You and your lab partner are studying the rate of a reaction, A + B --> C. You m
ID: 1023747 • 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 0.3 1.8
2 0.6 1.8
(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] = 0.3 and [B] = 5.4
[A] = 1.2 and [B] = 1.8
[A] = 1.5 and [B] = 1.8
[A] = 0.6 and [B] = 1.8
[A] = 0.9 and [B] = 1.8
[A] = 0.3 and [B] = 3.6
[A] = 0.6 and [B] = 5.4
[A] = 0.6 and [B] = 3.6
(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 4, doubling the concentration of B has no effect on the rate, and doubling the concentration of C increases the rate by a factor of 4. 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) 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.
Note that we are using Method of Initial Rates:
where:
r = k [A]^a [B]^b for any point
so choosing any 2 points we could do this:
r1 / r2 = /k [A]1^a [B]1^b( / (k [A]2^a [B]2^b)
if we were to make [A]1 = [A]2 in a set of expierments, we will be able to get rid of A, and make only B a case for analysis.
Similar, if we choose [B]1 = [B]2; then we could get rid of B, and analyse A
So, knowing this...
Choose any point that has either 0.3 or 0.6 M since we need to have the SAME initial concentrations in order to calculate the rate law order. Note that we must avoid "B" being the specific 1.8 M value (so we can compare different values)
[A] = 0.3 and [B] = 5.4 --> excellent since we can cancle [A] and compar [B]
[A] = 0.6 and [B] = 1.8--> Not recommended, since [B]new = [B]old
[A] = 0.3 and [B] = 3.6--> excellent since we can cancle [A] and compar [B]
[A] = 0.6 and [B] = 5.4--> excellent since we can cancle [A] and compar [B]
[A] = 0.6 and [B] = 3.6--> excellent since we can cancle [A] and compar [B]
(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 4
double means, 2x... if 2x --> 4, then we are talking about SECOND POWER relationship with respect to A
doubling the concentration of B has no effect on the rate
if (x)^n = (2x)^n; then we know n = 0, there is no effect of B, therefre B is zeroth order, or must not be included
and doubling the concentration of C increases the rate by a factor of 4
(x)^c = (2x)^2
then
x = 2
the reaction is 2nd order with respect to C
Then...
ignore all rates with [B]
Rate = k[A]^2 [C]^2 must be the rate
(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 _____?
if we reduce to a factor of 2... this means 1/2
so
Rate(old) = K*(1)^2 * (1)^2 X = K*X
Rate(new) = K*(1/2)^2 * (1/2)^2 X = 1/4*K*X
Compare
K*X / ( 1/4*K*X)
the rate is --> 1/4 changed or affected by a 25% or a factor multiplied by 1/4
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