Review Session-7

Review Session-7 - REVIEW
SESSION‐
7
...

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Unformatted text preview: REVIEW
SESSION‐
7
 Chemical
Kine6cs
 14.8)
You
study
the
effect
of
temperature
on
the
rate
of
reac8ons
and
graph
 the
ln(k)
vs
1/T.
How
do
the
two
graphs
compare
(a)
if
Ea
of
second
reac8on
 higher
than
Ea
of
first
reac8on,
but
both
reac8ons
have
same
frequency
factor
 (b)
if
Frequency
factor
of
second
reac8on
>
first
reac8on,
but
have
same
Ea
 14.8)
You
study
the
effect
of
temperature
on
the
rate
of
reac8ons
and
graph
 the
ln(k)
vs
1/T.
How
do
the
two
graphs
compare
(a)
if
Ea
of
second
reac8on
 higher
than
Ea
of
first
reac8on,
but
both
reac8ons
have
same
frequency
factor
 (b)
if
Frequency
factor
of
second
reac8on
>
first
reac8on,
but
have
same
Ea
 Ea1
 ln(k)
 Ea2
 ln(k)
 Ea1
 Ea2
 1/T
 Ea2
>
Ea1
 Same
frequency
factor
 1/T
 Ea2
=
Ea1
(same
slope)
 Different
frequency
factor
 14.51)
The
gas‐phase
reac8on
Cl+HBrHCl+Br
has
an
overall
enthalpy
change
 of
‐66kJ.
The
Ea
of
this
reac8on
is
7kJ.
(a)
Sketch
the
energy
profile
for
the
 reac8on,
label
Ea
and
ΔE
(b)
What
is
the
Ea
for
the
reverse
reac8on?
 14.51)
The
gas‐phase
reac8on
Cl+HBrHCl+Br
has
an
overall
enthalpy
change
 of
‐66kJ.
The
Ea
of
this
reac8on
is
7kJ.
(a)
Sketch
the
energy
profile
for
the
 reac8on,
label
Ea
and
ΔE
(b)
What
is
the
Ea
for
the
reverse
reac8on?
 Ea
=7kJ
 For
the
reverse
reac8on:
Ea=
(66+7)
=
73kJ
 ΔE=
‐66kJ
 14.52)
For
the
elementary
process
N2O5
NO2+NO3,
the
Ea
and
overall
ΔE
are
 154kJ/mol
and
136
kJ/mol.
(a)
Sketch
the
energy
profile
for
the
reac8on,
label
 Ea
and
ΔE
(b)
What
is
the
Ea
for
the
reverse
reac8on?
 14.52)
For
the
elementary
process
N2O5
NO2+NO3,
the
Ea
and
overall
ΔE
are
 154kJ/mol
and
136
kJ/mol.
(a)
Sketch
the
energy
profile
for
the
reac8on,
label
 Ea
and
ΔE
(b)
What
is
the
Ea
for
the
reverse
reac8on?
 For
the
reverse
reac8on:
Ea=
(154‐136)
=
18kJ
 Ea
=154kJ/mol
 ΔE=
136kJ/mol
 14.41)The
reac8on
SO2Cl2

SO2+Cl2
is
first
order
in
SO2Cl2.
Using
the
 kine8c
data,
determine
the
magnitude
of
first
order
rate
constant:
 
 
 
Time
(s)
 
 
 
Pressure
(SO2Cl2)
(atm)
 
 
 
0 
 
 
 
 
1.0
 
 
 
2500 
 
 
 
0.947
 
 
 
5000 
 
 
 
0.895
 
 
 
7500 
 
 
 
0.848
 
 






10000 
 
 
 
0.803



 14.41)The
reac8on
SO2Cl2

SO2+Cl2
is
first
order
in
SO2Cl2.
Using
the
 kine8c
data,
determine
the
magnitude
of
first
order
rate
constant:
 
 
 
Time
(s)
 
 
 
Pressure
(SO2Cl2)
(atm)
 
 
lnP
 
 
 
