## 15.17

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

Mia Navarro 1D
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

### 15.17

Why, when solving for the rate, is [B] squared?

Caroline LaPlaca
Posts: 30
Joined: Sat Jul 22, 2017 3:01 am

### Re: 15.17

No, cause its not like the K concentration thing.

Johann Park 2B
Posts: 51
Joined: Thu Jul 27, 2017 3:01 am
Been upvoted: 1 time

### Re: 15.17

With respect to [B], the reaction is second-order. You use the different values for the rates and concentrations to solve for the exponents (x, y, z) on the concentrations.

OliviaShearin2E
Posts: 37
Joined: Fri Sep 29, 2017 7:05 am

### Re: 15.17

Johann is right but to be clearer, first order in B means that the rate is proportional to $[B]^{1}$, second order in B means that the rate is proportional to $[B]^{2}$. and third order in B means that the rate is proportional to $[B]^{3}$.

Sandhya Rajkumar 1C
Posts: 50
Joined: Fri Jun 23, 2017 11:40 am

### Re: 15.17

The reaction is second order with respect to B because when the amount of B is doubled (increased by a factor of 2), the rate increases by a factor of 4, so instead of it being a 1:1 ratio, which would make it first order, it's a 1:2 ratio, which makes it second order.