Chap14_Part2

Chap14_Part2 - 1
 Concentra)on
and
Rate:
 NH4(aq)

 

NO2‐(aq)



N2(g)

 
2
H2O(l)
(a) 

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Unformatted text preview: 1
 Concentra)on
and
Rate:
 NH4+(aq)

+

NO2‐(aq)



N2(g)

+
2
H2O(l)
 (a)  These
data
are
ini7al
rates.


 (b)  For
constant
NO2‐,
the
rate
is
propor7onal
to
the
NH4+
concentra7on.


 (c)  For
constant
NH4+,
the
rate
is
propor7onal
to
the
NO2‐
concentra7on.
 + − Rate = k [ NH 4 ][ NO2 ] k
=
propor7onality
constant
 2
 "What
frac7on
of
the
molecules
in
the
origina7ng
compartment
is
 travelling
via
this
process
per
unit
7me”‐
Rate
constant
 + − Rate = k [ NH 4 ][ NO2 ] From
the
table,
 € Ini7al
rate
=
5.4
X
10‐7M/s
 [NH4]+
=
0.0100
M
 [NO2]‐
=
0.200
M
 
5.4X10‐7
M/s
 k

=

 =
2.7
X
10‐4
M‐1s‐1
 [0.01
M]
[0.20M]
 Once
we
have
both
rate
law
and
rate
constant
for
a
reac7on,
we
can
 calculate
the
rate
of
a
reac7on
fro
any
set
of
concentra7ons.
 3
 Using
the
table
14.2,
determine
the
rate
law,
rate
constant
and
rate
 of
the
reac7on
when
the
concentra7ons
of
both
reactants
raised
to
 [0.1M]
 + − Rate = k [ NH 4 ][ NO2 ] € k=
2.7X10‐4
M‐1s‐1
 (See
previous
slide)
 4
 Using
the
table
14.2,
determine
the
rate
law,
rate
constant
and
rate
 of
the
reac7on
when
the
concentra7ons
of
both
reactants
raised
to
 [0.1M]
 + − Rate = k [ NH 4 ][ NO2 ] € k=
2.7X10‐4
M‐1s‐1
 Rate
=
[2.7
X
10‐4M‐1s‐1][0.10M]
[0.10M]
=
2.7
X
10‐6
M/s


 5
 Rate
Laws
 Consider
a
general
reac7on
with
two
reactants
and
two
products:
 aA + bB → cC (a)  The
rate
law
is
 + dD Rate = k[ A]m [ B]n (b)
The
rate
depends
only
on
the
reactant
concentra7ons.
 (c)
The
propor7onality
constant
k
is
called
the
rate
constant.
 € (d)
The
values
of
m
and
n
are
usually
small
integers.
 6
 Exponents
in
the
Rate
Laws
 Rate = k[ A]m [ B]n [C ] p ... The
exponents
are
called
reac7on
orders.


 € This
reac7on
is
 
mth
order
in
[A]
 
 
 
 
 
nth
order
in
[B]
 
 
 
 
 
pth
order
in
[C]
 The
overall
reac7on
order
is
m
+
n
+
p
+
…
 The
reac7on
orders
must
be
determined
experimentally.
 7
 Reac)on
Orders:

Examples
 Rate = ± k[ A] 1st
order
in
[A],
overall
order
one
 Rate = ± k[ A] 2 2nd

order
in
[A],
overall
order
one
 Rate = ± k[ A][ B] 1st
order
in
[A],
1st
order
in
[B],
overall
order
two
 Rate = ± k[ A][ B][C ] Rate = ± k[ A][ B] 2 1st

order
in
[A],
1st
order
in
[B],
 1st
order
in
[C],
overall
order
three
 1st

order
in
[A],
2nd
order
in
[B],

 overall
order
three
 8
 2
ClO2(aq)
+
2
OH‐(aq)

ClO3‐(aq)
+
ClO2‐(aq)
+
H2O(l)
 Experiment
 
[ClO2]
(M) 
[OH‐]
(M) 
Rate
(M/sec)
 1
 
 
0.060 
 
0.030 
 
0.0248
 2
 
 
0.020 
 
0.030 
 
0.00276
 3
 
 
0.020 
 
0.090 
 
0.00828
 (a)
Determine
the
rate
law
for
the
reac7on.
 (b)
Calculate
the
rate
constant.
 9
 2
ClO2(aq)
+
2
OH‐(aq)

