Kinetics “How fast a reaction occurs” Factors 1.Concentration 2.Temperature 3.Surface area...

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Kinetics

“How fast a reaction occurs”Factors1.Concentration2.Temperature3.Surface area4.Catalyst

Measuring the Rate of a Reaction

Disappearance of a reactant

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Draw a graph of the following data (x=time)

Time(s) [C4H9Cl] (M)

0.0 0.100050.0 0.0905

100.0 0.0820150.0 0.0741200.0 0.0671300.0 0.0549400.0 0.0448500.0 0.0368800.0 0.0200

Rate is the Slope y = m x + b[C4H9Cl]= (Rate)(t)

Rate = [C4H9Cl] t

Rate = d[C4H9Cl] dt

(Note how rate decreased with decreasing concentration)

The problems on this page refer to the data table for C4H9Cl

a. Using the data in the table, calculate the rate of disappearance of C4H9Cl from 0 to 50 s.

b. Using the data in the table, calculate the rate of disappearance of C4H9Cl from 50 to 100 s.

Now graph the datac. Using the graph, calculate the rate at 100 s

(-1.64 X 10-4 M/s)d. Using the graph, estimate the rate at 0 seconds

and 300 s. (-2.0 X 10-4 M/s, -1.1 X 10-4 M/s)

Rates and Stoichiometry

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Rate = -[C4H9Cl] = [C4H9OH]

Rate of disappearance = Rate of appearance(only for 1:1 stoichiometry)

2HI(g) H2(g) + I2(g)

Rate = -1 [HI] = [H2]= [I2]

2 t t t

aA + bB cC + dD

Rate = -1 [A] = -1 [B] = 1 [C] = 1 [D ]

a t b t c t d t

Stoichiometry: Ex 1

How is the rate of the disappearance of ozone related to the appearance of oxygen according to:

2O3(g) 3O2(g)

Rate = -1 [O3] = [O2]

2 t 3 t-[O3] = 2 [O2] or RateO3 = -2/3 RateO2

Stoichiometry: Ex 2

If the rate of appearance of oxygen at some instant is 6.0 X 10-5 M/s, calculate the rate of disappearance of ozone.

[O3] = -2 [O2]

t 3 t [O3] = -2 (6.0 X 10-5 M/s)

t 3 [O3] = -4.0 X 10-5 M/s

Stoichiometry: Ex 3

The rate of decomposition of N2O5 at a particular instant is 4.2 X 10-7 M/s. What is the rate of appearance of NO2 and O2?

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

Rate = -1 [N2O5] = 1 [NO2] = [O2]

2 t 4 t t-[N2O5] = 1 [NO2]

2 t 4 t

[NO2] = – 4 [N2O5]

t 2 t[NO2] = – 2 [N2O5] = (-2)(4.2 X 10-7 M/s)

t t[NO2] = 8.4 X 10-7 M/s

t-[N2O5] = [O2]

2 t t-[O2] = (½)(4.2 X 10-7 M/s) = 2.1 X 10-7 M/s

The Rate Law

aA + bB cC

Rate = k[A]m[B]n

k = Rate constant[A] and [B] = Initial concentrationsm and n = exponents

k, m, and n must be determined experimentally

Rate Law: Ex 1

What is the rate law for the following reaction, given the following rate data:

NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

Experiment Initial NH4+

Concentration (M)Initial NO2

Concentration (M)Initial Rate (M/s)

1 0.0100 0.200 5.4 X 10-7

2 0.0200 0.200 10.8 X 10-7

3 0.200 0.0202 10.8 X 10-7

4 0.200 0.0404 21.6 X 10-7

Rate = k[NH4+]m[NO2

Divide Experiment 1 and 2Rate2 = k[NH4

+]m[NO2-]n

Rate1 = k[NH4+]m[NO2

10.8 X 10-7 = k[0.0200]m[0.200]n

5.4 X 10-7 = k[0.0100]m[0.200]n

2= [0.0200]m m = 11 =[0.0100]m

Rate = k[NH4+]m[NO2

Divide Experiment 3 and 4Rate6 = k[NH4

+]m[NO2-]n

Rate5 = k[NH4+]m[NO2

21.6 X 10-7 = k[0.200]1[0.0404]n

10.8 X 10-7 = k[0.200]1[0.0202]n

2= [0.0404]n n = 11 =[0.0202]n

Calculating kRate = k[NH4

+]1[NO2-]1

Consider Experiment 1 (can pick anyone)Rate1 = k[NH4

+]1[NO2-]1

5.4 X 10-7 = k[0.0100]1[0.200]1

k = 2.7 X 10-4 M-1s-1

Rate Law: Ex 2

Calculate the rate law for the following general reaction: S + O2 SO2

Experiment [S] (M) [O2] (M) Initial Rate (M/s)

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 16.0 X 10-5

ANS: Rate = (4.0 X 10-3 M-1s-1)[S]2

What is the rate of the reaction when [S] = 0.050 M and [O2] = 0.100 M?

