chem 4a: general chemistry with quantitative...

OverviewReaction Rates

Rates and Time

Initial Rates

First-Order Reactions

Half-life of First Order Reactions

Radioactive Decay

Carbon-14 Dating

Second-Order Reactions

Half-life of Second Order Reactions

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Slow vs. fast chemical reactions

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Slow vs. fast chemical reactions

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Microscopic View

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Ozone_Destruction.movMedia File (video/quicktime)

Rate of Reaction

Average rate =change in concentration

time interval

Average rate = [reactant ]final [reactant ]initial

tfinal tinitial

Average rate =[product ]final [product ]initial

tfinal tinitial

Average rate = [reactant ]t

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Rate of reactionFor the reaction

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

Average ratebetween 100 and200 s

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Rate of reaction

For the reaction


Ave. rate =

(0.00596 0.00846) mol L1

(200 100) s= 2.5 105 mol L1 s1

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Rate of reaction

For the reaction


Instantaneous rate = limt0

[reactant ]t

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Rate of reactionFor the reaction


Instantaneousrate at t=0 and att=200- Slope of thetangent at thosepoints

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Concentrations and rateIn the case of a reaction like

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

we need to account for the stoichiometry of the reaction when settingup the expressions for the reaction rate.

So that we have the rate expressed as a unique positive number forthe reaction, we need to write:

rate = [O2]t

= 12

[NO]t

=12

[NO2]t 11 / 1

Rates and TimeFor the reaction

N2O5(g) 2NO2(g) +12

O2(g)

the followingdata wascollected at318 K

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N2O5(g) 2NO2(g) +12

O2(g)

Graphically, thismeans

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N2O5(g) 2NO2(g) +12

O2(g)

Observe that therate of change inthe concentrationchanges withtime

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N2O5(g) 2NO2(g) +12

O2(g)

Observe that therate of change inthe concentrationchanges withtime

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Rates and TimeAgain back to the model system

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

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Concentrations and Rate

Experiment Initial Reactant Initial rateconcentration molL1

(T = 600 K) [NO] [O2] molL1s11 1.30 102 1.10 102 3.21 1032 1.30 102 2.30 102 6.40 1033 2.60 102 1.10 102 12.8 1034 1.30 102 3.30 102 9.60 1035 3.90 102 1.10 102 28.8 103

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(T = 600 K) [NO] [O2] molL1s11 1.30 102 1.10 102 3.21 1032 1.30 102 2.30 102 6.40 1033 2.60 102 1.10 102 12.8 1034 1.30 102 3.30 102 9.60 1035 3.90 102 1.10 102 28.8 103

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(T = 600 K) [NO] [O2] molL1s11 1.30 102 1.10 102 3.21 1032 1.30 102 2.30 102 6.40 1033 2.60 102 1.10 102 12.8 1034 1.30 102 3.30 102 9.60 1035 3.90 102 1.10 102 28.8 103

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Concentrations and RatesModel system:

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

rate [O2]

rate [NO]2

rate [O2] [NO]2 or rate = k [O2] [NO]2

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Concentrations and RatesModel system:

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

rate [O2]

rate [NO]2

rate [O2] [NO]2 or rate = k [O2] [NO]2

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Concentrations and Rates

Once we know the rate law

rate = k [O2][NO]2

We can calculate the value of the rate constant from the experimentaldata

k =rate

[O2][NO]2

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Concentrations and Rates

Once we know the rate law

rate = k [O2][NO]2

We can calculate the value of the rate constant from the experimentaldata

k = 12.8103molL1s1

[1.10102 molL1][2.60102molL1]2

= 1.72 103 mol2 L2 s1

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General form of rate laws

rate = k [reactant1]x [reactant2]y ...

According to this rate law, the reaction described is:

order x with respect to reactant 1 order y with respect to reactant 2 overall order: x + y

The order of a reaction with respect to eachreactant must be determined EXPERIMENTALLY

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Calculation of k

Units of the rate constant:

k : mol(order1)L(order1)s1

So that when multiplied by molorder Lorder , we get

molorder(order1)L(order1)order s1

or molL1s1

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Examples of reaction order

rate = k[(CH3)3CBr]

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

CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) rate = k [CHCl3 ] [Cl2]12

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Exercise

For each of the following reactions, use the given rate law todetermine the reaction order with respect to each reactant and theoverall order

2H2O2(aq) + 3I(aq) + 2H+(aq) I3 (aq) + 2H2O(`)rate = k [H2O2] [I]

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

CH3CHO(g) CH4(g) + CO(g) rate = k [CH3CHO]32

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Integrated Rate Laws:First-Order Reactions

For the reaction2N2O5(g) 4NO2 + O2(g)

the rate can is describe by the rate law

d [N2O5]dt

= k [N2O5]

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that can be rearranged as

d [N2O5][N2O5]

= kdt

and integrated between the initial and final conditions

[N2O5][N2O5]0

d [N2O5][N2O5]

= t

0kdt

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to give

ln(

[N2O5][N2O5]0

)= kt

or

ln [N2O5] = kt + ln [N2O5]0

Integrated first order rate law30 / 1

An alternative form of the integrated rate lawWe can also write

[N2O5][N2O5]0

= ekt

or

[N2O5] = [N2O5]0 ekt

Notice that the fraction of N2O5 remaining after a time interval t isalways ekt

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Graphically:(integrated first order rate law)

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

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Half-life of First Order Reactions

The half-life for a first-order reaction is independent of the initialconcentration.