0 
 
 
 
 
1.0 
 
 
 
 
 
 
0 

 
 
 
2500 
 
 
 
0.947 
 
 
 
 
 
‐0.0545
 
 
 
5000 
 
 
 
0.895 
 
 
 
 
 
‐0.111
 
 
 
7500 
 
 
 
0.848 
 
 
 
 
 
‐0.165
 
 






10000 
 
 
 
0.803 
 
 
 
 
 
‐0.219



 14.41)The
reac8on
SO2Cl2

SO2+Cl2
is
first
order
in
SO2Cl2.
Using
the
 kine8c
data,
determine
the
magnitude
of
first
order
rate
constant:
 
 
 
Time
(s)
 
 
 
Pressure
(SO2Cl2)
(atm)
 
 
lnP
 
 
 
0 
 
 
 
 
1.0 
 
 
 
 
 
 
0 

 
 
 
2500 
 
 
 
0.947 
 
 
 
 
 
‐0.0545
 
 
 
5000 
 
 
 
0.895 
 
 
 
 
 
‐0.111
 
 
 
7500 
 
 
 
0.848 
 
 
 
 
 
‐0.165
 
 






10000 
 
 
 
0.803 
 
 
 
 
 
‐0.219



 ‐k
=
Slope
=
2.19X10‐5s‐1
 lnP
 T
 14.55)A
certain
first
order
reac8on
has
a
rate
constant
of
2.75X10‐2
s‐1
at
20oC.
 What
is
the
value
of
k
at
60oC
of
(a)
Ea=
75.5kJ/mol?
 14.55)A
certain
first
order
reac8on
has
a
rate
constant
of
2.75X10‐2
s‐1
at
20oC.
 What
is
the
value
of
k
at
60oC
of
(a)
Ea=
75.5kJ/mol?
 ln(k1/k2)
=
‐(Ea/R)
(1/T1‐1/T2)
 (a) Ea=
75.5kJ
=
75500
J/mol
 






R
=
8.314
J/mol
 
k1
=
2.75X10‐2s‐1
 
T1
=
273+20
=
293K
 
T2
=
273+60
=
333K

 
k2
=
?
 
k2
=
1.14s‐1

 14.65)(a)
Based
on
the
following
reac8on
profile,
how
many
intermediates
are
 formed
in
the
reac8on
AD
(b)
How
many
transi8on
states
are
there?
(c)
 Which
step
is
fastest?
(d)
is
the
reac8on
endothermic
or
exothermic?
 C
 B
 A
 D
 14.65)(a)
Based
on
the
following
reac8on
profile,
how
many
intermediates
are
 formed
in
the
reac8on
AD
(b)
How
many
transi8on
states
are
there?
(c)
 Which
step
is
fastest?
(d)
is
the
reac8on
endothermic
or
exothermic?
 
2

intermediates

(B
and
C)
 3
transi8on
states
 CD
is
fastest
step
 C
 B
 A
 D
 Endothermic
reac8on
 Two
step
mechanism
with
a
Fast
Ini6al
Step
 •  A
proposed
mechanism
is
 Step
1:

NO
+
Br2
 NOBr2




(fast)
 Step
2:

NOBr2
+
NO
→
2
NOBr




(slow)
 Step
1
includes
the
forward
and
reverse
reac8ons.
 Two
step
mechanism
with
a
Fast
Ini6al
Step
 •  The
rate
of
the
overall
reac8on
depends
upon
 the
rate
of
the
slow
step.
 •  The
rate
law
for
that
step
would
be
 Rate
=
k2
[NOBr2]
[NO]
 •  But
how
can
we
find
[NOBr2]?
 Two
step
mechanism
with
a
Fast
Ini6al
Step
 •  NOBr2
can
react
two
ways:
 –  With
NO
to
form
NOBr
 –  By
decomposi8on
to
reform
NO
and
Br2
 •  The
reactants
and
products
of
the
first
step
 are
in
equilibrium
with
each
other.
 •  Therefore,
 Ratef
=
Rater
 Two
step
mechanism
with
a
Fast
Ini6al
Step
 •  Because
Ratef
=
Rater
,
 k1
[NO]
[Br2]
=
k−1
[NOBr2]
 •  Solving
for
[NOBr2]
gives
us
 k1
 [NO]
[Br2]
=
[NOBr2]
 k−1
 Two
step
mechanism
with
a
Fast
Ini6al
Step
 
