surface of a solid catalyst. surface of a solid catalyst.

 Adsorption – collection of one Adsorption – collection of one

substance on the surface of substance on the surface of

another substance. another substance.

Heterogeneous catalyst

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Steps in heterogeneous catalyst

reaction

1.

1. adsorption and activation of the adsorption and activation of the

reactants. reactants.

2.

2. migration of the adsorbed reactants migration of the adsorbed reactants

on the surface. on the surface.

3.

3. reaction of the adsorbed reaction of the adsorbed

substances. substances.

4.

4. escape, or desorption, of the escape, or desorption, of the

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 Exists in the same phase as the Exists in the same phase as the reacting molecules.

reacting molecules.

 Enzymes are natureEnzymes are nature’_’s catalysts.s catalysts.

Homogeneous catalyst

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End of lesson