Chemistry 1B, Fall 2013
Lectures 21-22
Chemistry 1B
on-line kinetics 1 !!!
Fall 2013
Chemistry 1B Fall 2013
Lectures 21-22
Chemical Kinetics
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2
STRUCTURE and DYNAMICS
on-line kinetics 1 !!!
Chemistry 1B so far: STRUCTURE
of atoms and molecules
Chemistry 1B Fall 2013
Chemistry 1B now: DYNAMICS
• introduction to kinetics
• differential rate expressions
chemical kinetics
3
thermodynamics (chem 1C, 163B) and kinetics (chem 1B, 163C)
4
thermodynamics (chem 1C, 163B) and kinetics (chem 1B, 163C)
HOWEVER:
Thermodynamics: Whether a reaction will occur
spontaneously
??
and “how far” it will proceed
[equilibrium conditions]
O2
+
H2
nada (no reaction)
almost forever
2H2(g)+O2(g) ô 2H2O(g)
very exothermic
Kinetics: How fast a reaction proceeds, and
the molecular steps involved in a
reaction [the mechanism of a reaction].
equilibrium
chem 1A-1C
‘all’
products at
equilibrium
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6
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Chemistry 1B, Fall 2013
Lectures 21-22
chemical kinetics (chapter 15)
thermodynamics (chem 1C, 163B) and kinetics (chem 1B, 163C)
CN-
BUT:
O2
+
H2
CH3CN
+
I-
H3C-I
Kinetics: How fast a reaction proceeds, and
the molecular steps involved in a
reaction [the mechanism of a reaction].
7
specific objectives for final exam material
http://www.bluffton.edu/~bergerd/classes/CEM221/sn-e/SN2_alternate.html
8
measuring how fast a reaction goes
BrO3-(aq) + 5Br-(aq) +6H+ (aq) ô 3 Br2 (aq) +3 H2O (aq)
HW9 #71, S19-S20
BrO3- + H+
• Numerical Problem Solving
 Lecture notes
 HW9 Z15.17, Z15.20, Z15.57,
Z15.73, Z15.82
HW9 #68,#69,#72,
#73,#75
Br - + H+
HW9 #70,#74
spectrophotometer
measures Br2 concentration (absorbance)
HW#9 DUE WEDNESDAY, 4th DECEMBER (last WebAssign)
9
http://www.chm.davidson.edu/vce/kinetics/BromateBromideReaction.html
[Br2] (absorbance) vs time
(for set of initial [BrO3-], [Br -], [H+]) and fixed Temperature
BrO3-(aq) + 5Br-(aq) +6H+ (aq)
ô 3 Br2 (aq) +3 H2O (aq)
A (absorbance)
related to [Br2] (Beer’s Law)
more [Br2] ï
greater absorbance
[Br2] increasing as
product appears
time (t=0 when reactants mixed)
10
[Br2] (absorbance) vs time
(for set of initial [BrO3-], [Br -], [H+]) and fixed Temperature
11
A (absorbance)
related to [Br2] (Beer’s Law)
BrO3-(aq) + 5Br-(aq) +6H+ (aq)
→
N∫C-CH3 + I-
Kinetics: How fast a reaction proceeds, and
the molecular steps involved in a
reaction [the mechanism of a reaction].
• Graphical interpretations
 Lecture notes
 HW9 Z15.36a, Z15.82
CH3I
N∫C-
H2 O
• Concepts and definitions
 Lecture notes
 HW9 Z15.55, Z15.52, Z15.68
+
ô 3 Br2 (aq) +3 H2O (aq)
more [Br2] ï
greater absorbance
[Br2] increasing as
product appears
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2
Chemistry 1B, Fall 2013
Lectures 21-22
[Br2] (absorbance) vs time
(for set of initial [BrO3-], [Br -], [H+]) and fixed Temperature
BrO3-(aq) + 5Br-(aq) +6H+ (aq)
Zumdahl Sec.15.1
ô 3 Br2 (aq) +3 H2O (aq)
A (absorbance)
related to [Br2] (Beer’s Law)
more [Br2] ï
greater absorbance
2NO2(g) ô 2NO(g) +O2(g)
[Br2] increasing as
product appears
13
Zumdahl section 15.1
14
DIFFERENTIAL rate of chemical reaction (appearance of product)
2NO2(g) ô 2NO(g) +O2(g)
2NO2(g) ô 2NO(g) +O2(g)
REACTANT DISAPPEARS
“initial rate” at t=0 (0ö 50 s)
PRODUCTS APPEAR
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reaction rate changes as reaction proceeds (Z fig.15.1, Table 15.2)
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relations among rates of chemical reaction based on different components
2NO2(g) ô 2NO(g) +O2(g)
slope=tangent to curve
in each t=50s interval
- [NO2] = [NO] = 2[O2]
disappearance
NO2
-0.021
appearance
NO
O2
0.021
mol L-1 at t=0
0.011
?
