You and your lab partner are studying the rate of a reaction, A + B --> C. You m
ID: 735389 • Letter: Y
Question
You and your lab partner are studying the rate of a reaction, A + B --> C. You make measurements of the initial rate under the following conditions:Experiment [A] (M) [B] (M) Rate (M/s)
1 0.4 1.9
2 0.8 1.9
(a) Which of the following reactant concentrations could you use for experiment 3 in order to determine the rate law, assuming that the rate law is of the form, Rate = k [A]x [B]y? Choose all correct possibilities.
[A] = 0.8 and [B] = 5.7
[A] = 0.4 and [B] = 5.7
[A] = 2.0 and [B] = 1.9
[A] = 0.4 and [B] = 3.8
[A] = 0.8 and [B] = 3.8
[A] = 1.6 and [B] = 1.9
[A] = 1.2 and [B] = 1.9
[A] = 0.8 and [B] = 1.9
(b) For a reaction of the form, A + B + C --> Products, the following observations are made: doubling the concentration of A increases the rate by a factor of 2, doubling the concentration of B has no effect on the rate, and tripling the concentration of C increases the rate by a factor of 9. Select the correct rate law for this reaction from the choices below.
Rate = k[A][B][C]
Rate = k[A][C]
Rate = k[A]2 [C]
Rate = k[A][C]2
Rate = k[A]2 [C]2
Rate = k[A]3 [C]
Rate = k[A][C]3
(c) By what factor will the rate of the reaction described in part (b) above change if the concentrations of A, B, and C are all halved (reduced by a factor of 2)?
The rate will be the original rate multiplied by a factor of .
Explanation / Answer
(a) we changed concentration of A in the first two while keeping concentration of B constant. Now we should keep [A] fixed and vary [B] There are eight choices. All of the ones that have [A] back to what it was in exp 1 or 2 and have [B] changed would be fine so, first and second choices are good, third choice changes [A] and keeps [B] what it was so won't help us find effect of change in [B], fourth and fifth are good, sixth and seventh change [A] and keep [B] what it was, so not helpful and the eighth is the same as Exp2, so no good. (b) [B] has no effect so we want it to the zero power, which is the same as leaving it out since anything the the zero poower equals one. [A] changes rate linearly, so we want it to the first power. Tripling [C] makes rate go up nine times. This is three squared, so we want [C]^2 all together we get k[A][B]2 which is the middle choice (c) cutting [A] in half halves rate. Cutting [B] in half does nothing. Cutting [C] in half makes it 1/4 as fast since (1/2)^2 = 1/4. A half times a fourth makes it 1/8 as fast So the original rate is multiplied by a factor of 1/8
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