Lect w2 152 - rate laws_alg

General Chemistry IIGeneral Chemistry IICHEM 152CHEM 152

Week 2

UA GenChem

Week 2 Reading Assignment

Chapter 13 – Sections 13.2 (rate), 13.3 (rate law), 13.4 (integrated

rate law)

UA GenChem

A more refined descriptionThe average reaction rate gives us information about the speed of the

reaction over a certain period of time. But what if we are interested in a more

detailed description.How do we

calculate the instantaneous reaction rate

at any particular

time?

0

0.001

0.002

0.003

0.0040.005

0.006

0.007

0.008

0.009

0.01

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Time, min

Conc

entra

tion,

mol

/L

UA GenChem

START

END

Different Rates During a Reaction

Overall averagerate

InitialRate

InstantaneousRate

(slope of the lineAt a point)

UA GenChem

Br2(aq) + HCOOH(aq) 2Br-(aq) + 2H+(aq) + CO2(g)

time

.

From your lab experience, how could we follow

the course of this reaction? Measure rates? What’s changing? What could we measure?

Measuring RatesConsider this reaction:

UA GenChem


time

393 nmlight

Detector

[Br2] Absorption3

93 n

m

Br2 (aq)

UA GenChem


Average rate = [Br2]t

= -[Br2]final – [Br2]initial

tfinal - tinitial

slope oftangent

slope oftangent slope of

tangent

Instantaneous rate = rate for specific tiny instance in time

UA GenChem

Br2(aq) + HCOOH(aq) HBr(aq) + CO2(g)

RATE

rate [Br2]

rate = k [Br2] + 0.0

k = rate[Br2]

= rate constant

= 3.50 x 10-3 s-1

Y=mX+b

Constant = Slope = k

How do the RATES

change with [Br2]?

Rate = k [Br2]1 this is called a 1st order reaction

Rate LawThe rate law for a

reaction tells us how rate varies with

concentration of the reactants.

UA GenChem

SOLUTION:

For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.(a) 2NO(g) + O2(g) 2NO2(g); rate = k[NO]2[O2](b)CH3CHO(g) CH4(g) + CO(g);

rate = k[CH3CHO]3/2

(c) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) +

2H2O(l); rate = k[H2O2][I-]

(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.

Your TurnYour Turn

UA GenChem

SOLUTION:

For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.(a) 2NO(g) + O2(g) 2NO2(g); rate = k[NO]2[O2](b)CH3CHO(g) CH4(g) + CO(g);

rate = k[CH3CHO]3/2

(c) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) +

2H2O(l); rate = k[H2O2][I-]

(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.

Your TurnYour Turn

UA GenChem

Consider the reaction:

2A + 2B + C → 2D + E

If the rate law for this reaction is Rate = k[A][B]2

What is the effect on the rate if the concentration of A is doubled?

1. No change

2. x23. x44. x85. x16

UA GenChem

What is the effect on the rate if the concentration of A is doubled?

Rate = k[A][B]2

1. No change2. x23. x44. x85. x16

UA GenChem

What is the effect on the rate if the concentrations of A, B & C are all

doubled?

Rate = k[A][B]2

1. No change2. x23. x44. x85. x16

UA GenChem

Measured Reaction Rates

• The rate of reaction changes with time, as the concentrations of reactants and products change.

• The rate constant, but not the rate, is independent of time.

• The instantaneous rate is the rate at any given time and is equal to the slope of a line tangent to the plot of concentration vs. time.

UA GenChem

Determining Reaction Orders

• The rate law for a given reaction must be determined experimentally.

• The order must be found for each species involved.

• The two major methods are1. Initial Rate Method

2. Integrated Rate Law Method

UA GenChem

Rate LawsIn general, forIn general, for

a A + b B a A + b B x X x X with a catalyst C with a catalyst C

Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp

The The exponentsexponents m, n, and p are the m, n, and p are the reaction order with respect each reaction order with respect each

reactantreactant

OverallOverall reaction order= m + n + p reaction order= m + n + p

These numbers can be 0, 1, 2 or These numbers can be 0, 1, 2 or fractions and fractions and must be determined by must be determined by

experimentexperiment!!

They are They are NOTNOT related to the related to the stoichiometric stoichiometric coefficientscoefficients a,b,x a,b,x

UA GenChem

Experiment

Initial Reactant Concentrations (mol/L) Initial Rate

(mol/L*s)

1

2

3

4

5

O2 NO

1.10x10-2 1.30x10-2 3.21x10-3

1.10x10-2 3.90x10-2 28.8x10-3

2.20x10-2

1.10x10-2

3.30x10-2

1.30x10-2

2.60x10-2

1.30x10-2

6.40x10-3

12.8x10-3

9.60x10-3

Your TurnYour Turn

UA GenChem

Determining Reaction Orders Using Initial Rates

Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate.