ANS: 1.0 X 10-5 M/s

Rate Law: Ex 3

A particular reaction varies with [H+] as follows. Calculate the Rate Law:

[H+] (M) Initial Rate (M/s)0.0500 6.4 X 10-7

0.100 3.2 X 10-7

0.200 1.6 X 10-7

Rate = 3.2 X 10-8 M2s-1[H+]-1

Rate Law: Ex 4

Calculate the rate if the [H+] = 0.400 M

Rate = 8 X 10-8 M/s

Calculate the rate law for the following reaction:

BrO3-(aq) + 5 Br-(aq) + 8 H+(aq) ----> 3 Br2(l) +

H2O(l)

Trial [BrO3-] [Br-] [H+] Initial

rate (mol/Ls)

1 0.10 0.10 0.10 8.0 x 10-

2 0.20 0.10 0.10 1.6 x 10-3

3 0.10 0.20 0.10 1.6 x 10-3

4 0.10 0.10 0.20 3.2 x 10-

Reaction Order

• m and n are called reaction orders• m + n + …. = overall reaction order• Units of k

– k must always have units that allow rate to have a unit of M/s

Reaction Order: Ex 1

What is the overall reaction order for the following reaction. What unit will the rate constant (k) have?

CHCl3 + Cl2 CCl4 + HCl

Rate=k[CHCl3][Cl2]1/2

Zeroth Order

• Reaction order can be zero• Concentration does not affect the rate• 2NH3 (g) N2(g) + 3H2(g)

Rate = k[NH3]0

Rate = k

Recognizing Zeroth Order

Concentration and Time: First Order

• First Order Reaction – reaction whose rate varies with the concentration of a single reactant to the first power

• Often the decay(chemical or nuclear) or decomposition of one substance

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

y = mx + b

First order eqns - linear ln[Conc] vs time graph

Example of a First Order Reaction

• First order in CH3NC

• ln[CH3NC]t = -kt + ln[CH3NC]0

• Can use this equation to calculate the concentration at any time.

This graph is not linear This graph is[A]t = [A]0e-kt ln[A]t = -kt + ln[A]0

First Order: Ex 1

The first order rate constant for the decomposition of an insecticide is 1.45 yr-1. If the starting concentration of the compound in a lake is 5.0 X 10-7 g/cm3, what will be the concentration the following year? In two years?

ln[A]t = -(1.45 yr-1 )(1 yr) + ln(5.0 X 10-7 g/cm3)

ln[A]t = -15.96

[A]t = e-15.96

[A]t = 1.2 X 10-7 g/cm3

or[A]t = [A]0e-kt

[A]t = (5.0 X 10-7 g/cm3)e-(1.45 yr-1 )(1yr)

[A]t = 1.2 X 10-7 g/cm3

([A]2yr = 2.8 X 10-8 g/cm3)

First Order: Ex 2

How long will it take the concentration to drop to 3.0 X 10-7 g/cm3?

First Order: Ex 3

The decomposition of dimethyl ether is a first order process with k=6.8 X10-4s-1. If the initial pressure is 135 torr, what is the partial pressure after 1420s?

(CH3)2O(g) CH4(g) + H2(g) + CO(g)

(ANS: 51 torr)

First Order Half-life

• Half-life – time required for the concentration of a reactant to drop to one half the initial value– Medicine in the body– Nuclear decay

• [A]t = ½[A]0

t½ = 0.693

Half-Life: Ex 1From the figure below, calculate the half-life and k

for the reaction of C4H9Cl with water.

Sodium-24 (used in some medical tests) has a half-life of 14.8 hours. What is the rate constant for its decay?

(ANS: 0.0468 hr-1)

Half-Life• Half-life - The time during which one-half of a

radioactive sample decays – Ranges from fraction of a second to billions of years.– You can’t hurry half-life.

Carbon-14 dating

• 14C in traces of once-living organisms– E.g., a 5730 years after death, only half of the 14C remains.

• ~ 50,000 years.• 15% margin of error• Mummies, the Dead Sea Scrolls, Shroud of Turin

Half-Life

Isotope Half-life

Uranium-238 4.51x109 years

Lead-210 20.4 years

Polonium-214 1.6x10-4 seconds

Half-life: Example 1

Carbon-14 has a half-life of 5730 years and is used to date artifacts. How much of a 26 g sample will exist after 3 half-lives? How long is that?

