20 points. Problem 6. Breaking Bad, and Going to Rehab, Graduating from Rehab, t
ID: 702713 • Letter: 2
Question
20 points. Problem 6. Breaking Bad, and Going to Rehab, Graduating from Rehab, then Breaking Bad Again... it's a vicious Cycle with a Steady State Solution say we have a fixed total concentration of people [G] + [B] + [B')-C where C is some fixed total number of people per square mile, then consider the simplified social scheme kgrad where G is a Good person that can break bad with rate constant k B is a Bad person that can go into therapy with rate constant k B' is "Bad person in therapy/rehab", and can graduate with rate constant Kgrad a. Considering standard type of rate laws for "first order reactions", write the three rate equations, i.e., (I'lI do the first one for you) b. Give the versions of the three equations for the condition of steady state. For steady state, only two of the equations are independent. Show that this is true by taking any two and developing the third c. There are three concentration unknowns, [G), [B), and [B']. If k 0.0004/day kr 0.0002/day rad 0.003/day, and if the third equation needed to solve for the three unknowns is [G+(B] + [B' 1000/mi 2Explanation / Answer
a)
d[G]/dt = Kgrad[B`] - KB[G]
d[B]/dt = KB[G] - KT[B]
d[B`]/dt = KT[B] - Kgrad[B`]
b)
at Steady State: d[ ]/dt =0
Thus
d[G]/dt = 0, Kgrad[B`] = KB[G] ----------------(1)
d[B]/dt = 0, KB[G] = KT[B] -------------------------(2)
d[B`]/dt =0, KT[B] = Kgrad[B`] -----------------------------(3)
From equation (1) and (2), KT[B] = Kgrad[B`] which is same as equation (3).
Equation (3) can be obtained from equation (1) and (2).
So there are only 2 independent equation.
From equation (2) and (3) we can get equation (1).
From equation (1) and (3) we can get equation (2).
From any of two equations, we can get the third one.
c)
KB = 0.0004/day , KT = 0.0002/day, Kgrad = 0.003/day
From equation (1), 0.003[B`] = 0.0004[G]
that results 30[B`] = 4[G]
From equation (2), 0.0004[G] = 0.0002[B]
that results 2[G] = [B]
Given: [G]+[B]+[B`]=1000/mi2
[G]+2[G]+(4/30)[G] = 1000
[G] = 319.148/mi2
[B] = 2[G] = 638.2978/mi2
[B`] = (4/30)[G] = 42.553/mi2
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