Top Banner
Chapter 13 Chemical Kinetics
81

Chapter 13 Chemical Kinetics

Jan 16, 2016

Download

Documents

Holland

Chapter 13 Chemical Kinetics. The study of reaction rate is called chemical kinetics. Reaction rate is measured by the change of concentration (molarity) of reactants or products per unit time. Molarity of A: [A], e.g. [NO 2 ]: molarity of NO 2. −. −. reactants  products,. - PowerPoint PPT Presentation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Chapter 13 Chemical Kinetics

Chapter 13

Chemical Kinetics

Page 2: Chapter 13 Chemical Kinetics

The study of reaction rate is called chemical kinetics.

Reaction rate is measured by the change of

concentration (molarity) of reactants or products

per unit time.

Page 3: Chapter 13 Chemical Kinetics

Δt

reactantΔreactantreactantr

if

if

tt

Unit:

Molarity of A: [A], e.g. [NO2]: molarity of NO2

Δt

productΔproductproductr OR

if

if

tt

reactants products,

− −

[reactant]↓ and [product]↑

mol·L−1·s−1 ≡ M·s−1

Page 4: Chapter 13 Chemical Kinetics

5 122

NO 0.0079 M 0.0100 Mr NO 4.2 10 M s

50 s 0 s-

t

0 s → 50 s:

5 122

NO 0.0065 M 0.0079 Mr NO 2.8 10 M s

100 s 50 s-

t

50 s → 100 s:

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

− = r(NO2)

rate is a function of time

Page 5: Chapter 13 Chemical Kinetics

5 1NO 0.0035 M 0.0021 Mr NO 2.8 10 M s

100 s 50 s-

t

5 122

O 0.0018 M 0.0011 Mr O 1.4 10 M s

100 s 50 s-

t

50 s → 100 s:

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

50 s → 100 s:

= r(NO2)

rate is a function of time

Page 6: Chapter 13 Chemical Kinetics

1

2

Δn

Δn

V

V

]Δ[O

]Δ[NO

Δt]Δ[O

Δt]Δ[NO

)r(O

)r(NO

2

2

2

2

O

NO

O

NO

2

2

2

2

2

2

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

Page 7: Chapter 13 Chemical Kinetics

a A + b B c C + d D

r(A) a

r(B) b

r(A) a

r(C) c

d

r(D)

c

r(C)

b

r(B)

a

r(A) r =

r does not depend upon the choice of species

r(A) r(B) ,

a b

r(A) r(C)

a c

Page 8: Chapter 13 Chemical Kinetics

r(N2O5) = 4.2 x 10−7 M·s−1

What are the rates of appearance of NO2 and O2 ?

2N2O5(g) 4NO2(g) + O2(g)

Page 9: Chapter 13 Chemical Kinetics

Consider the following balanced chemical equation:

H2O2(aq) + 3 I–(aq) + 2 H+(aq) I3–(aq) + 2 H2O(l)

In the first 10.0 seconds of the reaction, the concentration of I– dropped from 1.000 M to 0.868 M.

(a)Calculate the average rate of this reaction in this time interval.

(b) Predict the rate of change in the concentration of H+ (that is, [H+]/t) during this time interval.

Example 13.1. page 567

Page 10: Chapter 13 Chemical Kinetics

Consider the general reaction aA + bB cC and thefollowing average rate data over some time period Δt:

1Δ[A]0.0080 M s

Δt 1Δ[B]

0.0120 M sΔt

1Δ[C]0.0160 M s

Δt

Determine a set of possible coefficients to balance thisgeneral reaction.

Page 11: Chapter 13 Chemical Kinetics

All the rates in this table are average rates.

