Chemical reaction notation in differential equations

It's common to see chemical reactions in this notation:

$A \overset{k}{\longrightarrow} \emptyset$

which means $A$ is degraded with rate $k$. What's the proper way to convert this to an ordinary differential equation? does the notation suggest that $A$ is degraded in linear proportion to its concentration:

$\frac{dA}{dt} = -k[A]$

or does it mean that the degradation rate is constant irrespective of $[A]$?

$\frac{dA}{dt} = -k$

or is the original arrow notation perhaps ambiguous? references would be helpful.

• Typically, if the reaction is just a single elementary step, the coefficient of the reactant gives the order with respect to that reactant. With this being the case (ie you are working with an elementary reaction step), it can't be zero order. This question is related:chemistry.stackexchange.com/questions/263/zero-order-reactions – Tyberius Apr 17 '17 at 1:03
• This wouldn't qualify as computational chemistry. – Tyberius Apr 17 '17 at 4:02

In general, there is no way to know exactly how a net reaction should be turned into a differential equation. Or, perhaps it's more accurate to say that you can't know which differential equation should describe the reaction.

In your case, since you're talking about a decay process, this will always be first-order with respect to concentration and should thus be written as,

$$\frac{dA}{dt} = -k[A].$$

Some other examples are:

$\textbf{Elementary Reactions:}$

If a reaction consists of only a single step, or the rate is known to be affected by only a single step, then the rate can be written down similar to how you guess above. One example are decay processes, which are always first-order. Consider, $$\ce{H2O2(l) ->[k]H2O +\frac{1}{2}O2}$$ Because we know this is a decay process, we can write down,$$\frac{-d[\ce{H2O2}]}{dt}=k[\ce{H2O2}]$$ $\textbf{Reactions with Simple Physical Models:}$

I'm thinking specifically of the derivation of the Michaelis-Menten Equation. I'll provide a sketch of this.

The goal is to describe the kinetics of enzyme catalysis. Because the reactant is relatively localized (i.e. it's trapped in an enzyme), there can't be that many different states. At most, we can have this sequence of steps: $$\ce{E +S<=>[k_1][k_{-1}]ES<=>[k_2][k_{-2}]EP<=>[k_3][k_{-3}]E +P}$$where $\ce{E}$ is the enzyme, $\ce{ES}$ is the enzyme-substrate complex, and $\ce{EP}$ is the enzyme-product complex.

On nothing but physical intuition, we make two assumptions:

• The rate of change of $\ce{ES}$ is zero (i.e. constant concentration)
• The reverse step involving $\ce{EP}$ can be ignored, assuming $k_3\gg k_{-3}$

From these assumptions, and equilibrium reactions, we can write down a number of differential equations which give us a very useful and quite accurate equation.

The differential equations one can write down abide by the law of mass-action, which basically just says if we write down all the places some mass can go, then we can know the rate of change for a particular step.

For instance, $$\frac{d[\ce{ES}]}{dt}=k_1[\ce{E}]+k_1[\ce{S}]+k_{-2}[\ce{EP}]-k_{-1}[\ce{ES}]-k_2[\ce{ES}]$$

As you can see, you basically just pick a step and tally up all the ways things are reacting to form a molecule at this step and all the ways things are leaving this step.

$\textbf{Takeaway:}$

Pretty much, in order to write a differential equation describing rate of changes of reactions, we must have some physical intuition about what is going on. Once we have this intuition, we can almost always get somewhere meaningful.