## first order

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

annikaying
Posts: 94
Joined: Sat Sep 14, 2019 12:16 am

### first order

What does it mean for a reaction if it is first order versus second order or any other order?

MinuChoi
Posts: 100
Joined: Wed Sep 18, 2019 12:15 am

### Re: first order

For a first-order reaction, the rate is directly proportional to the concentration (e.g. if the concentration of a reactant was doubled, the rate of the reaction would be doubled.)
For a second-order reaction, the rate is proportional to the second power (square) of the concentration (e.g. if the concentration of a reactant was doubled, the rate of the reaction would be quadrupled).
This power patter applies to the other orders, including a zeroth-order reaction (which would mean that the rate is unaffected if the concentration of a reactant is changed).

Myka G 1l
Posts: 100
Joined: Fri Aug 30, 2019 12:17 am

### Re: first order

In a first order reaction, the rate of reaction is directly proportional to the concentration of a single reacting substance. Whereas in a 2nd order reaction the rate depends on the concentration of 2 reacting first order substances or a single reactant with a squared concentration. Because it depends on a single substance, the graph of a first order reaction is a straight line.

Lindsey Chheng 1E
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

### Re: first order

annikaying wrote:What does it mean for a reaction if it is first order versus second order or any other order?

0 order: Rate = k
k = mol/L x min
1st order: Rate = k[A]
k = 1/min
2nd order: Rate = k[A][B] or k[A]^2
k = L/mol x min
Last edited by Lindsey Chheng 1E on Sun Mar 01, 2020 2:37 pm, edited 1 time in total.

TimVintsDis4L
Posts: 104
Joined: Sat Aug 17, 2019 12:17 am

### Re: first order

First order reactions and second order reactions are way more common than third and forth, those are very unlikely because it requires 3 or 4 separate entities to perfectly come together, which is very unlikely.