# Solving a kinetic rate law

Here is a problem I just don't see how to get around... The following equation describes the oxidation by $\ce{O_2}$ of a phenolic compound, catalysed by an enzyme $\ce{E}$.

$\ce{4R–OH + O_2 ->[r] 4R–O. +H_2O}$

$r=k\frac{\ce{[R-OH]}}{\ce{K_M +[R-OH]}}[\ce{E}][\ce{O_2}]$

Rate $r$ is the speed at which $[\ce{R–OH}]$ is consumed. It tells us that our reaction follows a Michaelis-Menten kinetic law with respect to our substrate $[\ce{R–OH}]$ and first order kinetics with respect to oxygen.

First, we are asked to find a way by which the rate equation could be simplified by approximation.

We are told that initially, $[\ce{R–OH}] = 10 mol/m^3$ and that we want a $95\%$ conversion of that reactant. We also know that $K_M =0.04 mol/m^3$. Therefore $[\ce{R–OH}]$ can vary between being $250$ times and $12.5$ times greater than $K_M$, so we could approximate the problem by getting rid of $K_M$ which would leave us with the following rate: $r=k[\ce{E}][\ce{O_2}]$. Since $[\ce{E}]$ is constant, we could write $r=k'[\ce{O_2}]$. (This sort of trick has been used before in my chemical engineering class.)

The next question is the one I am struggling with.

We are asked to use our new kinetic rate law to write down the material balance equation expressing how $[\ce{R–OH}]$ would changes with time in a batch reactor (perfectly stirred, no input or output), then solve for $[\ce{R–OH}]$.

So I would be inclined to write $\frac{d[\ce{ROH}]}{dt}=-r=-k'[\ce{O_2}]$ but then I don't see how this can be solved since we have two different variables.

From the stoichiometry we can see $\frac{d[\ce{ROH}]}{dt}=4\frac{d[\ce{O_2}]}{dt}$ therefore $\frac{d[\ce{O_2}]}{dt} = -\frac{1}{4}k'[\ce{O_2}]$ and this can be integrated to give us $[\ce{O_2}]=[\ce{O_2}]_0e^{-k't/4}$. But how can the rate law be solved for $[\ce{R–OH}]$ ?

Any insight into this problem would be hugely appreciated.

• Hello, and Welcome to StackExchange. Looks like a pretty good question for a first post. You can write chemical equations using \ce. For example, $$\ce{NH4OH <=>[k_1][k_2] NH4+ + OH-}$$ would give: $$\ce{NH4OH <=>[k_1][k_2] NH4+ + OH-}$$ Good luck! – Pritt says Reinstate Monica May 22 '17 at 14:55
• In your simplification of the rxn. equation, you are assuming that the change in [R-OH] negligibly impacts the rate, so your mass balance would reflect that with the new rate law. Have you tried writing an expression for the concentration of $[\ce{O_2}]$ in terms of R-OH, knowing that overall mass is conserved? – J. Ari May 22 '17 at 15:38
• Is there a reason why you cannot relate the concentration of oxygen with the concentration of the alcohol? – CoffeeIsLife May 23 '17 at 6:38