Find rate constant by plugging numbers into integrated rate law.
Point out that often k is found in one equation and used in another equation.
39mod.
A first-order reaction is 38.5% complete in 480 s.
(a) Calculate the rate constant.
ln[C]
t
= -kt + ln[C]
o
ln[0.615
*
]
t
= (-k x 480) + ln[1]
k = 1.0128 x 10
-3
s
-1
* Note that 38.5% complete means that at time “t” 61.5% of the reactant is still present.

Chem 162-2016 Chapter 20 Electrochemistry
63
Determination of the rate constant from the half-life formula
.
The half life of a first order reaction is 2.64s.
What is the rate constant?
t
1/2
= 0.693/k
2.64s = 0.693/k
k = 0.2625s
-1

Chem 162-2016 Chapter 20 Electrochemistry
64
CHEM-2010 FINAL EXAM + ANSWERS
CHAPTER 12 - KINETICS
RATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIVES CONCEPTS
INTEGRATED RATE LAW
Determination of the rate constant from the slope of the integrated-rate-law line.
Since a plot of ln[A] vs time gives a straight line, then the reaction is first order.
ln[A]
t
= -kt + ln[A]
o
Y
=
mX +
b
slope = m = -k
slope = (Y
2
-Y
1
)/(X
2
-X
1
) = (-3.60 –(-1.47))/(1195-0) = -1.78 x 10
-3
s
-1
-k = slope
k = -slope = 1.78 x 10
-3
s
-1
B
= Answer
(X
2
Y
2
)
(X
1
Y
1
)
ln concentration

Chem 162-2016 Chapter 20 Electrochemistry
65
Finding other things after reaction order and k are found
2
Chem 162-2014 Exam I
H&P Chapter 13A – Chemical Kinetics
RATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIVES CALCULATIONS
The second-order reaction, 2Mn(CO)
5
→
Mn
2
(CO)
10
has a rate constant
equal to 3.0 × 10
9
M
-1
s
-1
at 25°C. If the initial concentration of Mn(CO)
5
is
1.0 × 10
-5
M, how long will it take for 90.% of the reactant to disappear?
A.
3.3 × 10
-16
s
B.
3.7 × 10
-15
s
C.
3.0 × 10
-4
s
D.
3.0 × 10
3
s
E.
1.0 × 10
-3
s
ET Note: Percentages only work in place of concentrations for 1
st
order reactions.
A second order reaction, given two concentrations and time, calls for the
integrated rate law:
1/[C]
t
=
kt + 1/[C]
o
k = 3.0 × 10
9
M
-1
s
-1
[C]
o
= 1.0 x 10
-5
M
90% of the reactant disappearing means that there is 10% remaining at time
t.
(1/(0.10 x (1.0 x 10
-5
))) = ((3.0 x 10
9
) x t) + (1/(1.0 x 10
-5
))
t = 3.0 x 10
-4
s

Chem 162-2016 Chapter 20 Electrochemistry
66
ARRHENIUS EQUATION FORMULAS
transparency
Rate equation:
Rate = k[X]
m
[Y]
n
“k” is the rate constant
The “k” in the rate law is the same as the “k” in the Arrhenius equation.
Arrhenius equation:
k = Ae
(-Ea/RT)
ln k = -E
a
/RT + ln A
Note: y = mx + b; y=ln(k); m=(-Ea/R); x=(1/T); b=ln(A)
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
)
ln (k
2
/k
1
) = -((E
a2
- E
a1
)/R) x (1/T)
This last formula probably will not be on the exam formula sheet so should be memorized.
IMPORTANT “RATE” FACTORS
Rate = k[X]
m
[Y]
n
• Concentration of reactants
Reaction order
• k (Arrhenius rate constant)
IMPORTANT “k” FACTORS
ln k = (-E
a
/R) x 1/T + ln A
• Collision frequency (part of A)
• Steric factor (also called orientation factor [part of A])
• Energy of activation
Catalyst
• Temperature
ln k = (-E
a
/R) x 1/T + ln A
ln (k
2
/k
1
) = -(E
a
/R) x [(1/T
2
) - (1/T
1
)]

Chem 162-2016 Chapter 20 Electrochemistry
67
THE ARRHENIUS EQUATION --
COLLISION THEORY
Rate = k[X]
m
[Y]
n
“k” is the rate constant
The “k” in the rate law is the same as the “k” in the Arrhenius equation.
Arrhenius equation: k = Ae
(-Ea/RT)
ln k = (-E
a
/RT) + ln A
E
a
= energy of activation
R = 8.314Jdeg
-1
mol
-1

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