Let's say we have:

$$ \begin{align} \ce{A &-> B}&\quad&\text{(fast equilibrium)}\label{rxn:1}\tag{1}\\ \ce{2 B + C &-> D}&\quad&\text{(slow)}\label{rxn:2}\tag{2}\\ \ce{D + E &-> F}&\quad&\text{(fast)}\label{rxn:3}\tag{3} \end{align} $$

Find the rate expression for the formation of $\ce{F}$.

Since I am writing the rate law for $\ce{F}$, should I write it as

$$\text{rate} = k_3[\ce{D}][\ce{E}]$$

because that's the reaction step that actually includes $\ce{F}$ as the product, or do I always have to write is in terms of the slow step i.e.

$$\text{rate} = k_2[\ce{B}]^2[\ce{C}]?$$

What happens with the leftover $[\ce{B}]$ and $[\ce{A}]$ in the steps \eqref{rxn:2} and \eqref{rxn:1}, respectively — do I include those concentrations as well?


I don't know how you call it in English but you can apply this approximation $\frac{d[D]}{dt} = 0$ because (3) is fast and $D$ have a really short lifetime, so you can say that


This approximation is why the teacher is always saying that only the slow reaction is important

Fast equilibrium means that the equilibrium is quickly governed by thermodynamics laws and so the constant

$$K = \frac{[B]}{[A]}$$

is not variating, so you can express $B$ in function of $A$

$$\frac{d[F]}{dt} = k_2K^2[A]^2[C]$$

I don't know how far you need to go but if you really want to solve for $[F]$ you need to write an equation of conservation

$$[N] = [A] + [B] + [C] + [D] + [E] + [F] = cst$$ $$d[A] + d[B] + d[C] + d[D] + d[E] + d[F] = 0$$

$$d[D] = 0$$ $$d[E] = 0$$

You can also say that $d[C] \approx 0$ because it looks like it's just a partner of collision, (it's the solvent) because the solvent is in much larger quantity you can say the concentration stay pretty the same

$$d[A] + d[B] = - d[F]$$

$$[A] - [A]_0 + [B] = -[F]$$

$$[A](1+K) = [A]_0 - [F]$$

$$[A]^2= \frac{([A]_0 - [F])^2}{(1+K)^2}$$

$$\frac{d[F]}{dt} = K_0\frac{([A]_0-[F])^2}{(1+K)^2}$$

with $K_0 = k_2K^2[C]_0$

| improve this answer | |
  • 2
    $\begingroup$ Steady state approximation $dD/dt=0$ $\endgroup$ – porphyrin May 2 '19 at 7:07
  • $\begingroup$ thank you for the precision, when you solve you obtain the equation of $F(t)$, you can also change the constant to see the influence desmos.com/calculator/7pq68olwtj $\endgroup$ – Tom May 2 '19 at 10:03
  • $\begingroup$ To second @porphyrin, a situation like the advancement of your reaction (3) being determined by the slowest reaction (2) of the three is referred as Bodenstein principle, too (en.wikipedia.org/wiki/Max_Bodenstein). $\endgroup$ – Buttonwood May 11 '19 at 18:33

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