chem 1310: introduction to physical chemistry part...

Chem 1310: Introduction to physical chemistry

Part 2b: rates

Kinetics

How fast does a reaction go? (and why?)

Solution of crystal violet poured into solution of NaOH.

What is crystal violet?

Large organic molecule, deep purple.Also known as "Gram's stain", used to distinguish

types of bacteria ("Gram-negative" and "Gram-positive").

Disinfectant and toxic! Do not get on skin!

What is crystal violet?

+

CV+

How does it react with OH- ?

++

-

CV+ OH-

How does it react with OH- ?

++

-

CV+ OH- CVOH

deep purple colorless

Following the progressof the reaction

• Visually: color changes:purple pink colorless

• Quantitative: colorimetry

measure transmitted light

Colorimetry

Law of Lambert-Beer

more convenient:use Absorbance A:

I0 It

ΔxZct eII ][

0

tIIA 0log AkZ ][

Following the progressof the reaction

t [CV+]s mol/L

0.0 5.00E-05

10.0 3.68E-05

20.0 2.71E-05

30.0 1.99E-05

40.0 1.46E-05

50.0 1.08E-05

60.0 7.93E-06

80.0 4.29E-06

100.0 2.32E-06

0.00E+00

1.00E-05

2.00E-05

3.00E-05

4.00E-05

5.00E-05

6.00E-05

0.0 20.0 40.0 60.0 80.0 100.0

t (s)

[CV+

] (m

ol/L

)

What is a rate ?The rate of a chemical reaction is the speed at which

it transforms reagents into products.It is a measure of how fast things change with time.Compare with the speed of a car: measures how fast

its position changes with time.

In chemistry we look at concentrations and changes in them.

txspeed

tZrate

][

A heterogeneous reaction

Calculating rates

0.00E+00

1.00E-05

2.00E-05

3.00E-05

4.00E-05

5.00E-05

6.00E-05

0.0 20.0 40.0 60.0 80.0 100.0

t (s)

[CV+

] (m

ol/L

)

0.00E+00

1.00E-05

2.00E-05

4.00E-05

5.00E-05

0.0 20.0 40.0 60.0

t (s)[C

V+](

mol

/L)

}{t

[CV]

tCVrate

][

t large: average rate over interval tt very small: instantaneous rate

note the "-" sign!

average rate overt = 10 .. 20 s

Decrease of concentration follows a smooth curve.

At each point, the rate is the (negative of the) slope of the curve.

Average and instantaneous rates

0.00E+00

1.00E-05

2.00E-05

3.00E-05

4.00E-05

5.00E-05

6.00E-05

0.0 20.0 40.0 60.0 80.0 100.0

t (s)

[CV+

] (m

ol/L

)

Average and instantaneous rates

We cannot measure rates directly.We can only measure concentrations.Rates can be estimated• by drawing a smooth curve and estimating the

slopes (tangents)• by using average rates over real (but small) time

intervals

Average and instantaneous rates

The instantaneous rate is givenby the slope (tangent)

t (s)

{[CV]

}

t t0.00E+00

3.00E -05

4.00E -05

5.00E -05

6.00E -05

0.0 60.0 80.0 100.0

[CV+

] (m

ol/L

) {[CV]

}

11-6

5

sLmol10*10.16.3310*70.3][)0.10(

tCVrate

11-7

5

sLmol10*4.10.8010*12.1][)0.80(

tCVrate

Average and instantaneous rates

The average rate over a larger intervaldiffers significantly from the instantaneous rates.

0.00E +00

1.00E -05

3.00E -05

4.00E -05

5.00E -05

6.00E -05

0.0 20.0 60.0 80.0 100.0

t (s)

[CV+

] (m

ol/L

) {[CV]

}

t

11-6

5

sLmol10*10.16.3310*70.3][)0.10(

tCVrate

11-7

5

sLmol10*4.10.8010*12.1][)0.80(

tCVrate

1-1-7

5

sLmol10*7.40.7010*27.3][)0.80..0.10(

tCVrateav

Following the progressof the reaction

t [CV+] av rates mol/L mol L-1s-1

0.0 5.00E-051.32E-06

10.0 3.68E-059.70E-07

20.0 2.71E-057.20E-07

30.0 1.99E-055.30E-07

40.0 1.46E-053.80E-07

50.0 1.08E-052.87E-07

60.0 7.93E-061.82E-07

80.0 4.29E-069.85E-08

100.0 2.32E-06

0.00E+00

1.00E-05

2.00E-05

3.00E-05

4.00E-05

5.00E-05

6.00E-05

0.0 20.0 40.0 60.0 80.0 100.0

t (s)

[CV+

] (m

ol/L

)

Rates depend on concentrationsFrom the table for Crystal Violet

we can plot rate vs concentration:

Looks pretty linear.The rate law is:

rate = k [CV+]with:

k = 2.68*10-2 s-1

(the rate constant).0.00E+00

2.00E-07

4.00E-07

6.00E-07

8.00E-07

1.00E-06

1.20E-06

1.40E-06

1.60E-06

1.80E-06

0.00E+00 1.00E-05 2.00E-05 3.00E-05 4.00E-05 5.00E-05 6.00E-05 7.00E-05

[CV+] (mol/L)

av ra

te (m

ol L

-1 s

-1)

Another rate law determination

It is not always this easy

Here we could get a set of rates from a single experiment.

