## 2nd order rates

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

804994652
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

### 2nd order rates

Just to clarify:
If a 2nd order rate looked like rate = k [OH-] [H3Br]
the rate is second order for the reaction
but the rate is first order for each reactant?

Nathan Tran 4K
Posts: 92
Joined: Fri Sep 28, 2018 12:16 am

### Re: 2nd order rates

Yes, that is true. The total rate order involves summing the exponents of each reactant.

Brian Chang 2H
Posts: 65
Joined: Fri Sep 28, 2018 12:17 am

### Re: 2nd order rates

That's perfectly fine.

The total order of the reaction is calculated by summing up the exponents the individual reactants. (1+1=2)

Ahmed Mahmood 4D
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

### Re: 2nd order rates

Yes. In order to get overall order, you simply add the powers of the reactants--so, two reactants of first order will form a second order rate law for the overall reaction.

Ashley P 4I
Posts: 64
Joined: Wed Nov 15, 2017 3:04 am

### Re: 2nd order rates

Ahmed Mahmood 4D wrote:Yes. In order to get overall order, you simply add the powers of the reactants--so, two reactants of first order will form a second order rate law for the overall reaction.

Thank you! I’m actually starting to understand this. Thanks to your comment. Do you think we will be expected to know 3rd orders? Or would be just need to be able to identify them?

Ahmed Mahmood 4D
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

### Re: 2nd order rates

I’d look through the textbook/notes and memorize concepts and formulas relating to third order rxn. Anything that isn’t mentioned probably isn’t necessary.

Jack Hewitt 2H
Posts: 67
Joined: Fri Sep 28, 2018 12:27 am

### Re: 2nd order rates

Yes, because total rate order is found by adding the exponents of the reactants.