## Integrated Rate Laws [ENDORSED]

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

William Lan 2l
Posts: 73
Joined: Fri Sep 29, 2017 7:07 am

### Integrated Rate Laws

How do we relate -d[A]/dt with k[A]? Basically, why are they equal to each other? And what's the difference between integrated rate laws and just rate laws in general?

Sohini Halder 1G
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
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### Re: Integrated Rate Laws  [ENDORSED]

Say A is your reactant. As the reaction progresses, A decreases (if first order, it decreases exponentially). Therefore, the rate is the negative change of the concentration of A over time, given by -d[A]/dt. This is equivalent to saying that the rate of the reaction is proportional (with a proportionality constant of k) to the concentration of A, given by k[A]. These inherently say the same thing so we can relate them by an equal sign. Once you integrate both sides, you get your integrated rate law.

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