15.19a

Abigail Urbina 1K
Posts: 102
Joined: Thu Jul 27, 2017 3:01 am

15.19a

Can someone explain to me how they found the order of reactant B? I know to look at experiments 2 and 3 to solve this. I'm just confused because the solutions manual says that increasing the concentration of B by the ratio 3.02/1.25 increases the rate by (3.02/1.25)^2, so the reaction is then second order. I'm not really sure how they came up with that logic. Could someone clarify?

Silvino Jimenez 1A
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

Re: 15.19a

I believe they made a very generous estimation and rounded the increase in rate between experiments 2 and 3. The change in [B] increased from 1.25 to 3.02 which is approx an increase of a multiple of 2. The change in the concentration is approx. an increase of a multiple of 3. Here is where I was confused too.

Here I am assuming they rounded the increase i concentration rate to a multiple of 4. This would allow us to conclude that the concentartion was squared thus making it a second order reaction

Alvin Tran 2E
Posts: 39
Joined: Fri Sep 29, 2017 7:06 am

Re: 15.19a

To find the order of [B] you want to compare reaction 1 and 3 because the concentrations of A and C stay the same.
So we have $\frac{Rate 3}{Rate 1} = \frac{50.8}{8.7} = \frac{k(1.25)^{N}(3.02)^{M}(1.25)^{L}}{k(1.25)^{N}(1.25)^{M}(1.25)^{L}}$
5.84 = 2.416M
Solve for M and you get 2.
So B is second order.

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