chapter 13 chemical kinetics. the study of reaction rate is called chemical kinetics. reaction rate...

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.

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

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

−

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

1

2

Δn

Δn

V

nΔ

V

nΔ

]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)

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

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)

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

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.

−

All the rates in this table are average rates.

− = r(NO2)

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

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

Reaction rate is a function of time

Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst

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

(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]

aA + bB cC +dD

r = k [A]m [B]n

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

Units

(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]

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

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?

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

(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



(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

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



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

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)

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

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

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.

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

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.

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

[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

Read a similar Example 13.3 on page 575

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

Practice on Example 13.4 on page 576 and check your answer

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

The half-life for first order reaction:

kt

2ln2/1

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

[N2O5] (M) Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

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

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

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

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

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


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

The half-life for second order reaction:

02/1 ]A[

1

kt

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

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

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)

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]


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

The half-life for zero order reaction:

kt

2

]A[ 02/1

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

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

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




Temperature

Catalyst

r = k [A]m [B]n

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

aA + bB cC +dD

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

RTEazezfk /

Another factor needs to be taken into account

molecular orientation

exptkk

Several Possible Orientations for a Collision Between Two BrNO Molecules

BrNO + BrNO 2NO +Br2

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 ↓

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

Try Example 13.8 on page 585 and check your answers

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

(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

Chemical Equation

Reactants Products

Reaction Mechanism

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

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

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

1 ― unimolecular2 ― bimolecular3 ― termolecular

What determines the rate of an overall reaction?

Team A

Team B

Slowest step: rate determining step

Ep

Reaction progress

Reactants

Products

∆E

intermediate

Which step is the rate determining step?

exothermic orendothermic?

Ea, 2

Ea, 1

2nd




Temperature

Catalyst

Catalyst is a substance that speeds up a reaction without

being consumed itself.

Catalyst changes the reaction mechanism through a lower

activation energy pathway.

Energy Plots for a Given Reaction

Ea

Ea

catalyst

homogeneous

heterogeneous

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

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

Heterogeneous catalyst: different phase from reactants.

These Cookies Contain Partially Hydrogenated Vegetable Oil

C = C + H2 − C − C −

− −−−

milk sugar = lactose

chapter 13 chemical kinetics. the study of reaction rate is called chemical kinetics. reaction rate...

Documents