## anonymous one year ago In the Thrall model, the reaction rate (with respect to time) is proportional to the amount x of the new substance at time t , and the reaction rate is also proportional to a - x where a is the original amount of the first substance. a. Why does this mean that dx/dt = K x (a - x) for some positive constant K ? b. Differentiate with respect to x your expression for dx/dt.

1. anonymous

I think I can probably handle b, but what do you think these variables mean? given a catalytic reaction. What variables would represent "first substance"? "the new substance"? K ? etc..

2. anonymous

Here's a bit more of that question ... if it helps at all. "R. M. Thrall and his University of Michigan colleagues (Report No. 40241-R-7, University of Michigan, 1967) gave the following crisp description of autocatalytic reaction of one substance into a new substance: "An autocatalytic reaction progresses in such a way that the first substance catalyzes its own formation."

3. hartnn

we won't need a variable to represent 'first substance', the reaction rate depends on the amount of 1st substance, which is already represented by 'a'. same goes for new substance, amount of new substance is represented by 'x' K is just a constant of proportionality! such constants come on all variation problems... ex: if x is proportional to y, x= ky, k= constant of proportionality.

4. Michele_Laino

I think that the differential equation: dx/dt = K x (a - x) is right!

5. anonymous

Is there any reason why I should assume here that a=x when t=0 ?

6. Michele_Laino

a is the initial concentration

7. hartnn

Direct variation: If $$x$$ is $$\text{directly proportional to} ~ y,$$ then we say x/y is a constant. this constant is the constant of proportionality = k (say) so x/y = k, or x =ky in your problem, reaction rate(dx/dt) is proportional to x and (a-x) hence, dx/dt = K x(a-x)

8. Michele_Laino

that is a second order process, which involves two substances

9. hartnn

a is the amount of 1st substance x is the amount of new substance at t=0, the reaction didn't happen yet, so at t = 0, x = 0

10. anonymous

awesome.. that's the piece had me confused..

11. anonymous

the equation makes sense now.

12. hartnn

had some doubt about part b. what you found in part a is dx/dt already! because is the reaction 'rate' Then how could we find dx/dt by differentiating the expression w.r.t x.... won't that give you 2nd derivative? d^2x/ dt^2

13. anonymous

well that was another thing I was wondering... if a 2nd differential was going to come into play here.

14. hartnn

oh, they are asking to differentiate the expression for dx/dt ! ok, got it :P

15. hartnn

yes, the answer for part b will indeed be d^2x/dt^2

16. hartnn

the wording were confusing! "for dx/dt" >>> i read it as 'to get dx/dt'

17. anonymous

oh right on.. I see .. differentiate the dx/dt .. all good.

18. anonymous

well if it confuses you, it must be bad ... lol.

19. anonymous

alright.. thanks again you guys.. I got this.. :)

20. hartnn

welcome ^_^

21. Preetha

Hey Hugh, give one of the Qualified Helpers a rating so they can get their share of OwlBucks!

22. anonymous

right on, sorry.. I usually always do that * eventually :)

23. Preetha

Thanks.

24. hartnn

Thanks hugh! much appreciated :)