O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n

Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant.

k [O2]2m[NO]2

n

k [O2]1m[NO]1

n=

rate2

rate1 =

[O2]2m

[O2]1m

=

6.40x10-3mol/L*s

3.21x10-3mol/L*s

[O2]2

[O2]1

m

=1.10x10-2mol/L

2.20x10-2mol/L m; 2 = 2m m = 1

Do a similar calculation for the other reactant(s).

UA GenChem

Experiment

Initial Reactant Concentrations (mol/L) Initial Rate

(mol/L*s)

1

2

3

4

5

O2 NO

1.10x10-2 1.30x10-2 3.21x10-3

1.10x10-2 3.90x10-2 28.8x10-3

2.20x10-2

1.10x10-2

3.30x10-2

1.30x10-2

2.60x10-2

1.30x10-2

6.40x10-3

12.8x10-3

9.60x10-3

Your TurnYour Turn

UA GenChem

Expected

Solution:

Another ExampleConsider this reaction inside a car engine:

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

How can we use this data to determine the rate law and the rate orders with respect to

each reactant?

Exp.

Initial Rate (mol/L*s)

Initial [NO2] Initial [CO]

123

0.00500.0800.0050

0.10

0.100.400.10

0.100.20

rate = k [NO2]m[CO]n

UA GenChem

0.080

0.0050

rate 2

rate 1

[NO2] 2

[NO2] 1

m=

k [NO2]m2[CO]n

2

k [NO2]m1 [CO]n

1

=

0.40

0.10=

m

16 = 4m and m = 2

The reaction is

2nd order in NO2.

First, choose two experiments in which [CO]

remains constant and the [NO2] varies.

Exp.



123

0.00500.0800.0050

0.10

0.100.400.10

0.100.20

One Variable at a Time

UA GenChem

Now, choose two experiments in which

[NO2] remains constant and the [CO]

varies.

Exp.



123

0.00500.0800.0050

0.10

0.100.400.10

0.100.20

One Variable at a Time

k [NO2]m3[CO]n

3

k [NO2]m1 [CO]n

1

[CO] 3

[CO] 1

n=

rate 3

rate 1=

0.0050

0.0050=

0.20

0.10

n; 1 = 2n and n = 0

The reaction is zero order in CO.

rate = k [NO2]2[CO]0 = k [NO2]2

Overall order

UA GenChem

Your TurnExp.



123

0.00500.0800.0050

0.10

0.100.400.10

0.100.20

rate = k [NO2]2

What is the value of the constant, k, in this reaction?

UA GenChem

What are the UNITS of k in this reaction?

1. s-1

2. L/(mol s)3. mol/(L s)4. L2/(mol2

s)

UA GenChem

Overall Reaction Order

Units of k (t in seconds)

1 1/s (or s-1)

0 mol/(L*s) (or mol L-1 s-1) M/s

2 L/(mol*s) (or L mol -1 s-1) 1/Ms

3 L2 /(mol2 *s) (or L2 mol-2 s-1) 1/M2s

Units: Rate Constant k

UA GenChem

Concentration-Time Concentration-Time RelationsRelations

Consider a Consider a FIRST ORDER REACTION:FIRST ORDER REACTION: AA BB

The rate law isThe rate law is Rate= k[A] Rate= k[A]

RATE= RATE= -d[A]/dt-d[A]/dt= = k[A]k[A]

How can we predict the concentration of reactants at any moment in time?

How do you find How do you find [A][A] = = f(t)f(t)??

UA GenChem

First-Order Rate Law

A Bk

k[A]dt

d[A]

[A]

[A]

t

00

kdt[A]d[A]

or kdt[A]d[A]

kt0

0

e[A] [A] or kt[A][A]

ln

UA GenChem

Integrating Integrating - (d[A]/d t)- (d[A]/d t) = = k [A],k [A], we we getget

[A] = - k tln[A]o

naturallogarithm [A] at time = 0

[A] = - k tln[A]o

naturallogarithm [A] at time = 0

[A] / [A][A] / [A]00 =fraction remaining after =fraction remaining after time t has elapsed.time t has elapsed.

[A] / [A][A] / [A]00 =fraction remaining after =fraction remaining after time t has elapsed.time t has elapsed.

