Chem 1310: Introduction to physical chemistry Part...
Transcript of 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