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1aa3-2017-tut2-a.pdf
1aa3-2017-tut2-a.pdf
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1aa3-2017-tut2-a.pdf-CHEMISTRY 1AA3 Week of JANUAR...
1aa3-2017-tut2-a.pdf-CHEMISTRY 1AA3 Week of JANUARY 16,
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1aa3-2017-tut2-a.pdf-CHEMISTRY 1AA3 Week of JANUARY 16,
Page 5
5
Again, for
NH
4
+
+
H
2
O
NH
3
+
H
3
O
+
pH
= p
K
a
+
log ([NH
3
]/[NH
4
+
])
8.74 = 9.255
2
+ log ([0.30]/[Y])
and we assume x is small
- 0.5152 = log ([0.30]/[Y])
This solves to give Y = 0.9825 M. When multiplied by volume, this also gives us 1.4
74
mol
of NH
4
Br, which will lead to the same mass value.
Any of these three approaches works, and with sufficient extra digits in our calculation will
lead us to the same final answer.
Kinetics
- Chapter 14
6.
Consider the reaction:
N
2
(g) + 3 H
2
(g)
à
2 NH
3
(g)
(a)
Write expressions for the rate of the reaction in terms of the changes of
concentrations of N
2
, H
2
, and NH
3
.
(b)
Suppose that at some moment the rate of consumption of H
2
is 0.074 M
-1
s
-1
.
What is the rate at which N
2
is being consumed?
SOLUTION:
(a)
Reaction rates are defined in terms of the reaction stoichiometry,
as written
.
rate
=
−
Δ
[N
2
]
=
−
1
Δ
[H
2
]
=
+1
Δ
[NH
3
]
Δ
t
3
Δ
t
2
Δ
t
(b)
Rate of consumption or production is defined in terms of the specified reactant or
product.
-
Δ
[H
2
] =
0.074 M
−
1
s
−
1
= 3 (-
Δ
[N
2
] )
Δ
t
Δ
t
-
Δ
[N
2
]
=
-1
Δ
[H
2
]
= 1 (0.074 M
−
1
s
−
1
)
=
0.025 M
1
s
−
1
Δ
t
3
Δ
t
3
Page 6
6
7.
Consider the reaction:
2 NO(g) + O
2
(g)
→
2 NO
2
(g)
(a)
Write expressions for the rate of the reaction in terms of the changes of
concentrations of NO, O
2
, and NO
2
.
(b)
Suppose that at a particular moment the rate of reaction with respect to NO is
0.066 M s
-1
. What is the rate at which O
2
is being consumed?
SOLUTION:
(a)
rate
=
−
1
Δ
[NO]
=
−Δ
[O
2
]
= +1
Δ
[NO
2
]
2
Δ
t
Δ
t
2
Δ
t
(b)
Δ
[NO] =
0.066 M
−
1
s
−
1
Δ
t
−
1
Δ
[NO]
=
−Δ
[O
2
]
2
Δ
t
Δ
t
Δ
[O
2
]
=
(0.066 M s
−
1
) =
0.033M s
−
1
Δ
t
2
8.
What are the overall orders of the reactions to which the following rate laws apply?
(a)
rate = k[Cl
2
][H
2
]
(b)
rate = k[NO
2
]
2
[O
2
]
(c)
rate = k[H
2
]
½
[C
2
H
4
]
SOLUTION:
The overall order of a reaction comes from the sum of the exponents in the rate law,
i.e.,
for
rate = k[A]
x
[B]
y
overall order = x + y
(a)
1 + 1 =
2
(b)
2 + 1 =
3
(c)
0.5 + 1 =
1.5
9.
The rate for the reaction:
NH
4
+
(aq) + NO
2
-
(aq)
→
N
2
(g) + 2 H
2
O (l)
Page 7
7
is given by the rate law:
rate = k[NH
4
+
][ NO
2
-
]. At 25
°
C, the rate constant for the
reaction is 3.0
×
10
-4
M
-1
s
-1
. Calculate the rate of the reaction at 25
°
C if [NH
4
+
] = 0.26
M and [NO
2
-
] = 0.080 M.
SOLUTION:
rate = k [NH
4
+
][ NO
2
−
]
and
k = 3.0
×
10
−
4
M
−
1
s
−
1
∴
rate
=
(3.0
×
10
−
4
M
−
1
s
−
1
)(0.26 M)(0.080 M)
rate
=
6.24
×
10
−
6
M/s
( =
Δ
[N
2
]
/
Δ
t )
10.
The rate law for the reaction:
F
2
(g) + 2 ClO
2
(g)
→
2 FClO
2
(g)
is
rate = k [F
2
][ClO
2
]. At 25
°
C, the rate constant for the reaction is 1.2 M
−
1
s
−
1
.
Calculate the rate of the reaction at 25
°
C if [F
2
] = 0.010 M and [ClO
2
] = 0.020 M.
SOLUTION:
rate = k [F
2
][ClO
2
]
and
k = 1.2 M
−
1
s
−
1
∴
rate
=
(1.2 M
−
1
s
−
1
)(0.010 M)(0.020 M)
rate
=
2.4
×
10
−
4
M/s
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