# 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?

• Do you have a citation for this? I though the molar coefficients are taken into account by the multiplication of the reactant side. IE, that's where the squaring comes from. Jun 2, 2015 at 13:39
• Don't confuse stoichiometry with mechanism. Mechanism determines the order of a reaction, i.e. the exponents in the rate law. "Elementary" reactions like 2X -> P with two reactant molecules will indeed have a second-order rate law. But that's just the exponent. The stoichiometry is another matter. If two molecules of X disappear to make one molecule of P, then the rate laws you write down better reflect that. A citation for this could be any book on chemical kinetics. I recommend Fogler. Extra thought question: what would the rate laws look like for an "elementary" reaction 2X -> 2P? Jun 2, 2015 at 15:18
• 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. Dec 10, 2016 at 21:31
• @CurtF. Thanks .Correct me for the following :1-i.e. $\frac{dP}{dt}\propto X^2$. What would the rate laws for this alternate reaction be($\ce{X->P}$)? $\frac{dP}{dt}=\frac{-dX}{dt} =kX^2$ 2- what would the rate laws look like for an "elementary" reaction 2X -> 2P? $\frac{dP}{dt}=\frac{-dX}{dt} =2kX^2$ Jan 26 at 21:11
• @porphyrin Thanks. Correct my answer: ** what would the rate laws look like for an "elementary" reaction A + A-> 2P?** $\frac{1}{2}\frac{dP}{dt}=\frac{-dA}{dt} =kA^2$ Jan 26 at 21:18