Chemistry 102, Test Chapter 13-14 Worksheet With Answers - 1997 Page 2
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1
2
3
4
5
6
7
8
9
10
11+
[NH
]
0.015
0.030
Rate
4
2
2
=
= [1.5]
[1.5] =
=
+
[NH
]
0.010
0.020
Rate
4
1
1
The change in concentration = 1.5. 1.5 must be raised the 1st power to obtain the rate
+
change of 1.5 Therefore, the reaction is 1st order in [NH
] .
4
-
Determine the order with respect to NO
:
2
-
[NO
]
Rate
2
0.020
0.020
+
0.010
0.005
Hold [NH
] constant at 0.010 M
4
−
[NO
]
0.020
0.020
Rate
2
2
2
=
= 2
4 =
=
−
[NO
]
0.010
0.005
Rate
2
1
1
The change concentration = 2 (the concentration was doubled); 2 must be raised to the 2nd
2
power to obtain the rate change of 4 : 2
= 4. Therefore, the reaction is 2nd order with respect
-
+
-
2
to [NO
].
Rate = k[NH
][NO
]
2
4
2
2
4.(1) For a reaction which has the rate law,
Rate = k[A]
,
a plot of which of the
following is a straight line?
1. [A] vs time
2. ln [A] vs time
3. ln 1/[A] vs time
2
4. 1/[A] vs time
5. 1/[A]
vs time
Answer: 4. 1/[A] vs time
Rate = k[A]2 indicates a second order reaction.
1
1
Integrated rate law:
-
= kt
where y = 1/[A] and x = t
[ A]
[A]
o
Plotting 1/[A] vs. t will result in a straight line with a slope = k.
5.(9) Consider the reaction
2-
-
HgCl
(aq) + 1/2 C
O
(aq)
Cl
(aq) + CO
(g) + 1/2 Hg
Cl
(s)
2
2
4
2
2
2
which has the rate expression:
2-
2
-3
2
Rate = k[HgCl
][C
O
]
and a rate constant = 7.8 x 10
/M
min.
2
2
4
2-
When [HgCl
] = 0.085 M and [C
0
] = 0.068 M, what is the rate of reaction?
2
2
4
-5
-6
1. 4.5 x 10
M/min
2. 3.1 x 10
M/min
-4
-7
3. 3.9 x 10
M/min
4. 6.4 x 10
M/min
-6
Answer: 2. 3.1 x 10
M/min
Note: This is Problem 24(a) from Chapter 11
2-
2
-3
Rate = k[HgCl
][C
O
]
k = 7.8 x 10
/M2min
2
2
4
[HgCl
] = 0.085 M
2
2-
[C
O
] = 0.068 M
2
4
Rate = ?
-3
2
2
Rate = (7.8 x 10
/M
min)[0.085][0.068]
Rate = 3.1 x 10-6 M/min
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