# Kinetics ODE Solutions

1. Mar 7, 2013

I've got this problem, with relation to rates. [A] is the concentration of one substance, is the concentration of another, etc. [R] is a general reactant, [P] is a general product, is a general substance (which thus relates to both reactants and products). But you won't actually need chemistry for this, just thought it might help conceptualize.

-d[A]/dt=k*[A]n1*n2*[C]n3 ...

n1...3 are orders of reaction, and they are there with respect to every , a different n for each substance. This may take any real value including 0, fractional or negative values. These orders are constant, as is k.

However, because of the rates changing, each substance will be dependent on time t, and by the definition of the rate as above, on the concentration of each other substance.
Typically I have seen this rearranged to:

-d[A]/([A]n1*n2*[C]n3 ...)=k*dt

And then "solved". But I don't know what this solving means! Can anyone help?
There is another system which helps along the solution, though again I don't know how. This is: if we are finding the rate of loss for a reactant, then for each and every reactant we can write [R]=[R]0-((vA/vR)*([A]0-[A])), where vA and vR are once again constants (stoichiometric coefficients, if you're interested - they will be real and positive), but there is a different vR for every reactant (as with the orders - e.g. there is vC, where [C]=[C]0-((vA/vC)*([A]0-[A])), etc.). And for each product, we can write similarly: [P]=[P]0+((vA/vP)*([A]0-[A])), the only difference being the plus I have placed in bold.

So, to summarize:

Variables: [A], , [C], [D], ..., t
Constants: k, n1 (pertains to [A]), n2 (pertains to ), n3, ..., vA (pertains to [A]), vB (pertains to ), vC (pertains to [C]), ..., [A]0 (initial value of [A]), 0, [C]0, [D]0, ...

Equations:

-d[A]/([A]n1*n2*[C]n3 ...)=k*dt
[R]=[R]0-((vA/vR)*([A]0-[A]))
[P]=[P]0+((vA/vP)*([A]0-[A]))

Our final answer would be a solution for [A] in terms of all of the constants listed above and t.

Obviously I don't necessarily seek an exact, complete answer for the general case, just to learn the method.

2. Mar 8, 2013

### bigfooted

I think the 're-arranging and solving' step only works in special cases like when the other concentrations remain (almost) constant or when you have simple reactions.

When the other concentrations can be considered constant, introduce a new constant $K=k^{n2}[C]^{n3}$ and solve $\frac{d[A]}{dt}=K[A]^{n1}$ by integration.

Most reaction rate systems describe reactions between only 2-4 species, so you would have something like A+B -> C+D
which results in the simple system
d[A]/dt = -k [A]
d/dt = -k[A]
d[C]/dt = k[A]
d[D]/dt = k[A]

The last two equations can be solved independently after the first two have been solved.
We know from subtracting the second equation from the first that:
[A]- = constant,
= [A] + constant
d[A]/dt = -k[A]([A] + constant)

This can be solved by integration and the other equations can then be solved as well.

3. Mar 8, 2013

### Staff: Mentor

In many situations involving batch reactors (typical of the differential equations you have written), there are multiple chemical reactions occurring, each involving some of the chemical species that are present in the mix. In order to predict how the concentrations of all the species are varying with time in the reactor, you need to write a mass balance equation for each of the species in the reactor, describing the rate of change of the concentration of the species with time. To calculate this rate of change, you need to include in the mass balance bookkeeping every reaction that consumes the species as a reactant, and each reaction that generates the species as a product. For each reaction, you need to calculate the rate of the reaction as a function of the concentrations of the reactants. The net result of all this is an ordinary differential equation for each species as a function of the concentrations of all the species. Typically, you will need to integrate this set of coupled ordinary differential equations numerically.