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](http://en.wikipedia.org/wiki/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.


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