Often this cannot be done: you have to stop the reaction to analyze the results.

Start reaction, stop it after a short interval, analyze.Do this for a number of initial concentrations

obtain a number of initial rates.Need lots of experiments to get accurate curves!

Using initial ratesC0 = 5.00e-5 mol/L

0.00E+00

1.00E-05

2.00E-05

3.00E-05

4.00E-05

5.00E-05

6.00E-05

0 20 40 60 80

C0 = 3.00e-5 mol/L

0.00E+00

1.00E-05

2.00E-05

3.00E-05

4.00E-05

5.00E-05

6.00E-05

0 20 40 60 80

C0 = 1.00e-5 mol/L

0.00E+00

1.00E-05

2.00E-05

3.00E-05

4.00E-05

5.00E-05

6.00E-05

0 20 40 60 80

Initial conc Conc after 1 s Initial rate5.00E-05 4.89E-05 1.14E-06

4.00E-05 3.91E-05 9.10E-073.00E-05 2.93E-05 6.82E-072.50E-05 2.44E-05 5.68E-072.00E-05 1.95E-05 4.55E-071.50E-05 1.47E-05 3.41E-071.00E-05 9.77E-06 2.27E-07 0.00E+00

2.00E-07

4.00E-07

6.00E-07

8.00E-07

1.00E-06

1.20E-06

1.40E-06

1.60E-06

0.00E+00 2.00E-05 4.00E-05 6.00E-05 8.00E-05

C0 (mol/L)

initi

al ra

te (m

ol L

-1 s

-1)

More complicated rate laws

2 NO + O2 2 NO2

expt [NO] [O2] initial rate

1 0.02 0.01 0.028

2 0.02 0.02 0.057

3 0.02 0.04 0.114

4 0.04 0.02 0.227

5 0.01 0.02 0.014

More complicated rate laws

2 NO + O2 2 NO2

expt [NO] [O2] initial rate

1 0.02 0.01 0.028

2 0.02 0.02 0.057

3 0.02 0.04 0.114

4 0.04 0.02 0.227

5 0.01 0.02 0.014

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0 0.01 0.02 0.03 0.04 0.05 0.06

[O2]

initi

al ra

te

linear in [O2]

etcraterate

)1(2

)2(2

]O[]O[

04.2028.0057.0

)1()2(

rate [O2]

More complicated rate laws

2 NO + O2 2 NO2

expt [NO] [O2] initial rate

1 0.02 0.01 0.028

2 0.02 0.02 0.057

3 0.02 0.04 0.114

4 0.04 0.02 0.227

5 0.01 0.02 0.014

0

0.05

0.1

0.15

0.2

0.25

0 0.01 0.02 0.03 0.04 0.05

[NO]

initi

al ra

te

definitely not linear in [NO] !

etcraterate

2

)2(

)4(

]NO[]NO[

98.3057.0227.0

)2()4(

quadratic ???

More complicated rate laws

2 NO + O2 2 NO2

expt [NO] [O2] initial rate

1 0.02 0.01 0.028

2 0.02 0.02 0.057

3 0.02 0.04 0.114

4 0.04 0.02 0.227

5 0.01 0.02 0.014

quadratic (second-order) in [NO]

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 0.0005 0.001 0.0015 0.002 0.0025 0.003

[NO]^2

initi

al ra

te

rate [NO]2

Putting it all together

2 NO + O2 2 NO2So we have

rate [O2] ([NO] constant)rate [NO]2 ([O2] constant)

Combining these givesrate = k [O2][NO]2

withk = 7069 L2mol-2s-1

Reaction is:•first-order in O2

•second-order in NO•third-order overall

expt [NO] [O2] initial rate est k1 0.02 0.01 0.028 70002 0.02 0.02 0.057 71253 0.02 0.04 0.114 71254 0.04 0.02 0.227 70945 0.01 0.02 0.014 7000

est k = rate/([O2][NO]2)

What do we mean by "the rate" ?

2 NO + O2 2 NO2

For every single molecule of O2, two molecules of NO are consumed and two molecules of NO2 are produced.

"The rate" of the reaction is defined as

contains every component divided by its coefficient in the balanced reaction equation.

tttrate

]NO[

21]O[]NO[

21 22

Stoichiometry and rate laws

Rate laws can not be deduced from the balanced reaction equation!

They must be measured experimentally.The results are not always intuitive.

NO2 + CO NO + CO2

rate = k [NO2]2

Consequences of a rate law

Stoichiometry and rate laws

You can even have non-integer orders:

C CCH3

CH3H

HC C

CH3

HH

CH3(cat I2)

trans-2-butene cis-2-butene

rate(cistrans) = k [cis-2-butene][I2]½

first-order in cis-2-butenehalf-order in I2 A more complicated rate law