− = r(NO2)

Page 12: Chapter 13 Chemical Kinetics

0 mileBarnesville

16 miles Griffin

56 miles Atlanta

2:00 pm 2:16 pm 3:16 pm

Average Speed from B to G = 16 miles ÷ 16 min = 1.0 mile/min

Average Speed from G to A = 40 miles ÷ 60 min = 0.7 mile/min

Δl

Δt

0limt

l d l

t d t

Instantaneous speed at green spot

Instantaneous speed contains more information

Page 13: Chapter 13 Chemical Kinetics

l

t

Δl

Δt

t

l

Instantaneous speed at the red point = slope of the red solid line =

Barnesville0 mile 0 min

Atlanta56 miles

Griffin20 miles

16 min 76 min

Page 14: Chapter 13 Chemical Kinetics

Reaction rate is a function of time

Page 15: Chapter 13 Chemical Kinetics

Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst

Page 16: Chapter 13 Chemical Kinetics

a A + b B c C + d D

r = k [A]m [B]n

Differential rate law: how r depends on concentrations

m, n: reaction order, mth order for A, nth order for B(m+n): overall reaction order

m and n must be measured from experiments. They canbe different from the stoichiometry.

k: rate constant: depends on temperature, but not concentrations

1 Δ[A] 1 d[A]

a Δt a dt

Page 17: Chapter 13 Chemical Kinetics

(1) 2N2O5(g) 4NO2(g) + O2(g) r = k[N2O5]

(2) CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) r = k[CHCl3][Cl2]1/2

(3) H2(g) + I2(g) 2HI(g) r = k[H2][I2]

Page 18: Chapter 13 Chemical Kinetics

aA + bB cC +dD

r = k [A]m [B]n

overall reaction order (m+n) unit of k ⇌

Units

Page 19: Chapter 13 Chemical Kinetics

(1) 2N2O5(g) 4NO2(g) + O2(g) r = k[N2O5]

(2) CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) r = k[CHCl3][Cl2]1/2

(3) H2(g) + I2(g) 2HI(g) r = k[H2][I2]

Page 20: Chapter 13 Chemical Kinetics

Experiment Number

Initial [NH4+] (M) Initial [NO2

−] (M) Initial Rate (M·s−1)

1 0.100 0.0050 1.35 x 10−7

2 0.100 0.010 2.70 x 10−7

3 0.200 0.010 5.40 x 10−7

r = k [NH4+]m [NO2

−]n

NH4+ (aq) + NO2

− (aq) N2(g) + 2H2O(l)

How to find the rate law by experiment: method of initial rates

Page 21: Chapter 13 Chemical Kinetics

A very common method to investigate how each factoraffects the whole system:

Change one thing at a time while keep the others constant.

z = f (x,y)

x

z

x

z

y

How does the change of x affect z?

y

z

y

z

x

How does the change of y affect z?

Page 22: Chapter 13 Chemical Kinetics

Experiment Number

Initial [NH4+] (M) Initial [NO2

−] (M) Initial Rate (M·s−1)

1 0.100 0.0050 1.35 x 10−7

2 0.100 0.010 2.70 x 10−7

3 0.200 0.010 5.40 x 10−7

r = k [NH4+]m [NO2

−]n

NH4+ (aq) + NO2

− (aq) N2(g) + 2H2O(l)

How to find the rate law by experiment: method of initial rates

Page 23: Chapter 13 Chemical Kinetics

(a) Determine the differential rate law

(b) Calculate the rate constant

(c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M

2NO(g) + 2H2(g) N2(g) + 2H2O(g)

Experiment Number

Initial [NO] (M) Initial [H2] (M) Initial Rate (M·s−1)

1 0.10 0.10 1.23 x 10-3

2 0.10 0.20 2.46 x 10-3

3 0.20 0.10 4.92 x 10-3

Page 24: Chapter 13 Chemical Kinetics

(a) Determine the differential rate law

(b) Calculate the rate constant

(c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M

2NO(g) + 2H2(g) N2(g) + 2H2O(g)

Experiment Number

Initial [NO] (M) Initial [H2] (M) Initial Rate (M·s−1)

1 0.10 0.10 1.23 x 10-3

2 0.10 0.30 3.69 x 10-3

3 0.30 0.10 1.11 x 10-2

again

Page 25: Chapter 13 Chemical Kinetics

Experiment Number

Initial [A] (M) Initial [B] (M) Initial Rate (M·s−1)

