CHEMICAL KINETICS

Goal of kinetics experiment is to measure concentration of a species at particular time during a rxn so a rate law can be determined

Rate of Rxn: describes how fast reactants used up & pdts formed

rates are obtained from concen. vs fct of time

Chem Kinetics: 1)study of rates, 2) factors that affect rxn rates, & 3) mechanisms (steps) by which rxns occur

From a chem eqn, rate can be determined by following the concenof any subst that is quantitatively detected

4 factors that affect chem rxns1) nature of reactants2) concen of reactants3) temp4) catalyst present

Write rate law for rxn to describe how rate depends on concen.

Order of rxn cannot be deduced from chemical eqn. of rxn

Rate law expressions - calculate rate of rxn from rate constant & reactant concen - convert into eqn to determine concen of reactants @ any time

Rate Law is deduced experimentally from how its rate varies w/ concen

Order of rxn cannot be deduced from chemical eqn. of rxn

exponents x & y- usually integers- value of x is the order of rxn w/ respect to A- y??

Values for k, x, & y have no relation to coeff of balanced chem eqn.,remember, must be determined experimentally

For rxn: A + B -----> pdts

general form: rate = k[A]x[B]y

Exponent Define 0 1 2

rate not depend on [reacts] rate is directly proportional to [reacts]rate is directly proportional to square of concen; [reacts]2

overall order of rxn = x + ySum of orders of reacts

order in NO: overall order:

- order in rate law may not match coeff. in balanced eqn- no way to predict rxn orders overall from balanced eqn- orders must be determined experimentally

Examples of observed rate laws for following rxns

3NO(g) ------> N2O(g) + NO2 (g) rate = k[NO]2

2NO2(g) + F2(g) ------> 2NO2F (g) rate = k[NO2][F2] order in NO2: order in F2:

overall order:

2nd 2nd

1st 1st

2nd

quick summary

rate law can be determined by 2 methods:

1) Method of Initial Rates (if time)

2) using Integrated Rate Eqn

ZERO ORDER Has a rate which is independent of concentration of reactant(s),therefore, increasing concen. of rxning species not speed up rate

Rate is:k

k[A] rate O

k

t

A-r

d

dRate is a CONSTANT

A -----> pdts

integration gives eqn called integrated zero-order rate law

[A] = -kt + [A]o

Initial concentrconcentr of chemical@ particular time

[A] = -kt + [A]o

eqn line: y = mx + b

time, t

[A]

calculate k from plot of graph;straight line plot of [A] vs time,t; slope = -k

Determine units:s

M

T

[A]- k

d

d

half-life describes time needed for half of reactant to be depleted

k2

A t O

21

FIRST ORDERDepends on concentration of only 1 reactant, if other reactants present buteach will be zero-order

eqn for first-order reaction A -----> pdts

1st order rate constant,units of 1/time

t

[A] rate

d

d

integration gives eqn called integrated first-order rate law

ln[A] = -kt + ln[A]o

rate is: [A] t

[A]- k

d

dknow: [A] rate k

t [A]

[A]ln O k

d

eqn line: y = m x + b

calculate k from plot of graph;plot of ln[A] vs time,t; gives straight lineslope = -k

time, t

[A]

time, t

ln[A]

Determine units:s

1

sM

M

t

1

[A]

[A]- k

dd

d

half-life describes time needed for half of reactant to be depleted

k

0.693

k

2Ln t

21

SECOND ORDERA. A ------ > pdts depends on concentration of 2nd-order reactant

second-order rate law

22 [A] t

]A[ ]A[rate k

d

dk

integrated in the form:

[A]O @ t = 0 & [A] @ t: [A]

1

]A[

1 t

t ]A[

]A[

O

2

k

d kd

B. or, A + B = pdts two 1st-order reactants:

k[A][B] t

]B[

t

]A[

d

d

d

d

rate is:

ln r = ln k + 2 ln[A]

Another way to represents rate laws, take ln of both sides:

Plot 1/[A] vs time,t; slope = 2nd-order rate constant; +k

[A]

1 t

]A[

1

O

k

eqn line: y = m x + b

time, t

1/[A]

half-life for 2nd order dependent on one 2nd order reactant: O21

Ak

1 t

Determine units:sM

1

sM

M

t

1

[A]

