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I have a box of 5 types of particles, $A$, $B$, $C$, $x$ and $y$. They react in the following way.

$\ce{$A$ + $B$ -> $C$ + $x$}$

and

$\ce{$C$ + $y$ -> $A$ + $B$}$

If only the first reaction was happening then $N(C) = k \cdot N(A)N(B)$. Here $N(I)$ is the number of particles of $I$ in the box. Can I write the total number of $C$ in the box as $\text{number of $C$ created} - \text{number of $C$ destroyed} = N(C) = k \cdot N(A)N(B) - k' \cdot N(C)N(y)$?

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    $\begingroup$ Not the total number but instead the number /time which is the rate $dN_c/dt$ . These equations then have to be integrated knowing the initial number present. Normally we do not use numbers in a box but equivalently and more conveniently concentration. In your scheme only x is present at long times given that A, B and y are present initially. $\endgroup$ – porphyrin Jun 19 '18 at 7:08
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Actually, the rate law for the species $\ce{C}$ is written as:

$$\frac{\mathrm{d}\ce{C}}{\mathrm{d}t} = k \ce{A} \ce{B} - k' \ce{C} \ce{y}$$

If you integrate this, you would get an equation in terms of the number of particles, as you require. However, this is a very typical integral, and there would be a number of other arbitrary constants, which would make the final result not equal to the one you've shown.

What you've written is a mistake on replacing the rate of formation of $\ce{C}$ with its number of particles itself. Both are not equivalent.

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