2
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18
3
Chemistry 1B, Fall 2013
Lectures 21-22
relations among rates of chemical reaction based on different components
2NO2(g) ô 2NO(g) +O2(g)
rate will (may) generally depend on concentrations
higher concentrations ï more collisions
more collisions ï more reactions occurs
in each t=50s interval
rate depends on concentrations
- [NO2] = [NO] = 2[O2]
disappearance
NO2
appearance
NO
O2
-0.021
mol L-1
0.021
at t=0
aA+bBöcC +dD
0.011
2
concentrations of reactants
each raised to a power
(usually integers)
rate constant for a given Temperature
19
generalized (differential) rate expression
kf
aA + bB ô cC + dD
kr
initial (differential) rate expression
k
f
aA + bB ô
cC + dD
forward reaction
k
aA + bB ò cC + dD
20
aA + bB òr cC + dD
reverse reaction
(products recombine)
general differential rate expression (can get more complicated):
forward reaction
reverse reaction
INITIAL RATE (initially only reactants present):
[A]o, [B]o ∫ 0
[C]o, [D]o = 0
• k is rate constant (for eqn written as disappearance of A)
forward rate (loss of [A])
kf rate constant of forward reaction
reverse rate (increase of [A])
kr rate constant of reverse reaction
21
• DEFINITION OF RATE ORDER rate is:
mth order in reactant [A]
nth order in reactant [B]
Note: Only in certain instances (discuss soon)
will the order of a reactant or
product (m,n) be the same as
its stoichiometric coefficient (a,b)
• overall rate order is [m +n ] th order (i.e. total order of reaction rate) 22
skills for HW and final (HW9 #68-#69)
rate laws: initial rate
2NO2(g) ô 2NO(g) +O2(g)
forward reaction
2NO2(g) ò 2NO(g) +O2(g)
reverse reaction
DIFFERENTIAL RATE EXPRESSIONS
• Write
rate expression in terms of reactant concentrations
• Determine order for each reactant from initial rate data
initial rate:
• only reactants present
• [NO]o=[O2]o=0
• no reverse reaction
• Determine overall rate order
• Determine rate constant
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Chemistry 1B, Fall 2013
Lectures 21-22
determining rate order from initial rate (table 15.4 p 723 Z)
determining rate order from initial rate (table 15.4 p 723 Z)
NH4+ (aq) + NO2- (aq) ö N2 (g) +2H20 ()
NH4+ (aq) + NO2- (aq) ö N2 (g) +2H20 ()
differential initial rate
(x rate)= [x conc]n
(2)=(2)1
to determine the order in a reactant (e.g. m or n):
identify two initial conditions (experiments) where
the concentration of only one reactant has changed
1 vs 2: [NH4+] const, double [NO2- ]
rate doubles
ï n=1
2 vs 3: [NO2-] const, double [NH4+ ]
rate doubles
ï m=1
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order of reaction
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given rate law: determine k (rate constant) HW7 #63-64
NH4+ (aq) + NO2- (aq) ö N2 (g) +2H20 ()
now that we know the order of the reaction
to get k use any ‘experiment’ (or average of all)
first order in [NH4+]
first order in [NO2-]
from exp 1
second order overall
from exp 3
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more complicated
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more complicated (see Table 15.5)
BrO3-(aq) + 5Br-(aq) +6H+ (aq) ô 3 Br2 (aq) +3 H2O (aq)
BrO3-(aq) + 5Br-(aq) +6H+ (aq) ô 3 Br2 (aq) +3 H2O (aq)
1
remember same reaction in stop-flow apparatus, earlier slide
[BrO3-]: exp 1-2
[Br-]: exp 2-3
[H+]:
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exp 1-4
1
[BrO3-] doubles, rate x2 ï n= 1
[Br-] doubles, rate x2 ï m= 1
2=[2]1
[H+]
4=[2]2
doubles,
rate x4 ï p= 2
2
(x rate)=[x conc]n
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Chemistry 1B, Fall 2013
Lectures 21-22
order of the reaction
after determining order, evaluate rate constant
BrO3-(aq) + 5Br-(aq) +6H+ (aq) ô 3 Br2 (aq) +3 H2O (aq)
first order in [BrO3-]
first order in [Br-]
second order in [H+]
fourth order overall
will be identical (within experimental) using any of the ‘experiments’
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skills for HW and final
END OF KINETICS 1
AND
differential rate expressions

• Write
rate expression in terms of reactant concentrations
onto more kinetics !!!
 • Determine order for each reactant from initial rate data
 • Determine overall rate order
 • Determine rate constant
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33
skills for HW and final
differential rate expressions

on-line kinetics 2 !!!
• Write
rate expression in terms of reactant concentrations
 • Determine order for each reactant from initial rate data
 • Determine overall rate order
Chemistry 1B Fall 2013
 • Determine rate constant
integral rate expression
• Determine order of reaction from plot of
‘concentration’ vs time
35
• Half-life of a reaction
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6
Chemistry 1B, Fall 2013
Lectures 21-22
integrated rate expression
math 11 or 19
differential rate gives change of reactant
or product concentrations with time
integrated rate gives value of reactant
or product concentrations with time
value
change
value
change
INTEGRATE !!!
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integrated rate expression for first-order reaction
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integrated first order reaction
2N2O5 (soln) ô 4NO2(soln) + O2(g)
differential rate law:
slope
intercept
integrated rate law:
first order:
How could we tell
if n=1 (first order in N2O5 )?
a plot of ln[A] vs t would be a
[ N2O5 ]
vs t
[not using multiple experiments with various
initial concentrations]
straight line with slope -k and intercept ln [A]o
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integrated rate law (first-order)
other integrated rate laws
(if reaction is first order)
zero-th order reaction; differential rate law
plot ln [N2O5] vs t
do we get straight line ??
(yes n=1, no n∫ 1)
40
zeroth order reaction;
integrated rate law
YES !!
n=1
it is first-order
plot [A] vs t gives straight line
slope = -k
t=0 intercept is ln [N2O5]o
second order reaction; differential rate law
second order reaction;
integrated rate law
plot [A]-1 vs t gives straight line
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Chemistry 1B, Fall 2013
Lectures 21-22
more complicated integrated rate laws
first order or second order ??? Example 15.5 and (HW9 #70)
(don’t fret)
2C4H6 ö C8H12
same rate data plotted two ways
more complicated to get integrated rate-laws for
rate expressions which depend on several reactant concentrations :

use [Br-]o and [H+]o >> [BrO3- ]o ; plot ln [BrO3- ] vs t
[Br-] and [H+] change relatively little (constant)
reaction ‘pseudo’ first-order; i.e rateº (k [Br-] [H+]2 ) [BrO3- ]1
ª constant = (k [Br-]o [H+]o2 )
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44
t½ half-life vs rate constant zeroth and 2nd order
t½ half-life vs rate constant
still the definition of half-life
first-order reaction (e.g. radioactive decay)
but:
larger k ï shorter t½
only for 1st order is the half the same
throughout the reaction (independent of
the ‘current’ concentration of [A])
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48
summary (also understand section 15.5; 1-5, 6*)
know how to use; all
needed formulas given on exam
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Chemistry 1B, Fall 2013
Lectures 21-22
initial rate (2 vs 3)
END OF on-line KINETICS 2
AND
onto more kinetics !!!
divide 3/2
49
(x rate)= [x conc]n
50
zeroth order reaction: fully occupied surface catalyst
initial rate (x rate)= [x conc]n
rate = ….. [A]0n
decomposition of N2O on platinum surface
nth order in A
double initial concentration of A i.e. [2 x]
rate
if the initial
concentration
of one reactant
doubles [2x] and
the reaction rate
(x rate)= [x conc]n
order n
(x2) doubles
2=[2]1
n=1
(x4) quadruples
4=[2]2
n=2
(x8) octuples
8=[2]3
n=3
(x rate)= [x conc]n also holds for other concentration multiples
e.g. [3x] triple initial concentration , n=2, rate 9 times faster
51
after surface covered, higher [N2O] leads to no greater rate of decomposition
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