[A]- k

22

d

d

FIRST ORDER REACTION

2 N2O5 (aq) --------> 4 NO2 (aq) + O2 (g)

ln[N2O5] 1/[N2O5], M-1

DATA Time, s [N2O5], M

0 0.0365600 0.02741200 0.02061800 0.01572400 0.01173000 0.008603600 0.00640

- 3.310- 3.597- 3.882- 4.154- 4.448- 4.756- 5.051

27.4 36.5 48.5 63.7 85.5116156

time,s

[N2O

5]

time,s

ln[N

2O5]

time,s

1/[N

2O5]

rate = k[N2O5]

time,s

ln[N

2O5]

slope = /s10*4.820- s 3000

(-3.310) - (-4.756)

tt

]Aln[]Aln[ 4-

03000

03000

SECOND ORDER REACTION

2 NO2 (g) --------> 2 NO (g) + O2 (g)

ln[N2O5] 1/[N2O5], M-1

DATA Time, s [N2O5], M

060120180240300360

- 4.605- 4.986- 5.263- 5.477- 5.655- 5.806- 5.937

100146193239286332379

0.01000.006830.005180.004180.03500.003010.00264

time,s

[NO

2]

time,s

ln[N

O2]

time,s

1/[N

O2]

rate = k[NO2]2

time,s

1/[N

O2]

slope = sL/mol 0.773 s 300

(100) - (332)

tt

[A]1 ]A[

1

0300

0300

ZERO ORDER REACTION

Can occur if:1) rate limited by [catalyst]2) photochemical rxn if rate determined by light intensity3) most often occur when subst as a metal surface or enzyme required for rxn to occur

2 N2O (g) --------> 2 N2 (g) + O2 (g)

N2O N2O N2O

N2O N2O N2O

N2O N2O N2O

N2O N2O N2O

N2O N2ON2O

N2O N2O N2O

N2ON2O

N2O N2O N2O

N2O N2O N2O

N2O N2O N2O

N2O N2O N2O

Pt metal surface

N2O N2O N2O

N2O N2O N2O

N2O N2O N2O

N2O N2O N2O

N2O N2ON2O

N2O N2O N2O

N2ON2O

N2O N2O N2O

N2O N2O N2O

N2O N2O N2O

N2O N2O N2O

Describe what is happening

Rxn occurs on a hot Pt surface, when surface completelycovered w/ N2O molecules, an increase of [N2O] has noeffect on rate, since only N2O molecules on the surface are reacting.

Therefore, the rate is constant because rsn is controlledby what happens on Pt surface rather than total [N2O].

time,s

[N2O

]

rate = k[N2O]0

k2

A t O

21

Determine: UNITS HALF-LIFE

0 ORDER

1ST ORDER

2ND ORDER

sL

mol

T

[A]- k

d

d

O21

Ak

1 t

smol

L

sM

M

t

1

[A]

[A]- k

22

d

d

s

1

sM

M

t

1

[A]

[A]- k

dd

d

k

693. t

21

Summary for reaction orders 0, 1, 2, & n

Zero-Order First-Order Second-Order nth-Order

Rate Law

IntegratedRate Law

Units ofRateConstant (k)

Linear Plotto determineky-intercept

Half-life

k

t

A-

d

d[A]

t

[A]- k

d

d 2[A]

t

[A]- k

d

d n[A]

t

[A]- k

d

d

[A] = [A]O - kt[A] = [A]Oe-kt

ln[A] = ln[A]O - ktt

[A]

1

]A[

1

O

k

order]1st [Except

t1)-(n [A]

1

]A[

11-n

O1n

k

sL

mol

s

1s

Lmol

11n

smol

L

[A] vs t -k

ln[A] vs t -k

k-

t vs]A[

1

order]1st [Except

t vs]A[

11n

k2

A t O

21

k

693. t

21 O2

1 Ak

1 t

[A]O ln[A]OO]A[

1

NOTES

Rate Rxn- describe rate rxn must determine concen of react/pdt at various times as rxn proceeds- devising methods is challenge for chemists-spectroscopic method: if 1 subst colored measure inc/dec in intensity of color

4 Factors: help control rates

Comparing the 2 experiments, [B] is ed by factor of:

1 1.0 * 10-2 M 1.0 * 10-2 M 1.5 * 10-6 M.s-1

2 1.0 * 10-2 M 2.0 * 10-2 M 3.0 * 10-6 M.s-1

3 2.0 * 10-2 M 1.0 * 10-2 M 6.0 * 10-6 M.s-1

METHOD OF INITIAL RATES

describing same rxn in each experiment, same rate law, form: rate = k[A]x[B]y

Notice, [A]O same in #1 & #2, what would affect the rxn rate?

deduce rate law from experimental rate data

Experiment [A]O [B]O initial rate

es in rxn rate due to diff initial concen of B

ratio [B] 2.0 10*0.1

10*0.22

2

rate es by factor of:

What order is rxn order in [B]?

rate ratio = ([B])yExponent y deduced from:

ratio rate 2.0 10*5.1

10*0.36

6

2.0 = (2.0)y solving, y = 1

rate = k[A]x[B]1

Experiments 1 & 3 show [B]O same but [A]O different

ratio [A] 2.0 10*0.1

10*0.22

2

[A] is ed by factor of:

rate es by factor of: ratio rate 4.0 10*5.1

10*0.66

6

What order is rxn order in [A]?

rate ratio = ([A])xExponent x deduced from:

4.0 = (2.0)xsolving, x = 2

rate = k[A]2[B]1

Rate constant, k, substitute data from any set of 3 sets into rate-law expression

or, rate = 1.5 M-2.s-1[A]2[B]rate1 = k[A]12[B]1

1