18. The volume is 1 liter so all molar concentrations are numerically equal to the number of moles. The
stoichiometry is 2:2:2 for the NO, the NH
3
and the H
2
O. Therefore the moles of NO reacted must be equal
to moles of NH
3
and H
2
O produced.
That number is given, amount of H
2
O at equilibrium=0.00294
moles. This is also same as [NH
3
]
eq
.
The moles [NO]
eq
= starting amount minus the amount reacted:
0.00345

0.00294
=
5.10 x 10

4
M. Calculate amount of H
2
reacted and subtract that value from the
original amount to find the amount left at equilibrium:
amt reacted =
NO
mol
2
H
mol
5
)
NO
mol
00294
.
0
(
2
=
0.00735 moles H
2
amt remaining:
0.0175 moles

0.00735 moles H
2
in the 1 Liter volume
=
0.0105 M
Summary of results:
[NO ]=
5.10 x 10

4
M;
[H
2
]= 0.0105 M;
[NH
3
] = 0.00294 M
19.
(a)
I use lines 1 and 2 to determine rate dependence on Br
2
under conditions where acetone is constant.
x
2
trial
2
x
1
trial
2
)
]
Br
([
)
]
Br
([
2
rate
1
rate
=
x
4
5
050
.
0
100
.
0
10
x
14
.
1
10
x
70
.
5
=


log
0.5
=
x log (2.00)
therefore x =

1
reaction rate is dependent on [Br
2
]

1
(b)
use lines 2 and 3 to determine rate dependence on acetone under conditions where Br
2
is constant.
x
3
trial
x
2
trial
)
]
acetone
([
)
]
acetone
([
3
rate
2
rate
=
y
4
4
400
.
0
300
.
0
10
x
97
.
2
10
x
14
.
1
=


log
(0.38383)
=
y log (0.75)
therefore y = 3.3
reaction rate is [acetone]
3.3
rate law:
rate
=
k [acetone]
3.3
[Br
2
]

1
(c)
Numerical value is calculated from data on any one line. I choose line 1.
rate
=
k [acetone]
3.3
[Br
2
]

1
(1)
5.70x10

5
=
k[0.300]
3.3
[0.100]

1
k =
3.15 x 10

4
The other lines give the same answer to three sig figs.
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 Spring '08
 LEMASTER
 Chemistry, Mole, Chemical reaction, Moles