Question

Using the concentration and rate data answer the following about the hypothetical reaction: A+2B+4C -> 2D+E

Answer 1

51)
a) I would expect this reaction to be a multi step reaction, based on the overall reaction because all the reactants isn't present in the rate equation.

b) Comparing lines 1 and 2: you doubled A and the rate is the same. So A is zero order.
Comparing lines 1 and 3, you doubled B and the rate is doubled. So B is first order.
Comparing lines 1 and 4, you doubled everything and the rate octupled (i.e. x8). We know doubling A has no effect, and doubling B would double the rate. This leaves a factor of 4 unaccounted for, which we conclude is the effect of doubling C. Since doubling C gives 4x the rate, C is second order.

c) The overall order of the reaction is 0+1+2 = 3.

d) The rate law is rate = k(B)(C)^2.

e) The rate law doesn't depend on A. This means A is not part of the rate-determining step of the reaction. So it must be involved in another step of the reaction. Therefore, there are at least 2 steps in this reaction.

f) Tripling A should have no effect according to the rate law. Doubling B would by itself double the rate, and tripling C would by itself multiply the rate by 9. The combined effect is to multiply the rate by 18 (= 2 x 9). The expected rate is 18 x 3.6x10^-2 = 0.648 = 6.5x10^-1(rounded to 2 significant figures).

g) This is the hardest part, definitely.
step 1: A + A <==> D
step 2: C + C <==> E
step 3: B + E ---> F
step 4: F + D <==> G the product
Note: other mechanisms are possible. I don't really like the idea of A reacting with itself to form D, but if I didn't include it, there is no step 1 and then the second step is rate-determining.

h)

Is what I got so far. Are they correct? also not sure how to answer part G+H.

Answer 2

can anyone help me with G and H

Answer 3

I've already done this for you, you just needed to fix the stoichiometry, the concept is abstract but that means you can pretty much do anything, you just have to be a little creative. But it's important that you keep the reactions simple (uni- or bimolecular).

g)

Step 1: \( A+C⇌AC\)
Step 2: \(AC+B→ACB\)
Step 3: \( B+2C⇌3D \) (rate-determining step)
Step 4: \( ACB+D⇌ACBD\)
Step 6: \( ACBD+C⇌ACBDC\)
Step 6: \(ACBDC+ heat→E\)

Look how the intermediates cancel out and you're left with the correct stoichiometry:

Step 1: \( A+C⇌\cancel {AC}\)
Step 2: \(\cancel {AC}+B→\cancel{ACB}\)
Step 3: \( B+2C⇌\cancel3 2D \) (rate-determining step)
Step 4: \( \cancel{ACB}+\cancel D⇌\cancel{ACBD}\)
Step 5: \(\cancel{ACBD}+C\rightleftharpoons \cancel{ACBDC} \)
Step 6: \(\cancel{ACBDC}+ heat→E\)

This adds up to: \(A+2B+4C→2D+E\)

As for h) i already answered it before in your previous question. Things to show are the reactants, activated complex, intermediates, and products.

Answer 4

so for the diagram, it means i have to have 5 hills?

Answer 5

also how do i determine how high or low the hills go?

Answer 6

yep, 5 in total. The rate-determining step would be the highest "hill" (because it's the slowest) the rest you can draw as your discretion, although none should be higher than the rate-determining step.

Answer 7

ok i drew it. Would this be right or am i not?

Answer 8

yeah, i would have the intermediates be of different energies than the starting materials, but it's good.

Answer 9

How would you draw the diagram?

Answer 10

the same, but i would have the intermediates have different energies (be at different heights)

Answer 11

how's this?

Answer 12

yeah thats better.