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CHEMISTRY 1622011
LECTURE 5
ANNOUNCEMENTS
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ATTENDANCE
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EXAMS
QUIZ
Chem 1622011 Lecture 5
1
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:
COMPLETE LECTURE 4
CHAPTER 13 (cont.)  CHEMICAL KINETICS
H&P 13.713.8
•
Collision theory
•
Effect of temperature on reaction rates
Chem 1622011 Lecture 5
2
FORMULAS
ET: Rate = speed.
Using a diagram with Y being 100 lbs and 110 lbs, and days 0 to 5, show that the increase in weight = slope = (Y
2
– Y
1
)/(X
2
– X
1
) = +2lb/day.
Decrease in weight = increase in weight = slope = (Y
2
– Y
1
)/(X
2
– X
1
) = 2lb/day.
ET: Point out that table in middle of page contains the key formulas in kinetics.
Discuss these formulas at beginning of recitation, focusing on 1
o
reaction.
C
12
H
22
O
11
(sucrose) + H
2
O → 2 C
6
H
12
O
6
(glucose)
Rate of sucrose disappearance = (∆ [sucrose])/(∆ time) =
([sucrose
f
]  [sucrose
i
])/(t
f
 t
i
)
Rate of glucose appearance = 2 x Rate of sucrose disappearance
aA + bB → cC + dD
General rate of reaction = (1/a)(∆[A]/∆t) = (1/b)(∆[B]/∆t) = (1/c)(∆[C]/∆t) = (1/d)(∆[D]/∆t)
Rate or a reaction may be written several ways.
Reaction: 2A + 3B + C → 2D
Rate = k[A]
m
[B]
n
[C]
p
∆C/∆t = k[A]
m
[B]
n
[C]
p
d[A]/dt = k[A]
m
[B]
n
[C]
p
Integrated
Reaction
Differentiated
rate law
k
Order
Reaction Rate**
RATE LAW***
(y
=
mx + b)
Halflife
*
units
0
Avg
Rate = (C
2
C
1
)/(t
2
t
1
)
Rate = k[C]
o
=k
[C]
t
= kt
+ [C]
o
t
1/2
= [C]
o
/2k
M
1
s
1
1
Avg
Rate = (C
2
C
1
)/(t
2
t
1
)
Rate = k[C]
1
ln[C]
t
= kt + ln[C]
o
t
1/2
= 0.693/k
M
o
s
1
2
Avg
Rate = (C
2
C
1
)/(t
2
t
1
)
Rate = k[C]
2
1/[C]
t
=
kt + 1/[C]
o
t
1/2
= 1/(k[C]
o
)
M
1
s
1
** Rate of appearance = +slope; rate of disappearance = rate of appearance = slope.
*** Differentiated Rate Law may be for more than one component, e.g., Rate = k[C]
1
[D]
2
*Halflives
:
For a zero order reaction, each successive halflife is ½ the time of the preceding one.
For a first order reaction, each successive halflife is equal in time to the preceding one.
For a second order reaction, each successive halflife is double time of the preceding one.
Arrhenius equation:
k = Ae
(Ea/RT)
A = frequency factor = combination of steric factor and collisional frequency
E
a
= energy of activation
ln k
2
 ln k
1
= (E
a
/RT
2
)  (E
a
/RT
1
)
Chem 1622011 Lecture 5
3
ln k = E
a
/RT + ln A
ln (k
2
/k
1
) = (E
a
/R)[(1/T
2
)  (1/T
1
)]
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View Full Document LECTURE 4 CONTINUED
PRIORITIES IN SOLVING KINETICS PROBLEMS
FIND REACTION ORDER
•
Use method of initial rates
 If initial rates are available*
*If initial rates are not available they can be calculated, but that is probably not worth the time on an exam.
•
Evaluation of halflife data
•
Identifying integrated rate law straight line graph
 From actual graph
 From description of graph
•
Recognizing k units
•
Plugging numbers into equations to get a constant “k”.
(the
“constantconstant”
method)
 Integrated rate law equation
 Differentiated rate law equation
 Halflife equation
FIND THE RATE CONSTANT
•
After identifying reaction order, plug numbers into
 Differentiated rate law equation or
 Integrated rate law equation or
 Halflife equation
•
Slope or negative slope of integrated rate law graph
Note: Diff. rate law equation contains conc. and rate terms.
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This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.
 Spring '11
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