- #1

- 112

- 5

1. Homework Statement

2. Homework Equations

rate law equation

3. The Attempt at a Solution

a) I would expect this reaction to be a multi-step reaction because there are six reactant molecules, as shown in the overall chemical equation. For this reaction to be one-step, all six molecules would have to collide at the same time, with proper orientation and sufficient energy. It is extremely unlikely that this would happen.

b)

i) Comparing lines 1 and 2: Doubling the [A] does not change the rate. The reaction is zero order, with respect to [A].

ii) Comparing lines 1 and 3: Doubling the (B) causes the rate to double. The reaction is first order, with respect to (B).

iii) Comparing lines 3 and 4: Doubling the [C] causes the rate to increase four times. The reaction is second order, with respect to [C].

c) Overall, it is a third-order reaction.

d) Rate law equation is: rate = k (B)[C]

e) The rate law equation confirms that this is a multi-step reaction because the exponents of the rate law do not match the coefficients of the hypothetical chemical equation.

f) Comparing lines 1 and 5: The (B) has been doubled, so the rate should double (first order). The [C] has been tripled, so the rate should change by a factor of 9 (second order). The rate increases twice, and then nine times, for a total increase of eighteen times. The rate is 6.5 × 10

g) This is the part I am having trouble on. Can someone please explain how I should go about writing the reaction mechanism.

h) This part can be done once part g) is complete.

Thanks.

2. Homework Equations

rate law equation

3. The Attempt at a Solution

a) I would expect this reaction to be a multi-step reaction because there are six reactant molecules, as shown in the overall chemical equation. For this reaction to be one-step, all six molecules would have to collide at the same time, with proper orientation and sufficient energy. It is extremely unlikely that this would happen.

b)

i) Comparing lines 1 and 2: Doubling the [A] does not change the rate. The reaction is zero order, with respect to [A].

ii) Comparing lines 1 and 3: Doubling the (B) causes the rate to double. The reaction is first order, with respect to (B).

iii) Comparing lines 3 and 4: Doubling the [C] causes the rate to increase four times. The reaction is second order, with respect to [C].

c) Overall, it is a third-order reaction.

d) Rate law equation is: rate = k (B)[C]

^{2}e) The rate law equation confirms that this is a multi-step reaction because the exponents of the rate law do not match the coefficients of the hypothetical chemical equation.

f) Comparing lines 1 and 5: The (B) has been doubled, so the rate should double (first order). The [C] has been tripled, so the rate should change by a factor of 9 (second order). The rate increases twice, and then nine times, for a total increase of eighteen times. The rate is 6.5 × 10

^{-1}mol/L•s.g) This is the part I am having trouble on. Can someone please explain how I should go about writing the reaction mechanism.

h) This part can be done once part g) is complete.

Thanks.

Last edited: