## Doubling Concentrations

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

David Sarkissian 1K
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### Doubling Concentrations

Does doubling concentrations double the order of a reaction or does it just increase the order by 1?

juliasloan_4g
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Joined: Fri Sep 28, 2018 12:28 am

### Re: Doubling Concentrations

You can only find orders based on experimental data so this depends on the reaction. If you double the concentration and nothing happens to the rate, this is a zero order reaction. If you double the concentration and the rate also doubles, this is a first order reaction. If you double the concentration and the rate is squared this is a second order reaction.

Joonsoo Kim 4L
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### Re: Doubling Concentrations

Changing the concentration in experiments can show you which order the reaction is, but it doesn't change the order of the reaction. In other words, in rate=k[A]^n, changing [A] does not change n.

MichelleRamirez_2F
Posts: 63
Joined: Fri Sep 28, 2018 12:28 am

### Re: Doubling Concentrations

If you are talking about a first order reaction the doubling of a reactant means the reaction rate is doubled. If it's a second order reaction then the reaction rate is increase by a factor of x^2, where x would be your reaction rate.