## 0 Order vs Independent Rate?

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

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NatBrown1I
Posts: 62
Joined: Fri Sep 28, 2018 12:17 am

### 0 Order vs Independent Rate?

What does 0 order mean versus independent of rate?

Gillian Murphy 2C
Posts: 30
Joined: Fri Sep 28, 2018 12:22 am

### Re: 0 Order vs Independent Rate?

I'm not positive, but I think they're the same thing. When a reaction is 0 order, the rate = k[A]^0, and since [A]^0 is equal to 1, the rate will just be rate = k. Since the rate is always the same regardless of the reactant concentration, the rate is independent of the concentration.

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