Reaction Kinetics and Stoichiometry — Mass conservation

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?

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.

• Should your third equation read $-\tfrac{1}{2}\tfrac{d[\mathrm{A}]}{dt}$? I can see how this works with one reaction, but A reacts with three other species as well; where does the factor of 1/2 fit in to the full differential equation? – SteelAngel Feb 2 '15 at 15:12
• I don't understand your comment. Third equation states that $\frac{d\xi}{dt} corresponds to$\frac 1 \nu_i \frac{d[i]} {dt}$for every specie$i$. The full differential equation is just the combination of the expressions for all the reagents in term of$\xi$so that the stoichiometric number is automatically take into account. I edited the answer adding the global expressions for each concentration variatiation. – Renato Feb 3 '15 at 9:54 • I have updated your post with some more chemistry markup. If you want to know more, please have a look here and here. Even if it is tempting, please do not use markup in the title field, see here for details. You might also want to revisit your comment, that got a little messed up. – Martin - マーチン Feb 3 '15 at 11:12 • Thank you for the more expansive explanation. What I talked about the third equation, you wrote:$\frac{\mathrm{d}\xi}{\mathrm{d}t} = -\frac{1}{2}\frac{[\ce{A}]}{\mathrm{d}t}$. The right hand side should be a derivative,$-\frac{1}{2}\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}\$. – SteelAngel Feb 3 '15 at 11:58
• Yes, you are absolutely right. I missed a "d". I edited my answer to correct the problem. – Renato Feb 3 '15 at 12:18