Called the Called the integratedintegrated first-order rate first-order rate lawlaw

Integrated Rate Laws

UA GenChem

Do you have to memorize all of the kinetics equations?

1. Yes2. No

UA GenChem

• Plot the experimental ln[A] vs. time• If the graph is linear, the reaction is

first-order.• The slope of this line = -k

y = mx + b

First-Order Rate Law

ln [A] = -kt + ln [A]0

UA GenChem

Examining a reaction

Consider the process in which methyl isonitrile is converted to acetonitrile.

CH3NC CH3CN

This reaction is suspected to be first order…

Design an experiment to determine the order of the reaction.

UA GenChem

First-Order Processes

• When ln P is plotted as a function of time, a straight line results.– The process is first-order.– k is the negative slope: 5.1 10-5 s-1.

UA GenChem

First-Order Reactions

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

ln[A]0

[A]0/2

k=t½

ln2k

=0.693

k=

What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln2k

=0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order?

units of k (s-1)

UA GenChem

A plot of [N2O5] vs. time for three half-lives

UA GenChem

Second-Order Rate Law2A Bk

2eff

2eff [A]k

dtd[A]

or [A]kR

[A]

[A]

t

0eff2eff2

0

dtk[A]d[A]

or dtk[A]d[A]

0effeff

0 [A]1

tk[A]1

or tk[A]

1[A]1

UA GenChem

• Plot the experimental 1/[A] vs. time• If the graph is linear, the reaction is

second-order.• The rate constant is k = slope

y = mx + b

0][

1

][

1

Ak t

A

Second-Order Rate Law

UA GenChem

Second-Order Half Life

0[A]1

kt[A]

12

1

21

0021 [A]

1kt

[A]1

21

0k[A]1

t2

1

UA GenChem

Zero-Order Reactions

A product rate = -D[A]Dt

rate = k [A]0 = k

k = rate[A]0

= M/sD[A]Dt

= k-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

t½ = t when [A] = [A]0/2

t½ =[A]0

2k

[A] = [A]0 - kt

UA GenChem

Consider the following data for the reaction:

N2O5 → 2NO2 + ½ O2

Graphical Analysis

UA GenChem

Summary

ln[A]t = -kt + ln[A]0

1/[A]t = kt + 1/[A]0

[A]t = -kt + [A]0

UA GenChem

At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction,

with the very high rate constant of 87 s-

1, to two molecules of ethylene (C2H4).

If the initial C4H8 concentration is 2.00M, what is the concentration after

0.010 s?

_____________ M

UA GenChem

At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4).

(b) What fraction of C4H8 has decomposed in this time?

UA GenChem

SOLUTION:

ln2.00

[C4H8] = -(87s-1)(0.010s)

[C4H8] = 0.84 mol/L

ln[C4H8]

0

[C4H8]t= -kt

(a)

(b) [C4H8]0 - [C4H8]t

[C4H8]0

=

2.00M - 0.84M 2.00M = 0.58

UA GenChem

PLAN:

Cyclopropane is the smallest cyclic hydrocarbon. Because its 600 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 10000C via the following first-order reaction: CH2

H2C CH2(g)

H3C CH CH2 (g)

The rate constant is 9.2s-1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value?

One-quarter of the initial value means two half-lives have passed.

Use t1/2 = ln2/k to find the half-life

UA GenChem

SOLUTION:

t1/2 = 0.693/9.2s-1 = 0.075s(a)

2 t1/2 = 2(0.075s) = 0.15 s(b)

UA GenChem

0th Order 1st Order 2nd Order

Plot for straight line

Slope, y-intercept

Half-life

Rate law

rate = k rate = k [A] rate = k [A]2

Units for k mol/(L*s) 1/s L/(mol*s)

Int. rate law (straight-line form)

[A]t =

-k t + [A]0

ln[A]t =

-k t + ln[A]0

1/[A]t =

k t + 1/[A]0

[A]t vs. t ln[A]t vs. t 1/[A]t vs t

-k, [A]0 -k, ln[A]0k, 1/[A]0

[A]0/(2k) (ln 2)/k 1/(k [A]0)

Overview

UA GenChem

Summary Activity:N2O5(g) NO3(g) + NO2(g)

Consider the following graphs and reaction data to predict the concentration of N2O5 at 275 sec.

Time (s)

[N2O5]

0 1.000

25 0.822

50 0.677

75 0.557

100 0.458

125 0.377

150 0.310

175 0.255

200 0.210

UA GenChem

What is the concentration of N2O5 after 275 sec?

______________ mol/L

Technology

Lect w2 152 - rate laws_alg