chemoleo
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Chemical engineering question - calculation of lag phase?
I have the following system:
I am monitoring the reaction of A to B with time, conversion is a normal exponential function.
B however, facilitates reaction of C, so while no B is present no C can form. The more B is present, the more C can form.
The conversion rate is in the range of many minutes, I guess it takes 30 min to get from A to 50% B.
C facilitates reaction of D.
D facilitates reaction of E.
And so on, each has (making an assumption here) a similar conversion rate).
The previous reactant is required to form the following product, and the product has no effect on the reactant (i.e. B reacts to completion
independent of formation of C). But C can only form as B is present.
I monitor B, C, D, E etc.
B is a perfect exponential, C (as it requires B to be present) however has a tiny lag, and then the exponential, and D has a greater lag, and E has a
lag of several hours before the reaction commences.
An infinite series of these reaction sequences would produce an infinite lag...
Anyway the question is - I have curves B, C, D, E, each with a slightly greater lag.
Let the lag be slight, so that I can't judge by eye which comes before which. How can I work out the sequence? Is there a number, or a constant, by
which I can describe the extent of the lag?
What equations (simple ones?) are there to put a lag into an exponential, which applies to this system, so I can do a curve fit to curves B, C, D, E
to work out rate constants and 'lag constants'?
[Edited on 15-6-2009 by chemoleo]
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JohnWW
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The conversion of A to B, if it follows a simple exponential curve, would be a first-order chemical reaction, with rate of conversion proportional to
the remaining A and negatively proportional to the molar concentration of the product B. Your next statement "B however, facilitates reaction of C, so
while no B is present no C can form", and "The more B is present, the more C can form", suggests that this is a subsequent first-order reaction; and
similarly the subsequent self-transformations to D and E.
The only real situation in which anything like such a chain of successive first-order reactions occurs is in the radioactive decay of a whole series
of radioactive nuclei heavier than Bi-209, especially that of U-238 and its products. The concentrations, and hence rates of decay, of each successive
product follow a skewed bell-shaped curve (the "lags"), being thus made to be effectively multi-order reactions. The rates of decay of each can be
found as the solutions of a system of simultaneous linear first-order differential equations.
[Edited on 16-6-09 by JohnWW]
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Magpie
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This seems like a strange mechanism. Not something familiar to me, except the exponential decay part. Perhaps this is something from biology?
I'm assuming, as did JohnWW that the first reaction is a 1st order exponential decay that can be represented as:
A = A(0)e(exp-kt) (sorry for the poor notation)
I'm not real confident about this but couldn't the equations for the remaining species, ie, B, C, D, etc, be expressed as follows:
B = A(0)e(exp-k(t-B))
C= A(0)e(exp-k(t-C))
where A(O) = starting concentration of A
e = natural log base
exp = exponent
k = a constant
B = time delay for specie B
C = time delay for specie C
Is the "paint" sketch here a reasonable represention of what you think is happening?
The single most important condition for a successful synthesis is good mixing - Nicodem
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12AX7
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The subsequent solutions are, I believe, of the form A * x^n * e^x. The first, [A], has n = 0. As A turns to B, B rises, but as B turns to C, C
rises and B falls. Depending on the ratio of rates, you will get some amount of "hump", where B rises, then falls. The same is true for all
subsequent concentrations, until the final product is reached (whose concentration obviously grows exponentially, after the initial lag period).
After several time constants, the relative concentrations of intermediate reactants are equal to the ratio of rates.
I made a Vensim program which illustrates this for a series of radionucleides. I'll go dig it up.
Tim
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woelen
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If the time constants are equal for all reactions in the sequence, then the answer of 12AX7 is what I expect.
I suggest you go to the Laplace domain. Each of the reactions has a rate 1/(1+s*tau), where tau is the rate constant for that reaction. Because there
is no backwards influence you simply can write the total reaction as 1/[(1+s*tau1)*(1+s*tau2)*(1+s*tau3)*.....*(1+s*tauN)]. Then apply the
back-transform, which is easy if all tau's are equal.
This reaction scheme has the same mathematics as a series of first order electronic filters (RC-combinations), which are decoupled by perfect
unit-amplifiers with infinite input resistance and zero output resistance and infinite gain-bandwitdth product.
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chemoleo
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Thanks, all!
From JohnWW:
| Quote: | | The only real situation in which anything like such a chain of successive first-order reactions occurs is in the radioactive decay of a whole series
of radioactive nuclei heavier than Bi-209, especially that of U-238 and its products. The concentrations, and hence rates of decay, of each successive
product follow a skewed bell-shaped curve (the "lags"), being thus made to be effectively multi-order reactions. The rates of decay of each can be
found as the solutions of a system of simultaneous linear first-order differential equations. |
This is precisely the scenario! Now imagine I am monitoring the decay of Product 3 (=C) - it'll have a lag depending on the decay rates of A, and B.
Magpie, what you state is what I mean, inc. the exponential decay function - but the time delay isn't just a value you can subtract - it is more like
this:
Ignore the units here - the exponential function of formation of product B would be the red curve, and the delayed one, say, of product D (which
depends on production of B, and C) looks like the blue curve (and if I monitor the decay/loss or reactants, it would be just y(max)-y(i) at any given
time point i.
12AX/Woelen, the rates of conversion are differnt - as, i.e. in the radioactive decay series. If you were to monitor the last product of radioactive
decay (i.e. one that doesn't decay further), its appearance with time would look like the blue curve.
Woelen, would you mind explaining htis a bit further, or do you have a link? It is interesting as the 1/(1+constant) frequently crops up in kinetics
questions...
Many thanks!
[Edited on 18-6-2009 by chemoleo]
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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