## first vs. second order

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

005199302
Posts: 108
Joined: Fri Sep 28, 2018 12:15 am

### first vs. second order

How is a second order reaction different than a first order reaction?

Lexie Baughman 2C
Posts: 30
Joined: Sat Oct 06, 2018 12:16 am

### Re: first vs. second order

In a first order reaction, there will be one reactant present in the rate law. For a second order reaction, you can either have a rate law with one reactant to the second order, or with two reactants both to the first order. I remember it by adding the exponents of all the reactants in the rate law - whatever the exponent is, that's the order of the rate law.

kateminden
Posts: 63
Joined: Fri Sep 28, 2018 12:24 am

### Re: first vs. second order

In first-order reactions, the reaction rate is directly proportional to the concentration of one of the reactants. These reactions usually take the form of A → products. In contrast, a second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. Second-order reactions often take the form of either A + B → products or 2A → products.