## 15.19a Order

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Kathleen Vidanes 1E
Posts: 62
Joined: Fri Sep 29, 2017 7:07 am

### 15.19a Order

Hi, I'm confused as to how we can find the order for B using experiments 2 and 3, especially since the initial concentration of A changes from 2.5 to 1.25. Could someone please explain how the order is found? Or if there is another method of determining the order for B? Thank you!

Beza Ayalew 1I
Posts: 58
Joined: Fri Sep 29, 2017 7:07 am

### Re: 15.19a Order

Just like in 17, you use the 2 rows where Bs concentration is changing, but As and Cs concentrations are not, which in this case would be experiments 1 and 3. Then you can divide the 2 different B concentrations to get a number to the power of n, and then divide the initial rates to solve for what the b concentrations to the power of n is equal to. In this case, I got2.416^n=5.839, so because n in that equation is equal to 2, you know the concentration of B is raised to the second power in the rate law

Beza Ayalew 1I
Posts: 58
Joined: Fri Sep 29, 2017 7:07 am

### Re: 15.19a Order

I just realized this isn't what the solutions manual did, but it's what made the most sense to me personally

Gwyneth Huynh 1J
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

### Re: 15.19a Order

I found the order for reactant B using experiments 1 and 3 instead because only the concentration of B is changed between those two reactions. I don't think you can compare experiments 2 and 3 because both the concentrations of A and B are changed. As illustrated by comparing experiments 1 and 2, changing the concentration of reactant A does affect the rate. Therefore, we can't compare experiments 2 and 3 because both A and B would be affecting the rate of the reaction.

Tiffany 1B
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

### Re: 15.19a Order

Yes, exactly what Beza was saying. You compare the two rate laws and their initial rates. Although the solutions manual does say 2&3 for the order of B you would actually use 1 and 3. So to solve for B it would look something like rate3/rate 1= [A]^l[B]^m[C]^n/[A]^l[B]^m[C]^n where everything would cancel except for the two [B]^m which you could then use to solve for the order of B. Hope that helps

Ashley Davis 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.19a Order

Just to understand what the soln manual is trying to say, how would we determine that increasing the concentration of B by the ratio 3.02/1.25 increases the rate by (3.02/1.25)^2? I'm not sure how we would figure this given how we've gone about solving these kind of problems so far. I personally prefer Beza's method.

Kathleen Vidanes 1E
Posts: 62
Joined: Fri Sep 29, 2017 7:07 am

### Re: 15.19a Order

Thank you so much for all of the help! I do prefer Beza’s method as well and believe that it much easier to understand, however I would also like to know how the solutions manual went about solving the problem. If anyone could explain, that would be great!