# Law of mass action

If you have the forward reaction

$$\ce{2X ->[K] P}$$

which of the following systems of differential equations would model the reaction's kinetics?

$$\begin{array}{rl} \dfrac{\mathrm{d}[\ce{X}]}{\mathrm{d}t} &= -2K[\ce{X}]^2\\\\ \dfrac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= K[\ce{X}]^2\end{array} \tag{system 1}$$

or

$$\begin{array}{rl} \dfrac{\mathrm{d}[\ce{X}]}{\mathrm{d}t} &= -K[\ce{X}]^2\\\\ \dfrac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= K[\ce{X}]^2\end{array} \tag{system 2}$$

The first answer is correct. If the reaction is $\ce{2X->P}$, then two units of $\ce{X}$ should disappear for every unit of $\ce{P}$ formed. Only the first possibility meets this criterion.
The rate law for the reaction could be anything. In your example you have apparently assumed that it is a second order reaction in $\ce{X}$, but we could make any other assumption. For example, product formation rate could be $\frac{dP}{dt}=k~X^{1.2}$ or $\frac{dP}{dt}=k~\frac{X}{K_m + X^{2.4}}$. But if the reaction you want to model is really $\ce{2X->P}$, then these rate laws would imply that $\frac{dX}{dt}=-2~k~X^{1.2}$ or $\frac{dX}{dt}=-2~k~\frac{X}{K_m + X^{2.4}}$.
Thus, if $\frac{dP}{dt}=k X^2$, then the rate of X depletion must be -2 times that, or $\frac{dX}{dt}=-2k X^2$.
Here is another way to see the problem with the rate law without the $-2$: suppose there was a reaction $\ce{X->P}$, i.e. only one molecule of $\ce{X}$ was needed to form $\ce{P}$. Suppose also that this reaction followed second-order kinetics in $\ce{X}$, i.e. $\frac{dP}{dt}\propto X^2$. What would the rate laws for this alternate reaction be?
• another way to look at it is that if the reaction is $\ce{A + B \rightarrow P}$ then we would have say 1 mol A and 1 mol of B. But in your case if we have 1 mol of A it behaves as 1/2 mol because its A + A reacting, so we need to double up. – porphyrin Dec 10 '16 at 21:31