Subs8tu8ng
this
expression
for
[NOBr2]
in
 the
rate
law
for
the
rate‐determining
step
 gives
 Rate
=
 k2k1
 k−1
 [NO]
[Br2]
[NO]
 =
k
[NO]2
[Br2]
 Consider
the
following
three‐step
mechanism:
 
Ce4+

+
Mn2+

→

Ce3+
+
Mn3+
 
Ce4+

+
Mn3+

→

Ce3+
+
Mn4+
 
Mn4+

+
Tl+

→

Mn2+

+
Tl3+

 (1)
Write
balance
equa8on
for
overall
reac8on?
 (2)
Intermediate(s)
?
 (3)
Iden8fy
catalyst(s):

 Consider
the
following
three‐step
mechanism:
 
Ce4+

+
Mn2+

→

Ce3+
+
Mn3+
 
Ce4+

+
Mn3+

→

Ce3+
+
Mn4+
 
Mn4+

+
Tl+

→

Mn2+

+
Tl3+

 (1)
Write
balance
equa8on
for
overall
reac8on?
 2
Ce4+
+
Tl+


2
Ce3+
+Tl3+

 (2)
Intermediate(s)
?
 (3)
Iden8fy
catalyst(s):

 Consider
the
following
three‐step
mechanism:
 
Ce4+

+
Mn2+

→

Ce3+
+
Mn3+
 
Ce4+

+Mn3+

→

Ce3+
+
Mn4+
 
Mn4+

+
Tl+

→

Mn2+

+
Tl3+

 (1)
Write
balance
equa8on
for
overall
reac8on?
 2
Ce4+
+
Tl+


2
Ce3+
+Tl3+

 (2)
Intermediate(s)
?
 Mn3+
and
Mn4+
 (3)
Iden8fy
catalyst(s):

 Consider
the
following
three‐step
mechanism:
 
Ce4+

+
Mn2+

→

Ce3+
+
Mn3+
 
Ce4+

+
Mn3+

→

Ce3+
+
Mn4+
 
Mn4+

+
Tl+

→

Mn2+

+
Tl3+

 (1)
Write
balance
equa8on
for
overall
reac8on?
 2
Ce4+
+
Tl+


2
Ce3+
+Tl3+

 (2)
Intermediate(s)
?
 Mn3+
and
Mn4+
 (3)
Iden8fy
catalyst(s):

 Mn2+
 1)
Assuming
complete
dissocia8on
of
the
solute,
how
many
grams
of

KNO3
 must
be
added
to
275

of
water
to
produce
a
solu8on
that
freezes
at
14.5
?
The
 freezing
point
for
pure
water
is
0.0

and
Kf
=
1.86
(Mastering
Chemistry)
 1)
Assuming
complete
dissocia8on
of
the
solute,
how
many
grams
of

KNO3
 must
be
added
to
275

of
water
to
produce
a
solu8on
that
freezes
at
14.5
?
The
 freezing
point
for
pure
water
is
0.0

and
Kf
=
1.86
(Mastering
Chemistry)
 ΔT
=
14.5
,
Kf
=
1.86
,
i=2
(not
given
any
info)
 
ΔT
=
Kf
.
m
.
i
 m
=
ΔT/Kf
.
I
 m
=
14.5/(1.86x2)
=
3.9molals
 Molality
=
number
of
moles/kg
of
solvent
 3.9
=
n/
(0.275kg)


(assump8on)
 n=
1.07
moles
of
KNO3
=>
108g
 ...
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