Chemistry 102, Test Chapter 13-14 Worksheet With Answers - 1997 Page 4
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1
2
3
4
5
6
7
8
9
10
11x
-1
= 120.863
x = 4.7 x 10
L/mol/.sec
-3
3.9 x 10
L / mol.sec
-2
9.(3) A reaction which is second order has a rate constant of 1.0 x 10
/M•s. If the initial
concentration of the reactant is 0.100 M, how long will it take for the concentration to
become 0.0300 M?
3
3
1. 2.3 x 10
sec
2. 1.0 x 10
sec
4
2
3. 1.7 x 10
sec
4. 6.2 x 10
sec
3
-2
Answer: 1. 2.3 x 10
s
Second order:
k = 1.0 x 10
/M.sec
[A]
= 0.100 M
o
[A] = 0.0300 M
1
1
-
= kt
t = ?
[ A]
[A]
o
1
1
-2
-
= (1.0 x 10
/M.sec)(t)
0.0300M
0.100M
-2
23.33 = (1.0 x10
/M.sec)(t)
3
2.3 x 10
sec = t
10.(11)A catalyst is effective because
1. it increases the number of molecules with energy equal to or greater than the
activation energy.
2. it lowers the activation energy for the reaction.
3. it increases the number of collisions between molecules.
4. it increases the temperature of the molecules in the reaction.
Answer: 2.
A catalyst provides a different pathway for a reaction. This new
pathway has a lower activation energy than the uncatalyzed pathway. More collisions
occur with this lower Ea. See Figure 11.9 page 319 for a diagram.
11.(12)A 10 degree increase in temperature often doubles the rate of a chemical reaction. This
effect is best attributed to
1. a decrease in activation energy
2. a more feasible pathway at the higher temperature.
3. more collisions with the required energy (activation energy).
4. a doubling of the collisions at the higher temperature.
Answer: 3.
At a higher temperature, the molecules have a greater kinetic energy
( K. E. α T). Collisions between more energetic molecules occur with more energy. So
more of the collisions that occur do so with the minimum energy required - the activation
energy. The activation energy is independent of temperature so it does not decrease with an
increase in temperature.
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