Tritium undergoes beta decay and has a half life of 12.33 years. How much of a 3.0 g sample of tritium remains after 24.66 years?

Cesium-137 has a half-life of 30 years. If you start with a 200 gram sample, and you now have 25 grams left, how much time has passed?

Rate Law: Ex 1

Uranium-238 has a half-life of 4.5 X 109 yr. If 1.000 mg of a 1.257 mg sample of uranium-238 remains, how old is the sample?

Rate Law: Ex 2

A wooden object is found to have a carbon-14 activity of 11.6 disintegrations per second. Fresh wood has 15.2 disintegrations per second. If the half-life of 14C is 5715 yr, how old is the object?

ANS: 2230 yr

After 2.00 yr, 0.953 g of a 1.000 g sample of strontium-90 remains.

a. Calculate k (0.0241 y-1)b.Calculate how much remains after 5.00 years.

(0.886 g)

A sample for medical imaging contains 18F (1/2 life = 110 minutes). What percentage of the original sample remains after 300 minutes?

ANS: 15.1%

Concentration and Time: Second Order

Second Order Reaction – reaction whose rate varies with the concentration of a single reactant to the second power

1 = kt + 1

[A]t [A]0

y = mx + b

2nd order eqns give a linear 1/[Conc] vs time graph

Second Order Half-life

t½ = 1

Second Order: Ex 1

The following data is for the decomposition of nitrogen dioxide: NO2(g) NO(g) + ½ O2(g)

Is the reaction first or second order?

Time (s) [NO2] (M)

0.0 0.0100050.0 0.00787

100.0 0.00649200.0 0.00481300.0 0.00380

Plot both ln[NO2] vs time and 1/[NO2] vs time

Which is linear?

Time (s) [NO2] (M) ln[NO2] 1/[NO2]

0.0 0.01000 -4.610 10050.0 0.00787 -4.945 127

100.0 0.00649 -5.038 154200.0 0.00481 -5.337 208300.0 0.00380 -5.573 263

Pick a convenient point 1 = kt + 1

[A]t [A]0

Calculate the slope.

Second Order: Ex 2

What is the half life for the above reaction?

t½ = 1

t½ ~ 1

(0.5 M-1s-1)(0.0100 M)t½ ~ 200 s

Time (s) [CH3NCl](M)

0 0.01652000 0.0115000 0.00598000 0.00314

12000 0.0013715000 0.00074

Is the following reaction first or second order? Draw two graphs, and calculate k.

Temperature and Rate

• Rate of most chemical reactions increases with temperature– Bread rising– Plants growing– Light sticks

• Rate = k[A]m[B]n…• k increases with temperature• Orders (m, n, etc..) do not change

Collision Model

For an effective collision to occur, molecules must collide:

1.Often (increases with temperature)2.Proper orientation3.Enough Activation Energy (increases with

temperature)

Proper vs. Improper Orientation

Insufficient vs. Sufficient Activation Energy

Activation Energy

• Energy needed to form activated complex or transition state

• Energy needed to start a reaction

a.Rank the following reactions from slowest to fastestb.Rank the reverse reactions from slowest to fastest

Reaction Mechanisms

Elementary Processes – only one step involved– Unimolecular – Bimolecular– Termolecular(rare)

Multistep Reactions

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

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

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

NO3(g) is an intermediate

Reaction Mechanisms: Ex 1

Ozone decomposes in the following steps. Describe the molecularity of each step and write the overall equation:

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

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

O3(g) O2(g) + O(g) unimolecular

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

2O3(g) 3O2(g)

O(g) is an intermediate

The following steps:Mo(CO)6 Mo(CO)5 + CO

Mo(CO)5 + P(CH3)3 Mo(CO)5P(CH3)3

Are proposed for the reaction:Mo(CO)6 + P(CH3)3 Mo(CO)5P(CH3)3 + CO

Is the proposed mechanism consistent with the reaction?