1 0.100 0.100 4.0 x 10−5

2 0.100 0.200 4.0 x 10−5

3 0.200 0.100 1.6 x 10−4

A + B C

(a) Determine the differential rate law

(b) Calculate the rate constant

(c) Calculate the rate when [A] = 0.050 M and [B] = 0.100 M

Page 26: Chapter 13 Chemical Kinetics

From the data, determine:

(a) the rate law for the reaction

(b) the rate constant (k) for the reaction

EXAMPLE 13.2 Determining the Order and Rate Constant of a Reaction

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

Page 27: Chapter 13 Chemical Kinetics

Use the data in table to determine

1) The orders for all three reactants 2) The overall reaction order3) The value of the rate constant

Page 28: Chapter 13 Chemical Kinetics

r = k [BrO3−]m [Br−]n [H+]p

r1 = k (0.10 M)m (0.10 M)n (0.10 M)p = 8.0 x 10−4 M · s−1

r2 = k (0.20 M)m (0.10 M)n (0.10 M)p = 1.6 x 10−3 M · s−1

r3 = k (0.20 M)m (0.20 M)n (0.10 M)p = 3.2 x 10−3 M · s−1

r4 = k (0.10 M)m (0.10 M)n (0.20 M)p = 3.2 x 10−3 M · s−1

Page 29: Chapter 13 Chemical Kinetics

one quiz after lab

Relationship among reaction rates as expressedby different species.

d

r(D)

c

r(C)

b

r(B)

a

r(A) r =

overall reaction order (m+n) unit of k ⇌

Method of initial rates:

table of experimental data rate order, k, rate at otherconcentrations.

Page 30: Chapter 13 Chemical Kinetics

a A + b B c C + d D

r = k [A]m [B]n

Differential rate law: how r depends on concentrations

Differential rate law: differential equation

[ ] [ ][ ] [ ]m nA d A

r k A Ba t a dt

How concentration changes as a function of time integrated rate law

Page 31: Chapter 13 Chemical Kinetics

A Products

Δ[A] d[A]r = r(A) k[A]

Δt dt

First order reaction differential rate law:

First order reaction integrated rate law:

0ln[A]ktln[A] kt[A]

[A]ln 0 or

integrated rate law: how concentration changes as a function of time.

Page 32: Chapter 13 Chemical Kinetics

0ln[A] kt ln[A]

Plot ln[A] vs. t gives a straight line

Slope = −k, intercept = ln[A]0

First order reaction integrated rate law:

y = mx + b

[A] is the molarity of A at t

Page 33: Chapter 13 Chemical Kinetics

[N2O5] (M) Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

N2O5(g) 2NO2(g) + ½ O2(g)

Use these data, verify that the rate law is first order in N2O5,and calculate the rate constant.

k = 6.93 x 10−3 s−1

Page 34: Chapter 13 Chemical Kinetics

Read a similar Example 13.3 on page 575

Page 35: Chapter 13 Chemical Kinetics

Using the data given in the previous example, calculate [N2O5]

at 150 s after the start of the reaction.

[N2O5] (M) Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

0.0354 M

Page 36: Chapter 13 Chemical Kinetics

Practice on Example 13.4 on page 576 and check your answer

Page 37: Chapter 13 Chemical Kinetics

The half-life of a reaction, t1/2, is the time required for a

reactant to reach one-half of its initial concentration.

[N2O5] (M) Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

Page 38: Chapter 13 Chemical Kinetics

The half-life for first order reaction:

kt

2ln2/1

The half-life for first order reaction does NOT depend onconcentration.

Page 39: Chapter 13 Chemical Kinetics

[N2O5] (M) Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

Page 40: Chapter 13 Chemical Kinetics

A Plot of [N2O5] versus Time for the Decomposition Reaction of N2O5

kt2 5 2 5 0[N O ] [N O ] e

Page 41: Chapter 13 Chemical Kinetics

A certain first order reaction has a half-life of 20.0 s.

a)Calculate the rate constant for this reaction.

b)How much time is required for this reaction to be 75 %

complete?

kt

2ln2/1 kt

[A]

[A]ln 0

a) k = 0.0347 s−1 b) k = 40.0 s

Page 42: Chapter 13 Chemical Kinetics

Try Example 13.6 and For Practice 13.6on page 579 and check your answers

Page 43: Chapter 13 Chemical Kinetics

A Products

2k[A]dt

d[A]

Δt

Δ[A]r(A)

Second order reaction differential rate law:

Second order reaction integrated rate law:

0[A]

1kt

[A]

1

integrated rate law: how concentration changes as a function of time.

