Units [ENDORSED]

[Lisa Lim 3G]

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

[EPerez1B]

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]

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]

Yes, it'll be always M/Sec

[Samuel_Vydro_1I]

Divide by Molarity for each order essentially

[Chew 2H]

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

[Dang Lam]

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]

Yes! M/s