Rate Determining Step

• Consider two toll plazas

Plaza A limits the Plaza B limits theflow of traffic flow of traffic

St 1: NO2 + NO2 NO3 + NO (slow)

St 2: NO3 + CO NO2 + CO2 (fast)

NO2 + CO NO + CO2

Since Step 1 is much slower, it controls the rate (rate determining step)

Rate = k[NO2]2

(agrees with experiment)

Mechanisms with an Initial Fast Step

Proposed mechanismSt 1: NO + Br2 NOBr2 (fast)

St 2: NOBr2 + NO 2NOBr (slow)

Rate = k[NOBr2][NO]

• NOBr2 is intermediate

• Quickly breaks up into NO and Br2

Rate = k[NOBr2][NO]

Rate = k[NO][Br2][NO]

Rate = k[NO]2[Br2]

Show that the following mechanism is consistent with the rate law: Rate = k[NO]2[Br2]

St 1: NO + NO N2O2 (fast)

St 2: N2O2 + Br2 2NOBr (slow)

Catalyst

• Catalyst – substance that increases the speed of a reaction without being used up in the process

• Can increase the rate of reaction by thousands of times

• Lowers the activation energy• Ex:2KClO3(s) 2KCl(s) + 3O2(g)

• KClO3 is stable

• Adding MnO2 allows the reaction to go fast

• Red line shows pressure during a decomposition without a catalyst (HCOOH CO2 + H2)

• Blue line shows with a catalyst

Ex: 2H2O2(aq) 2H2O(l) + O2(g)

• Br-(aq) acts as a catalyst• Brown Br2 formed in between

• Br- regenerated at the end

Catalytic Convertor

• heterogeneous (different phase) catalyst• Use Platinum and Rhodium ($$)• Speed up:

CO + O2 CO2

NO, NO2 N2

Enzymes

• Biochemical catalysts• Large protein molecules • Catalase – in liver to decompose H2O2

• Active Site – “lock and key” model• Turnover numbers of 1000 to 10 million per

second

2. B 2A16. 0.033 mol 8.3 X 10-4

0.055 mol 5.5 X 10-4

0.070 mol 3.8 X 10-4

0.080 mol 2.5 X 10-4

18.8.3 X 10-5 M/s 7.0 X 105 M/s6.9 X 10-5 3.5 X 105 M/s5.2 X 10-5

3.4 X 10-5

20. a) -[H2O]/2t = [H2]/2t = [O2]/t

b) -[SO2]/2t = [O2]/t = [SO3]/2t

c) -[NO]/2t = -[H2]/2t = [N2]/t = [H2O]/2t

22. RateCO2 = RateH2O = 0.050 M/s

24.a) rate = k[A][C]2 b) rate double c) no change d) 9X e) 27 times

26. a) rate = k[H2][NO]2 b) 1.1 M/s

c) 6.0 M/s28.a,b) rate = k[C2H5Br][OH-], k = 3.6X10-5 M-1s-1

c) ¼ of old rate ( ½ X ½ )

30. a) Rate = k[ClO2]2[OH-]

b) k = 2.3 X 102 M-2s-1

c) Rate = 0.12 M/s

32. a) Rate = k[NO]2[O2]

b and c) k = 7.11 X 103 M-2s-1

d) Rate = 0.400 M/se) Rate = 0.200 M/s

38.a) 2.56 s b) 0.0125 M40. a) 0.0032 mol N2O5 b) 2.2 min c) 1.69 min

42. K = 2.08 X 10-4 s-1 t = 3.33 X 103s

44. a) First order b) k = 0.0101 s-1

c) half-life = 68.7 s46. a) First order b) 3.68 X 10-3 min-1

52. a) Ea = 154 kJ b) 18 kJ

54. a) 70 kJ b) 45 kJ c) 45 kJ(b) and (c) are faster than (a)

56. Ea = 1.15 kJ/mol

58. Ea = 130 kJ/mol, A = 1.0 X 1010.

84. 1.75 X 10-7 (for S)2(1.75 X 10-7) = 3.5 X 10-7 for H+ and Cl-

87. time = 48 s 95. Ea = 150 kJ/mol

• GOGGLES• Do one trial for each of the three ratios• We will compile class data to get an average time• Color is more of a light orange than a yellow• Record the color every 30 seconds (rather than

just guessing at the change time)

The initial rate of the reaction has been measured at the reactant concentrations shown (in mol/L):

BrO3-(aq) + 5 Br-(aq) + 8 H+(aq) ----> 3 Br2(l) +

H2O(l)

Experiment [BrO3

-] [Br-] [H+] Initial rate (M/s) 1 0.10 0.10 0.10 8.0 x 10-4 2 0.20 0.10 0.10 1.6 x 10-3 3 0.10 0.20 0.10 1.6 x 10-3 4 0.10 0.10 0.20 3.2 x 10-3

According to these results what would be the initial rate (in M/s) if all three concentrations are:

[BrO3-]=[Br-]=[H+]=0.20 M?

Kinetics “How fast a reaction occurs” Factors 1.Concentration 2.Temperature 3.Surface area...

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