Page 44: Chapter 13 Chemical Kinetics

Plot 1/[A] vs. t gives a straight line

Slope = k, intercept = 1/[A]0

Second order reaction integrated rate law:

0[A]

1kt

[A]

1

y = mx + b

Page 45: Chapter 13 Chemical Kinetics

The half-life for second order reaction:

02/1 ]A[

1

kt

The half-life for second order reaction depends on initial concentration.

Page 46: Chapter 13 Chemical Kinetics

A certain reaction has the following general form: A BAt a particular temperature [A]0 = 2.80 x 10−3 M, concentration versus time data were collected for this reaction,and a plot of 1/[A] versus time resulted in a straight line with aslope value of 3.60 x 10−2 M−1·s−1

a)Determine the (differential) rate law, the integrated rate law, and the value of the rate constant.

b) Calculate the half-life for this reaction.

c) How much time is required for the concentration of A to decrease to 7.00 x 10−4 M ?

0[A]

1kt

[A]

1

02/1 ]A[

1

kt

Page 47: Chapter 13 Chemical Kinetics

For first order reaction, show that

from the integrated rate law

kt

2ln2/1

0ln[A] kt ln[A]

(show your work, do not copy the question)

Page 48: Chapter 13 Chemical Kinetics

A Products

kk[A]dt

d[A]

Δt

Δ[A]r(A) 0

Zero order reaction differential rate law:

Zero order reaction integrated rate law:

0[A]kt[A]

integrated rate law: how concentration changes as a function of time.

Page 49: Chapter 13 Chemical Kinetics

Plot [A] vs. t gives a straight line

Slope = −k, intercept = [A]0

Zero order reaction integrated rate law:

0[A]kt[A]

y = mx + b

Page 50: Chapter 13 Chemical Kinetics

The half-life for zero order reaction:

kt

2

]A[ 02/1

The half-life for zero order reaction depends on initial concentration.

Page 51: Chapter 13 Chemical Kinetics

The Decomposition Reaction 2N2O(g) 2N2 (g) + O2 (g) takes place on a Platinum Surface

Page 52: Chapter 13 Chemical Kinetics

The decomposition of ethanol on alumina surface C2H5OH(g) C2H4(g) + H2O(g) was studied at 600 K.Concentration versus time data were collected for this reaction, and a plot of [C2H5OH] versus time resulted in a straight line with a slope value of −4.00 x 10−5 M·s−1

a)Determine the (differential) rate law, the integrated rate law, and the value of the rate constant.

b) If the initial concentration of C2H5OH was 1.25 x 10−2 M, calculate the half-life of this reaction.

c) How much time is required for all the 1.25 x 10−2 M C2H5OH to decompose ? 0[A]kt[A]

kt

2

]A[ 02/1

Page 53: Chapter 13 Chemical Kinetics
Page 54: Chapter 13 Chemical Kinetics

Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst

Page 55: Chapter 13 Chemical Kinetics

r = k [A]m [B]n

k: rate constant: depends on temperature, but not concentrations

aA + bB cC +dD

Page 56: Chapter 13 Chemical Kinetics

A Plot Showing the Exponential Dependence of the Rate Constant on Absolute Temperature

Collision model: molecules must collide to react.

T ↑ v ↑ kinetic energy = ½ mv2 ↑

Not all collisions lead to products

Page 57: Chapter 13 Chemical Kinetics

Prentice Hall © 2003 Chapter 14

RTEaef /Fraction of molecules whose Ek > Ea is

T ↑ f ↑ Ea ↑ f ↓

activation energy

Page 58: Chapter 13 Chemical Kinetics

RTEazezfk /

Another factor needs to be taken into account

molecular orientation

exptkk

Page 59: Chapter 13 Chemical Kinetics

Several Possible Orientations for a Collision Between Two BrNO Molecules

BrNO + BrNO 2NO +Br2

Page 60: Chapter 13 Chemical Kinetics

212

1 11ln

TTR

E

k

k a

Arrhenius equation

RTEaAek /

RTEapzek / steric factor p ≤ 1

A: frequency factor How k depends on T

T ↑ k ↑ Ea ↑ k ↓

Page 61: Chapter 13 Chemical Kinetics

At 550 °C the rate constant for

CH4(g) + 2S2(g) CS2(g) + 2H2S(g)

