>> So, really thinking about how are these kinetics or abiotic reactions.

So, a key question is, how are they different?

How is abiotic different than biotic?

>> Well, let's take a look at this in terms of sort of the individual steps that

are involved.

Remember that we've got a biogeochemical process, and so we've got a biological,

geochemical as well as chemical reactions, and

we break those down in terms of the kinetic associated with each one of these.

So, this is part of a graphic that was borrowed from a Navy paper that came out.

You'd think that the first one, the biotic step,

the formation of basically the reduced iron and the reduced sulphites.

We're talking about rates on the order of days, and

then we move into the precipitation of the minerals.

So, that formation, for example,

of iron sulfide that takes place basically instantaneously.

So the reaction kinetics are really favorable in this case.

But then we move in to the actual abiotic degradation of the containment itself and

so here's an example of PCE being degraded by an iron sulfide and we're talking about

half flies in this case under sort of an ideal conditions on the order of 30 days.

>> And okay so several steps three steps here that are involved

>> That maybe the key point is

that it's that last one in this one, that abiotic reaction itself,

the degradation of the contaminant that's really the rate-limiting step.

This one says 30 days but I think that's under idealized conditions,

right Dave, in the field.

>> Sure, it's going to be longer?

>> Yeah, yeah, definitely could range out into the years.

>> Okay, let's go to the reaction models.

>> Okay, so there's a lot of different ways that you can describe this

kinetics and we'll deal with some of this first order and normalized first order or

second order.

There's also zero order,

which we're not necessarily going to deal with within this particular lecture.

>> So, zero on the zero for this lecture.

>> Zero on the zero order, exactly.

>> Let’s go to first order >> So first order,

this is a sort of kinetic formula that should be familiar to most people.

But again, we're talking about change in concentration overtime,

being first order with respect to the contaminant concentration, then there's

some first order rate coefficient, and we're going to call this k in this case.

That's really defining your rate.

So, the solution of this reaction is in exponential form.

Looking at C at any particular time being equal to C not,

exponential rate to this minus kt, and that t is the time in this case.

>> So, if you want to visualize this,

in terms of a graph, we've got a graph up here.

The y axis is really this natural log of this normalized concentration.

The concentration at any time, divided by that starting concentration,

that's on the y-axis.

On the x-axis, you just have time.

And this can be from a lab study or in the field, but it's just the number of days or

years or whatever you have on that x axis And then you got the slope of this thing,

and that slope is this rate, is that right too?

>> Yeah, and this is a what is essentially a pseudo first-order rate coefficient,

since the influence that the mineral has on, and

it's sort of built-in into that rate coefficient.

And some people like dealing with rate coefficients,

other people like dealing with half-lives, I know you're a half-life sort of a guy,

so the conversion for that is pretty straightforward and shown there.

And you get this rate coefficient and this half life,

you can compare it to other attenuation rates just with the idea that it may be

different especially if you're dealing with field data to differentiate between

the abiotic component and the biotic component.

>> Just a couple quick points if you have a k and

you want to think about it in terms of a half life use that formula on the right.

Side there t 1/2 = 0.693 / k.

If you have a tough life, you want to convert it to a k,

it’s the same formula but, where did the 0.693 come from?

>> That if I remember correctly,

it’s the natural log of 2 so >> Did I get that right.

>> You're a brilliant guy that sounds right.

>> Thanks. We then now just going to look at this as

normalized first-order reaction.

So the initial form of the reaction is the same where again,

looking at the changing concentration over time in this case we've got

a KN which is essentially a second-order rate coefficient

where we've got our rate expression having this S term in it.

So it's a reactive mineral.

Either concentration or surface area something like that that we're actually

building into our rate reaction.

So the solution of it is then shown here.

it's got a similar formula.

But we're then using the k value, the first order rate coefficient, and

we can divide by that reactive mineral concentration or

surface area, that S term.