I would like to calculate the pre-exponential factor of a surface reaction using transition state theory. I've seen two ways to calculate it but I get different answers for each method (and I don't understand why):
$A = \frac {k_b T} {h} \frac {q^{TS}} {q^{reactant}}$ ... (1)
$A = \frac {k_b T} {h} \exp \bigg({\frac {S^{TS} - S^{reactant}} {R}}\bigg)$ ... (2)
where $q^i$ is the partition function of species i, $S^i$ is the entropy of species i, $k_b$ is the Boltzmann constant, $T$ is temperature, $h$ is Planck's constant and $R$ is the molar gas constant.
I assumed the adsorbates only have vibrational contributions to $A$ and the vibrations can be described using the Harmonic approximation.
I am using the following equations to estimate $q^i$ and $S^i$:
$q^i_{vib} = \prod_j^{n_{vib}} \frac {1} {1-\exp \big(- \frac {\Theta_j} {T}\big) }$ ... (3)
$\frac {S^i_{vib}} {R} = \sum_j^{n_{vib}} \frac {\Theta_{V,j}} {T} \frac {\exp \big({- \frac {\Theta_{V, j}}{T}}\big)} {1 - \exp\big( - \frac{\Theta_{V, j}}{T}\big)} - \ln \bigg(1-\exp\big(-\frac{\Theta_{V,j}}{T}\big)\bigg)$ ... (4)
where $n_{vib}$ is the number of vibrational modes, $\Theta_{V,j}$ is the vibrational temperature of a mode (in K).
Inputs
T = 750 K
Reactant $\Theta_V$ in K:
0.42190081
0.08952052
0.08826819
0.06520753
0.05161751
0.05077054
Transition State $\Theta_{V}$ in K:
0.21401905
0.06477819
0.05798548
0.05276798
0.04599558
0.14870034$i$ (Imaginary frequency is ignored)
Results
$A_{eq 1} = 1.61\times 10^{13} s^{-1}$
$A_{eq 2} = 1.36\times 10^{13} s^{-1}$
Intermediate values
$q^{TS} = 9.743$
$q^{reactants} = 9.430$
$\frac {S^{TS}}{R} = 4.93$
$\frac {S^{reactant}} {R} = 5.07$
