Suppose we have have an ideal gas mixture and a reversible elementary reaction: $$ \ce{O + H_2 <=> H + OH} $$ Its forward reaction rate may be computed as follows: $$ \frac{d[\ce{H}]}{dt} = \frac{d[\ce{OH}]}{dt} = -\frac{d[\ce{O}]}{dt} = -\frac{d[\ce{H2}]}{dt} = k^{(f)}\left[\ce{O}\right]\left[\ce{H2}\right] = A T^n e^{\left(\frac{E_a}{RT}\right)}\left[\ce{O}\right]\left[\ce{H2}\right] $$ where $A, n$ and $E_a$ are known values. The exponents of concentrations $\left[\ce{O}\right]$ and $\left[\ce{H2}\right]$ are both equal to one due since the reaction is elementary.
Also, there is thermodynamic data for the species - heat of formation, entropy, heat capacity - for all species. For example, for H these values are: $$ \Delta H^{298K}_f\left(\ce{H}\right) = 52.1~\text{kcal/mol}, \quad S^{300K}\left(\ce{H}\right) = 27.4~\text{cal/mol K},\quad C_p(\ce{H}) = 4.97~\text{cal/mol K}. $$ For simplicity, heat capacity is assumed constant here.
Parameters $A, n$ and $E_a$ for backward reaction rate $\left(k^{(b)}\right)$ are unknown. According to Warnatz's et al "Combustion" book backward reaction may instead be computed from equilibrium constant, which, in turn, is determined by thermodynamic data: $$ K_c = \frac{k^{(f)}}{k^{(b)}}= exp\left(-\Delta_R\bar{A}^0/RT\right) $$
But how to compute $\Delta_R\bar{A}^0$?
I assume that it equals to an increment of free energy (Helmholtz function) $A = U - TS$ corresponding to consumption of one mole of each reactant, but not sure how to properly calculate it from thermo data above.