ClO3‐(aq)
+
ClO2‐(aq)
+
H2O(l)
 Experiment
 
[ClO2]
(M) 
[OH‐]
(M) 
Rate
(M/sec)
 1
 
 
0.060 
 
0.030 
 
0.0248
 2
 
 
0.020 
 
0.030 
 
0.00276
 3
 
 
0.020 
 
0.090 
 
0.00828
 (a)
Determine
the
rate
law
for
the
reac7on.
 Decreasing
[ClO2]
by
3
7mes

decreases
rate
by
9
7mes.
Means,
 increasing
[ClO2]
by
3
7mes

increases
rate
by
9
7mes:
Rate
α
[ClO2]2
 Look
at
entry
2
and
3:Increasing
[OH]‐
3
7mes
increases
rate
by
3
7mes
 Rate
α
[OH]‐
 Rate
=
k
[ClO2]2[OH]‐
 (b)
Calculate
the
rate
constant.
 10
 2
ClO2(aq)
+
2
OH‐(aq)

ClO3‐(aq)
+
ClO2‐(aq)
+
H2O(l)
 Experiment
 
[ClO2]
(M) 
[OH‐]
(M) 
Rate
(M/sec)
 1
 
 
0.060 
 
0.030 
 
0.0248
 2
 
 
0.020 
 
0.030 
 
0.00276
 3
 
 
0.020 
 
0.090 
 
0.00828
 (a)
Determine
the
rate
law
for
the
reac7on.
 Decreasing
[ClO2]
by
3
7mes

decreases
rate
by
9
7mes.
Means,
 increasing
[ClO2]
by
3
7mes

increases
rate
by
9
7mes:
Rate
α
[ClO2]2
 Look
at
entry
2
and
3:Increasing
[OH]‐
3
7mes
increases
rate
by
3
7mes
 Rate
α
[OH]‐
 Rate
=
k
[ClO2]2[OH]‐
 (b)
Calculate
the
rate
constant.
 k=
rate/[ClO2]2[OH]‐
=
(0.0248
M/s)/
(0.06M)2(0.03M)
=
230
M‐2s‐1
 Units
of
rate
constant
depends
on
overall
reac)on
order
of
the
rate
law.
 11
 2
ClO2(aq)
+
2
OH‐(aq)

ClO3‐(aq)
+
ClO2‐(aq)
+
H2O(l)
 Experiment
 
[ClO2]
(M) 
[OH‐]
(M) 
Rate
(M/sec)
 1
 
 
0.060 
 
0.030 
 
0.0248
 2
 
 
0.020 
 
0.030 
 
0.00276
 3
 
 
0.020 
 
0.090 
 
0.00828
 (a)
Determine
the
rate
law
for
the
reac7on.
 Decreasing
[ClO2]
by
3
7mes

decreases
rate
by
9
7mes.
Means,
 increasing
[ClO2]
by
3
7mes

increases
rate
by
9
7mes:
Rate
α
[ClO2]2
 Look
at
entry
2
and
3:Increasing
[OH]‐
3
7mes
increases
rate
by
3
7mes
 Rate
α
[OH]‐
 Rate
=
k
[ClO2]2[OH]‐
 (b)
Calculate
the
rate
constant.
 k=
rate/[ClO2]2[OH]‐
=
(0.0248
M/s)/
(0.06M)2(0.03M)
=
230
M‐2s‐1
 Units
of
rate
constant
depends
on
overall
reac)on
order
of
the
rate
law.
 Can
you
consider
entry
2
or
3
to
calculate
rate
constant?

 12
 (1)
Consider
the
reac7on
below
between
hydrogen
and
iodine
gas:
 H2(g)+I2
(g)

2HI
(g)
 Rate
=
k
[H2][I2]

 (a)
What
is
the
reac7on
order
of
the
reactant
[H2]
in
this
reac7on?
 (b)
What
is
the
overall
reac7on
order
of
this
reac7on?
 (c)
What
are
the
units
of
rate
constant?
 13
 (1)
Consider
the
reac7on
below
between
hydrogen
and
iodine
gas:
 H2(g)+I2
(g)

2HI
(g)
 Rate
=
k
[H2][I2]

 (a)
What
is
the
reac7on
order
of
the
reactant
[H2]
in
this
reac7on?
 1
 (b)
What
is
the
overall
reac7on
order
of
this
reac7on?
 (c)
What
are
the
units
of
rate
constant?
 14
 (1)
Consider
the
reac7on
below
between
hydrogen
and
iodine
gas:
 H2(g)+I2
(g)

2HI
(g)
 Rate
=
k
[H2][I2]

 (a)
What
is
the
reac7on
order
of
the
reactant
[H2]
in
this
reac7on?
 1
 (b)
What
is
the
overall
reac7on
order
of
this
reac7on?
 1
+
1
=2
 (c)
What
are
the
units
of
rate
constant?
 15
 (1)
Consider
the
reac7on
below
between
hydrogen
and
iodine
gas:
 H2(g)+I2
(g)

