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.


1 Answer 1


You are right that you can determine the backwards rate if you know the forward rate and the equilibrium constant. In fact, you have the right formula

$$ K_c = \frac{k^f}{k^b} $$

In general, the equilibrium constant can be calculated from the change of the Gibbs free energy during the reaction

$$ \Delta G^r = - R T \ln K_c$$

Therefore in principle you need to calculate the total Gibbs free energy of the reactants and the products and take their difference to get the equilibrium constant. How to do this depends on the nature of the data you have; it might involve quite complicated thermodynamic calculations.

Edit: I was mistaken previously, the equilibrium coefficient is expressed with the rate coefficients.

  • $\begingroup$ Two questions: 1. In the book eq. constant is given as the ratio of $k$'s, not $v$'s. Is it the same thing? 2. There are two constants in the book: $K_p = exp(-\Delta_RG/RT)$, where G is Gibbs energy, and $K_c = exp(-\Delta_RA/RT)$, where A is free energy. Which one do we need? $\endgroup$
    – omican
    Commented Aug 4, 2022 at 9:38
  • 1
    $\begingroup$ I made an edit to the post, it is indeed the ratio of the rate coefficients that you need to use. As per which equilibrium constant you need will depend on the nature of the exercise that you are trying to solve. $\endgroup$
    – Szgoger
    Commented Aug 4, 2022 at 9:44
  • $\begingroup$ I forgot to mention that the ideal gas model is used. It seems that $K_p$ corresponds to partial pressures of the mixture (it is not discussed in detail in the book and thus somewhat confusing) while $K_c$ corresponds to molar concentrations. Probably, $K_c$ is what I need. But it is calculated via free energy (A), not Gibbs energy (G). Suppose we have succesfully calculated $\Delta G$, is it possible to find $\Delta A$ from it? $\endgroup$
    – omican
    Commented Aug 4, 2022 at 9:58

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