Units in modified Arrhenius equation?

The modified Arrhenius equation is used to express the rate constant in a chemical mechanism model I'm working with. The equations is as follows: $$k_\mathrm{f} = A\times T^b\times\exp\left(-\frac{E_\mathrm{a}}{RT}\right)$$ The paper states that "Units are Moles, cm3, Seconds, K, and Calories/Mole" so what would be the final units of the rate constant? I'm a bit confused due to the presence of the exponential function. When I simply plug the values in as given in the paper the rate becomes huge when multiplied by the molar concentration, is this because it gives molecules per second? To give you and idea for the following reaction: $$\ce{C2H5OH + OH <=>C2H4OH + H2O}$$ The Arrhenius constants are as follows: $$A = 1.74E+11$$ $$b = 0.27$$ $$E_a = 600.0$$ I am yet to calculate the reverse reaction using Gibbs Free Energy, is it just equally as large and thus it all cancels out or are the final units really not in $\mathrm{mol\ s^{-1}}$?

1. The exponential must be dimensionless. That means if $E_a$ is in units of $\mathrm{cal\cdot mol^{-1}}$, then $R$ must be units of $\mathrm {cal\cdot mol^{-1}\cdot K^{-1}}$. Make sure you use the right value of $R$ for these units; $R=8.314~ \mathrm{J \cdot mol^{-1}\cdot K^{-1}}$, but $R = 1.987~ \mathrm{cal \cdot mol^{-1}\cdot K^{-1}}$. Temperature must obviously be in Kelvins.

2. The example reaction you gave is bimolecular, so I would think $k_f$ is meant to have units of $\mathrm{cm^3 \cdot s^{-1}\cdot mol^{-1}}$. That way, when you multiply $k_f$ by the molar concentration of both reactants, e.g. ethanol and hydroxyl, you get a reaction rate that is in units of $\mathrm{mol \cdot s^{-1}\cdot cm^{-3}}$.

3. The remaining terms must therefore combine to give the right units for $k_f$. Since the units of $T^b$ will have units of $\mathrm K^b$, then the units for $A$ will be the units for $k_f$ divided by $\mathrm K^b$, i.e. the units of $A$ are $\mathrm{{cm}^3 \cdot s^{-1}\cdot mol^{-1} \cdot K^{\it-b}}$.

• You additionally mentioned multiplying by the "molar" concentration, but the concentration in mol per cm^3 will be 1000-fold less than the concentration in mol per L, so if you really are using molar concentrations, make sure you divide them both by 1000 first. – Curt F. Jul 29 '15 at 21:26