I am not a chemist. I hope I will be specific enough.

Suppose there are two chemical species $\ce{A}$, $\ce{B}$ with the following properties:

  • at temperature $t < T_r$, no reaction occurs between $\ce{A}$ and $\ce{B}$ (in any combination).
  • at $t\ge T_r$, $\ce{A}$ interacts with itself to create $\ce{A_2}$, $\ce{B}$ reacts with itself to create $\ce{B_2}$, and $\ce{A}$ and $\ce{B}$ are reacting to create $\ce{AB}$.
  • $\ce{A_2}$, $\ce{B_2}$ and $\ce{AB}$ are never reacting.

In experiment, we first mix $\ce{A}$ and $\ce{B}$ in temperature $t<T_r$. Amounts of species mixed are $a$ for $\ce{A}$, $b$ for $\ce{B}$. Then, we add heat to obtain temperature $t\ge T_r$ and start the reaction.

  1. What amounts of $\ce{A_2}$, $\ce{B_2}$, and $\ce{AB}$ can be expected to be produced?
  2. To obtain the amounts, should probability theory be used? E.g., amount of $\ce{AB}$ equals to probability that species $\ce{A}$, $\ce{B}$ will interact ("collide" or similar interpretation).

Assume the rates of the reactions are equal.

  • $\begingroup$ I don't think you can get the relative amounts produced without the rates of the various reactions. $\endgroup$
    – Dan
    May 23 '12 at 20:10
  • $\begingroup$ Will it help to assume equality of the rates? $\endgroup$
    – Calikin
    May 23 '12 at 20:58
  • 1
    $\begingroup$ You'll also need the initial concentrations. See F'x's answer: you need to solve a system of differential equations. $\endgroup$
    – Dan
    May 23 '12 at 21:09

Well, if you assume the rates are known and the reactions' order follows from stoechiometry (e.g. if they are elementary reactions), you can put the chemical kinetics into simple equations:

$$\frac{\mathrm da}{\mathrm dt} = -k_1 a(t)^2 - k_3 a(t)b(t) $$ $$\frac{\mathrm db}{\mathrm dt} = -k_2 b(t)^2 - k_3 a(t)b(t) $$

($t$ here being time, not temperature).

Knowing initial amounts or concentrations $a_0=a(t=0)$ and $b_0=b(t=0)$, you can pretty much integrate the system to find out what happens.

Edit — solving this system for $k_1=k_2=k_3$ yields the quantities of AA, AB and BB at infinite time to be as follows:

$$aa = \frac{a_0^2}{a_0+b_0}$$ $$ab = \frac{a_0 b_0}{a_0+b_0}$$ $$bb = \frac{b_0^2}{a_0+b_0}$$

  • $\begingroup$ Thank you this is very interesting. I was expecting $ab = \frac{2a_0b_0}{(a_0+b_0)^2} \cdot (a_0+b_0) = \frac{2a_0b_0}{a_0+b_0}$, because this is the probability-approach hypothetical result. If we have a box with balls labeled A and B, then probability of drawing ball A twice, with replacing, is $\frac{a_0^2}{(a_0+b_0)^2}$, while p. of drawing ball A and B in any order is: $\frac{2a_0b_0}{(a_0+b_0)^2}$. Prob. of double-B is $\frac{b_0^2}{(a_0+b_0)^2}$. I wonder what happened with the "2" in $2a_0b_0$, in your result. $\endgroup$
    – Calikin
    May 23 '12 at 22:17
  • 1
    $\begingroup$ Because A and B are conserved, you need to have $aa+ab=a_0$, and also $bb+ab=b_0$, which is one way of seeing that your expressions can't work… $\endgroup$
    – F'x
    May 23 '12 at 22:43
  • $\begingroup$ Short test: $a_0 = 10,\space b_0 = 0$. Then, computed values are: $aa = \frac{10^2}{10+0}=10, \space ab=0, \space bb=0$. I think $aa$ should be 5? Five molecules $A_2$, each having two a's – so 10 a's total, like the $a_0$. $\endgroup$
    – Calikin
    May 24 '12 at 21:31
  • $\begingroup$ @Calikin "this is very interesting ...because this is the probability-approach hypothetical result. " It is not that surprising as you used it as an assumption for the kinetic equations. Off course, in reality it can be (very) different, but since you didn't give any extra info,Fx automatically made this assumption as most reasonable assumption. $\endgroup$
    – Greg
    Jun 27 '15 at 8:35

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