## Units [ENDORSED]

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

Lisa Lim 3G
Posts: 12
Joined: Fri Sep 25, 2015 3:00 am

### Units

Does the rate constant k of zero-order reactions always have the units of mole/L*sec?

EPerez1B
Posts: 27
Joined: Fri Sep 25, 2015 3:00 am

### Re: Units

Yes, other ways of writing the same units would be M (molarity)/s or Ms^-1 or mol(L^-1)(s^-1).

Carolyn Stephens 3H
Posts: 15
Joined: Fri Sep 25, 2015 3:00 am
Been upvoted: 1 time

### Re: Units  [ENDORSED]

An easy way to determine the units of k is to understand that the rate needs to have units of M/s so for 0 order rate=k and therefore k=M/s. for 1st order: rate=k[A] where [A] has the units of M and so when multiplied with k, k must have units of s^-1. For second order rate=k[A]^2 where [A]^2 has units of M^2 so in order to get rate = M/s, k must be M^-1 s^-1.

Michelle Kam 1F
Posts: 10
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Units

Yes, it'll be always M/Sec

Samuel_Vydro_1I
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

### Re: Units

Divide by Molarity for each order essentially

Chew 2H
Posts: 34
Joined: Fri Sep 29, 2017 7:05 am

### Re: Units

There's also one using kPA/sec. Does anyone know how kPA relates to the rate of reactions?

Dang Lam
Posts: 55
Joined: Thu Jul 27, 2017 3:01 am

### Re: Units

yes because for oth order, the rate law is written as:
rate = k[A]^0
which is equal to rate = k
since the unit for rate = mol/l*s, the unit for k is therefore mol/l*s

Jennie Fox 1D
Posts: 66
Joined: Sat Jul 22, 2017 3:01 am

Yes! M/s