2HI
(g)
 Rate
=
k
[H2][I2]

 (a)
What
is
the
reac7on
order
of
the
reactant
[H2]
in
this
reac7on?
 1
 (b)
What
is
the
overall
reac7on
order
of
this
reac7on?
 1
+
1
=2
 (c)
What
are
the
units
of
rate
constant?
 Rate
constant
(k)=
rate/(M)(M)
=
(M/s)/(M2)=
M‐1s‐1

 16
 (1)
Consider
the
reac7on
below
between
hydrogen
and
iodine
gas:
 X2(g)+Y2
(g)

2XY
(g)
 Rate
=
k
[X2]2[Y2]3

 (a)
What
is
the
reac7on
order
ofthe
reactant
[Y2]
in
this
reac7on?
 3
 (b)
What
is
the
overall
reac7on
order
of
this
reac7on?
 2
+
3
=5
 (c)
What
are
the
units
of
rate
constant?
 Rate
constant
(k)=
rate/(M)2(M)3
=
(M/s)/(M5)=
M‐4s‐1

 17
 First
Order
Reac)on
 k A→B First
order
reac7on
is
one
whose
rate
depends
on
the
concentra7on
of

a
single
 reactant
raised
to
the
first
power
 Rate
=
‐Δ[A]/Δt
=
k[A]
 Δ[A]/[A]
=
‐kΔt

(differen7al
rate
law)
 Integra7on
on
both
sides,

ln[A]t‐
ln[A]0
=‐kt
(Integrated
rate
law)

 The
concentra7on
of
A
as
a
func7on
of
7me
is
 A(t ) = A(0) exp(−kt ) or ln[ A(t )] = ln[ A(0)] − kt Recall
that
k
has
units
of
7me‐1.
 18
 Problem:
The
decomposi7on
of
a
certain
insec7cide
in
water
follows
first
order
 kine7cs
with
a
rate
constant
of
1.45
yr‐1.
If
the
concentra7on
of
the
insec7cide
is
 5.0X10‐7g/cm3,
(a)
what
is
its
concentra7on
aker

1year
(b)
how
long
will
it
take
for
 the
concentra7on
of
insec7cide
to
decrease
to
3.0X10‐7g/cm3

 19
 Problem:
The
decomposi7on
of
a
certain
insec7cide
in
water
follows
first
order
 kine7cs
with
a
rate
constant
of
1.45
yr‐1.
If
the
concentra7on
of
the
insec7cide
is
 5.0X10‐7g/cm3,
(a)
what
is
its
concentra7on
aker

1year
(b)
how
long
will
it
take
for
 the
concentra7on
of
insec7cide
to
decrease
to
3.0X10‐7g/cm3

 (a)
k
=
1.45
yr‐1
 [A]0
=
5X10‐7
g/cm3
 ln[A]t=1
yr
=
‐
kt+
ln[A]0
 




=
‐
(1.45
yr‐1)
(1
yr)
+
ln
(5X10‐7)
 




=
‐1.45+
(‐14.51)
=
‐15.96
 
[A]t=1yr
=
exp(‐15.96)
=
1.2X10‐7
g/cm3


 20
 Problem:
The
decomposi7on
of
a
certain
insec7cide
in
water
follows
first
order
 kine7cs
with
a
rate
constant
of
1.45
yr‐1.
If
the
concentra7on
of
the
insec7cide
is
 5.0X10‐7g/cm3,
(a)
what
is
its
concentra7on
aker

1year
(b)
how
long
will
it
take
for
 the
concentra7on
of
insec7cide
to
decrease
to
3.0X10‐7g/cm3

 (a)
k
=
1.45
yr‐1
 (b)
k
=
1.45
yr‐1
 [A]0
=
5X10‐7
g/cm3
 [A]0
=
5X10‐7
g/cm3
 ln[A]t=1
yr
=
‐
kt+
ln[A]0
 ln[A]t=1
yr
=
‐
kt+
ln[A]0
 




=
‐
(1.45
yr‐1)
(1
yr)
+
ln
(5X10‐7)
 ln(3X10‐7)=
‐
(1.45
yr‐1)
(t
yr)
+
ln
(5X10‐7)
 




=
‐1.45+
(‐14.51)
=
‐15.96
 




t=
0.35
yr
 
[A]t=1yr
=
exp(‐15.96)
=
1.2X10‐7
g/cm3


 21
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This note was uploaded on 09/13/2011 for the course CHEM 102 taught by Professor Todd during the Spring '08 term at UNC.

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