## 15.19 part a

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

Paula Dowdell 1F
Posts: 72
Joined: Tue Nov 15, 2016 3:00 am

### 15.19 part a

Hi! Can someone explain to me how they determined that [B] is second order?
I determined [A] by solving as I have previously seen: 1.25^a/2.5 = 8.7/17/4 => 0.5^a=0.5 (first order)
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Alvin Tran 2E
Posts: 39
Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.19 part a

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.