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### 15.17

Posted: Mon Feb 26, 2018 11:35 pm
Why, when solving for the rate, is [B] squared?

### Re: 15.17

Posted: Tue Feb 27, 2018 1:25 am
No, cause its not like the K concentration thing.

### Re: 15.17

Posted: Tue Feb 27, 2018 10:33 am
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.

### Re: 15.17

Posted: Tue Feb 27, 2018 5:36 pm
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 $[B]^{3}$.

### Re: 15.17

Posted: Tue Feb 27, 2018 10:03 pm
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.