6
$\begingroup$

I have a question about this proof that shows the relationship between the rate constant and the equilibrium constant. In this proof, the assumption is that the reaction is $(a+b)$th order in the forward direction and $(c+d)$th order in the reverse direction.

I had always thought reaction orders are empirically determined (and not necessarily whole numbers); can one simply add the stoichiometric coefficients like this? If not, how does one show that equilibrium constant is a ratio of the rate constants since the equilibrium constant requires the concentrations raised to the order of the species' stoichiometric coefficient?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

For a reaction step, the assumption that the rate is proportional to the product of the concentration of reactants raised to their respective powers is valid.

However, generally, reactions are made up of many reaction steps (e.g. SN1 reaction). For such reactions, you cannot use the assumption above. You have to understand the reaction mechanisms in order to determine the rate equation.

(For reactions that involve decimal powers, usually the steady state approximation is applied, and some mathematics is done along with empirical findings in order to determine the rate.)

Addendum: Yes. A reaction step means an elementary reaction. From Wikipedia: "An elementary reaction is a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state.". Therefore, there is only one reaction step.

Likewise, for the equilibrium constant,$K$, in your ratio derivation $K_c = k_f/k_r$, the chemical species must also be from that particular reaction step, and not the global reaction step. It makes sense because the equilibrium constant, K, is related to the free energy change, $\Delta{G}$ of a reaction step. In a two step reaction (e.g. SN1), there are two $\Delta{G}$ and therefore two equilibrium constants.

However, in practice, people do use the overall equilibrium constant, $\beta{}$, which is the product of the equilibrium constants at each reaction step. $\beta{} = K_1K_2K_3...$ (E.g.If the first step gives you 70% yield, and the second step gives you 70% yield, the global step would be 49% yield).(http://en.wikipedia.org/wiki/Chemical_equilibrium#Multiple_equilibria)

In short, for your rate constant,$k$ and your equilibrium constant $K$, in $K_c = k_f/k_r$, they must be of that of a single reaction step.

Improve this answer
$\endgroup$
24
  • $\begingroup$ Is a reaction step an elementary reaction? For an elementary reaction aA+bB->cC+dD, Kc = kf/kr = ([C]^c[D]^d)/([A]^a[B]^b), but for a global reaction qQ+pP->tT, Kc = [T]^x/([Q]^y[P]^z) where x,y,z are empirically determined and no longer equal to the stoichiometric coefficients (for H2O formation this would be Kc = [H2O]^x/([H2]^y[O2]^z))? So the fact that Kc = kf/kr is always true, but Kc is not necessarily a ratio of the concentrations raised to their stoichiometric coefficients? $\endgroup$
    – Yandle
    Commented Oct 11, 2014 at 1:35
  • $\begingroup$ I am confused now. I looked at the derivation for free energy here and it doesn't assume the reaction as elementary or global. This answer and an example in my combustion book for CO2 dissociation (which I do not believe is an elementary reaction) also uses stoichiometric coefficients for Kp. But in another chapter the book states that Rate = k[A]^x[B]^y where x and y are empirically determined. $\endgroup$
    – Yandle
    Commented Oct 11, 2014 at 5:07
  • 1
    $\begingroup$ @Yandle, I've edited my answer to explain further. $\endgroup$
    – t.c
    Commented Oct 11, 2014 at 5:44
  • $\begingroup$ So when we derive K from Gibbs free energy (like in the Wikipedia link), the assumption is that the reaction is an elementary reaction? My confusion: For global reaction (A+B->C+D), Rate = kf[A]^x[B]^y where x and y are not the stoichiometric coefficients and hence K=kf/kr would be ratios of the concentrations raised to empirically determined values x,y etc.. However, if I go B=K1K2... I still end up with concentrations raised to the stoichiometric coefficients of the global reaction as if the global reaction was elementary, since intermediate terms cancel. $\endgroup$
    – Yandle
    Commented Oct 11, 2014 at 6:18
  • 1
    $\begingroup$ 1. Yes your assumption that it is an elementary reaction is correct. 2. Yes for global reaction (A+B->C+D), you would not be able to use K = kf/kr. 3. While you can multiply the eqm constants to obtain the global eqm constant, you cannot multiply the rate constants to obtain the global rate constant (in fact there is no such thing since the overall rate is dependant on many rate constants, for example overall rate = k1/(k2-k1) [A]^3/2 [B]^1/2)). Therefore, you cannot use $\beta{}=kf_{overall}/kr_{overall}$ - it would not be mathematically equivalent. K = kf/kr must be of a reaction step. $\endgroup$
    – t.c
    Commented Oct 11, 2014 at 6:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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