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Half-life of a first-order reaction

Half-life (t 12) is the time it takes for the concentration of a reactant to

drop to one-half of its initial value.

Fast reaction = short half-life

For a first-order process:

ln(

[N2O5][N2O5]0

)= kt

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Half-life of a first-order reaction

Half-life (t 12) is the time it takes for the concentration of a reactant to

drop to one-half of its initial value.

Fast reaction = short half-life

For a first-order process:

ln

[N2O5]02

[N2O5]0

= kt 12

t 12

=ln 2k

=0.693

k

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Radioactive decay

Different types of radioactive decay:

decay: 23894 Pu 23492 U +

42 He

decay: 9038Sr 9039 Y +

01 e

decay: 60m27 Co 6027 Co +

00

occur according to first-order rate laws.

In the expression of the rate law, we can substitute concentration fornumber of atoms in a sample, since they are directly proportional.

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Kinetics of Radioactive Decay

The kinetics are usually describedin terms of the half-life of theprocess:

NN0

= exp

( ln2

t 12

t

)

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Exercise

Strontium-90 is a radioactive by-product of nuclear reactors thatbehaves biologically like calcium, the element above it in Group 2.When 90Sr is ingested by mammals, it is found in their milk andeventually in the bones of those drinking the milk. If a sample of 90Srhas an activity of 1.2 1012 Bq, what are the activity and the fractionof nuclei that have decayed after 59 yr (t 1

2of 90Sr = 29 yr, 1 Bq = 1

disintegration per second)

A = A0ekt

where k = ln2t 12

= 0.69329 yr = 0.0239 yr1

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Exercise

Strontium-90 is a radioactive by-product of nuclear reactors thatbehaves biologically like calcium, the element above it in Group 2.When 90Sr is ingested by mammals, it is found in their milk andeventually in the bones of those drinking the milk. If a sample of 90Srhas an activity of 1.2 1012 Bq, what are the activity and the fractionof nuclei that have decayed after 59 yr (t 1

2of 90Sr = 29 yr, 1 Bq = 1

disintegration per second)

A = A0ekt

where k = ln2t 12

= 0.69329 yr = 0.0239 yr1

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Exercise

then

A = 1.2 1012 Bq e0.0239 yr159 yr = 2.9 1011 Bq

The fraction that has decayed is1.2 1012 Bq 2.9 1011 Bq

1.2 1012 Bq= 0.76

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Carbon-14 Dating14C is produced at a steady rate in the atmosphere through thereaction:

147 N +

10 n146 C +11 H

It enters the biomass as 14CO2 and H 14CO3 and reaches anequilibrium distribution in living organisms similar to that of theatmosphere 12C/ 14C = 1012 :1

After metabolism ceases, the amount of 14C will decrease accordingto the decay:

146 C 147 N +01 e

with t1412

C = 5730 yr

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Sample problem

A skull believed to belong to an ancient human being is found to havea carbon-14 decay rate of 0.075 Bqg1. If the modern day standardhas a specific activity of 0.225 Bqg1, calculate the age of the skull.

k = ln2t 12

= 0.6935730 yr = 1.209 104 yr1 ln AA0

= 1.098

t =1.098

1.209 104 yr1= 9.0 103 yr1

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Sample problem


k = ln2t 12

= 0.6935730 yr = 1.209 104 yr1 ln AA0

= 1.098

t =1.098

1.209 104 yr1= 9.0 103 yr1

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Sample problem


k = ln2t 12

= 0.6935730 yr = 1.209 104 yr1 ln AA0

= 1.098

t =1.098

1.209 104 yr1= 9.0 103 yr1

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Sample problem


k = ln2t 12

= 0.6935730 yr = 1.209 104 yr1 ln AA0

= 1.098

t =1.098

1.209 104 yr1= 9.0 103 yr1

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For the reaction2NO(g) 2NO(g) + O2(g)

the rate can be described by the rate law

d [NO2]dt

= k [NO2]2

This is a second-order kinetic law

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that can be rearranged as

[NO2]2d [NO2] = kdt

and integrated between the initial and final conditions

[NO2][NO2]0

[NO2]2d [NO2] = t

0kdt

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to give

1[NO2]

=1

[NO2]0+ k t

the integrated second-order rate law.

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Half-life of Second Order Reactions

Notice that in this case

t 12

=1

k [NO2]0

is a function of the initial concentration of the reactant, and thereforeuseless as a characterization of the reaction!

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Graphically:(integrated second order rate law)

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

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Reaction RatesRates and TimeInitial RatesFirst-Order ReactionsHalf-life of First Order ReactionsRadioactive DecayCarbon-14 DatingSecond-Order ReactionsHalf-life of Second Order Reactions

chem 4a: general chemistry with quantitative...

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