In many textbooks, it is written that

For order (m + n), the rate constant, k has units of mol1−(m+n)·L(m+n)−1·s−1

However, by the Arrhenius equation, k can be expressed as

k = PZexp(-Ea/RT)

and its unit doesn't seem to change with the order.

I think if reaction velocity, v is defined as

v = k([A]/cref)m([B]/cref)n

like the definition of equilibrium, the unit of k becomes mol·L−1·s−1 and it looks more reasonable. Am I wrong?


The rate has units $\displaystyle \frac{concentration}{time}$ and is $\displaystyle \frac{d[C]}{dt}$ and the right hand side of the equation has always to have the same units as the left hand side, so for first order reaction consuming C;

$\displaystyle \frac{d[C]}{dt} = -k_1[C]$

and the rate constant $k_1$ therefore has units $\displaystyle \frac{1}{time} $ usually $\mathrm{s^{-1}}$ and for second order

$\displaystyle \frac{d[C]}{dt} = -k_2[C][C]$

the units of rate constant $ k_2$ are $\displaystyle\frac{1}{time \times\,concentration}$ which is $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$.

The units for other examples follows in the same way.

In the Arrhenius equation as the exponential is dimensionless the other terms must have the units of the rate constant, which means units of PZ in your notation. These units must always depend on the order of reaction as described above and if they don't then there is an error somewhere.


By the Arrhenius equation, $k$ (rate constant) can be expressed as,

$$ k = \rho Z \times \exp\bigg(-\frac{E_a}{RT}\bigg) $$

where $\rho$ (aka. steric factor, or probability factor) is defined as the ratio of $A$ (Arrhenius contant or pre-exponential factor) and $Z$ (collision frequency) (Ref. 1).

And, units of $A$ are same as that of units of $k$ (Ref. 2), hence the units of $\rho Z$ is same as the units of $A$, or $k$.


  1. Steric factor - Wikipedia: para 1
  2. Pre-exponential factor -Wikipedia: para 2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.