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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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..
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."
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I think that the differential equation: dx/dt = K x (a - x) is right!
Is there any reason why I should assume here that a=x when t=0 ?
a is the initial concentration
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)
that is a second order process, which involves two substances
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
awesome.. that's the piece had me confused..
the equation makes sense now.
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
well that was another thing I was wondering... if a 2nd differential was going to come into play here.
oh, they are asking to differentiate the expression for dx/dt ! ok, got it :P
yes, the answer for part b will indeed be d^2x/dt^2
the wording were confusing! "for dx/dt" >>> i read it as 'to get dx/dt'
oh right on.. I see .. differentiate the dx/dt .. all good.
well if it confuses you, it must be bad ... lol.
alright.. thanks again you guys.. I got this.. :)
welcome ^_^
Hey Hugh, give one of the Qualified Helpers a rating so they can get their share of OwlBucks!
right on, sorry.. I usually always do that * eventually :)
Thanks.
Thanks hugh! much appreciated :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question