Reaction Kinetics and Stoichiometry -- Mass conservation

Question

This question has been bothering me for some time, and I can't seem to find a good answer online.

Say I have four chemical species $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, and these four react in the following ways:

\begin{align}\ce{ A + A &-> B\\ A + B &-> C\\ A + C &-> D\\ A + D &-> B + C\\ }\end{align}

The kinetic reaction equations for these four species should be: \begin{align} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= -k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\ \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= +k_1[\ce{A}]^2 -k_2[\ce{A}][\ce{B}] +k_4[\ce{A}][\ce{D}]\\ \frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\ \frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\ \end{align}

where $k_i$ is the Arrhenius coefficient for that reaction.

In my mind, the stoichiometry does not seem to add up right. It seems that when the system undergoes $\ce{A + A -> B}$, the concentration of $\ce{A}$ should decrease by twice as much as a reaction like $\ce{A + B -> C}$. Similarly, the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.

Therefore, stoichiometrically, I want to write: \begin{align} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= -2k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\ \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= +k_1[\ce{A}]^2 -k_2[\ce{A}][\ce{B}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\ \frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\ \frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\ \end{align}

Is this wrong-headed? What conceptual issue am I missing?

Answer 1

Reaction rates should be defined in terms of the Extent of reaction ($\xi$) that corresponds to the number of moles (or the molarity for reactions in solution) of specie $i$ divided by the the stoichiometric number, $\nu_i$: $$\xi = \frac{[i]}{\nu_i}$$ Therefore, considering that the first reaction should be: $$\ce{A + A -> B ~=~ 2A -> B}$$ The corresponding reaction rates are: $$\frac{\mathrm{d}\xi}{\mathrm{d}t} = -\frac{1}{2} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$ In this way, the reaction rate is always the same (positive) number, independently on the specie we are referring to. The global rate equations are: \begin{align} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= −2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]−k_3[\ce{A}][\ce{C}]−k_4[\ce{A}][\ce{D}]\\ \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= 2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]+k_4[\ce{A}][D]\\ \frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\ \frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\ \end{align}

It is conceptually wrong to assume that

the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.

The reaction scheme you described is the "parallel reactions" model. The amount of $\ce{A}$ consumed by each reaction depends on the specific rate. The stoichiometry of the first step says that every time 1 molecule of $\ce{B}$ is produced 2 molecules of $\ce{A}$ are consumed. But maybe $k_1$ is very small and so all the other process will consume $\ce{A}$ much more rapidly than this one.

You cannot predict what will happen in a reaction system just looking at the stoichiometry of each step. If you have a large amount of $\ce{C}$, it will probably consume a lot of $\ce{A}$ in the third step, producing a lot of $\ce{D}$ that, in turn, will produce a lot of $\ce{B}$, etc. Stoichiometry is just one piece of information. But kinetics is much more that this.