is 1.1 M·s−1, and at 625 °C the rate constant is

6.4 M·s−1. Using these value, calculate Ea for this reaction.

212

1 11ln

TTR

E

k

k a

Ea = 1.4 x 105 J/mol

Page 62: Chapter 13 Chemical Kinetics

Try Example 13.8 on page 585 and check your answers

Page 63: Chapter 13 Chemical Kinetics

The ball can climb over the hill only if its kinetic energy is greaterthan Ep = mgh, where m is the mass of the ball, h is the heightof the hill, and g is gravitational acceleration.

h

Page 64: Chapter 13 Chemical Kinetics

(a) The Change in Potential as a Function of Reaction Progress (b) A Molecular Representation of the Reaction

BrNO + BrNO 2NO +Br2

O N Br

exothermic reaction

Page 65: Chapter 13 Chemical Kinetics

Chemical Equation

Reactants Products

Reaction Mechanism

Page 66: Chapter 13 Chemical Kinetics

NO2 + CO NO + CO2

Step 1: NO2 + NO2 NO3 + NO Step 2: NO3 + CO NO2 + CO2 +

NO2 + CO NO + CO2

NO3: intermediate, does notappear in overall reaction

Page 67: Chapter 13 Chemical Kinetics

Whole process is called the reaction mechanism.

Each single step is called an elementary reaction/step.An elementary reaction is a single collision.

NO2 NO2 NO3 NO

NO3 CO NO2 CO2

Page 68: Chapter 13 Chemical Kinetics

For an elementary reaction

aA + bB cC + dD

r = k[A]a[B]b

Not for overall reaction!

The number of molecules that react in an elementary reaction is called the molecularity of that that elementary reaction (not applicable to overall reaction).

Overall reaction: r = k[A]m[B]n , find m and n by experiments

Page 69: Chapter 13 Chemical Kinetics

1 ― unimolecular2 ― bimolecular3 ― termolecular

Page 70: Chapter 13 Chemical Kinetics

What determines the rate of an overall reaction?

Page 71: Chapter 13 Chemical Kinetics

Team A

Team B

Slowest step: rate determining step

Page 72: Chapter 13 Chemical Kinetics

Ep

Reaction progress

Reactants

Products

∆E

intermediate

Which step is the rate determining step?

exothermic orendothermic?

Ea, 2

Ea, 1

2nd

Page 73: Chapter 13 Chemical Kinetics

Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst

Page 74: Chapter 13 Chemical Kinetics

Catalyst is a substance that speeds up a reaction without

being consumed itself.

Catalyst changes the reaction mechanism through a lower

activation energy pathway.

Page 75: Chapter 13 Chemical Kinetics

Energy Plots for a Given Reaction

Ea

Ea

Page 76: Chapter 13 Chemical Kinetics

catalyst

homogeneous

heterogeneous

Page 77: Chapter 13 Chemical Kinetics

Homogeneous catalyst: same phase as reactants.

3O2(g) 2O3(g)

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

2NO2(g) 2NO(g) + 2O(g)

2O2(g) + 2O(g) 2O3(g)

light

+

3O2(g) 2O3(g)

NO(g): catalyst; NO2(g), O(g): intermediates

Page 78: Chapter 13 Chemical Kinetics

The Decomposition Reaction 2N2O(g) 2N2 (g) + O2 (g) takes place on a Platinum Surface

Heterogeneous catalyst: different phase from reactants.

Page 79: Chapter 13 Chemical Kinetics

These Cookies Contain Partially Hydrogenated Vegetable Oil

C = C + H2 − C − C −

− −−−

Page 80: Chapter 13 Chemical Kinetics

milk sugar = lactose

Page 81: Chapter 13 Chemical Kinetics