125-l degradation is a First Rate Chemical reaction?
This is what I know so far:
Half life time for 125-l insulin = 60 days
Original starting amount of 125-l insulin = 50 microliters
cpm is counts per minute
At expiry date = 20,000 counts per minute
4 months approximately 31 days, 30 days, or 29 days
4 months (30 days/ 1 month) = 120 days
This is what we are seeking for:
counts per minute at 4 months past the expiry day. That is beyond the 60 days half life.
We might need to make a graph to figure out the count at 120 days past the expiry date. The expiry date is not the same as the half life. This problem is based upon the first order chemical reaction rate laws. I will need a few nights to do it. I do not have your text book to study the examples. You are making it harder on the answerer than yourself. You would be better to study the examples in your book carefully than to asks me. It might be a problem that the class wrote. I have thought about a Michaelis-Menten kinetics, but we are not studying the reaction substrate and an excess of enzyme. This is not exactly an initial reaction rate and concentration problem in a fixed amount of reactants and resulting products. The question portrayed itself as a half life, first order chemical reaction in the degradation of 125-l insulin. It is one reactant degrading into another product, thus lowering the counts per minute. Half life reaction was discovered by experimentally measuring the counts per minute. There are many computational chemistry situations where it has to be experimentally measured, before using descriptive laws and mathematics techniques that I learned in computational chemistry.
It is often referred as chemistry & computers. It should be taken, before trying physical chemistry. It is viewed as difficult, but not true, just carefully read the problems and keep an open mind and try to solve it with some math and treat it like word problems and asked yourself does the answer relates? Do the units cancel out? It is relating to the real world occurances? I do not enjoy world problems, but it is as fun as puzzle solving. I enjoy the puzzle feeling that is why I am trying this.
The dinner was great at my corner. I am still fascinated with photography, therefore I am sane. A Ph.D, that is too much financial liability, what if I cannot defend my thesis? What could I invent that is so original that the whole world would come and see? I will own my own soap, sodium, and halogen factory. Oh, no I am still in earth science.
I cannot assure the correctness of this problem. Here are my findings.
I suspect this 125-l insulin degrades in a first order reaction. I have not done any studies on the mechanism of this degradation of 125-l insulin. I am ignorant. As the chemicool.com says, we will solve it for you. That is vague. Who? Will it be right? It is scary, I cannot answer you.
I did not write this in an equation editor, so you will have to follow as best as you can by text and ( ) grouping symbols for one term, denominator or numerator. * is the multiplication symbol, just like in computer programming.
R=kN
k=constant
R=rate
N=number, I will substitute this with counts per minute
ln(Nt/No)=-kt
t=time interval of decay
k=decay constant
Nt=remaining number or remaining counts per minute
Change of concentration with time is part of this first order reaction
Rate= - (delta [ A concentration]/delta time) = k[A concentration]
ln[A] t - ln[A] = -k*time
ln([A]t/[A]0) = - kt
The Half Life
ln (1/2[A]0/[A]0)=-k time
LN 1/2 = -kt
t=-(LN1/2)/k = 0.693/k
Half Life = 60 days
solve for K
60 days is the half life
t=-(LN1/2)/k=0.693/k
60 days = 0.693/k
60 days * k = 0.693
k=0.693/60 = 0.0116
k=0.0116
50 micro liters is extra information, because it was not used in the calculation. Even after expiration time, we still have 50 micro liters, but less counts per minute, because the 125-l insulin in this 50 micro liters have degraded. I do not know this for sure, because I am not in your class or in your job or project. I am doing this blindly. If there are errors, it is understandable. All I have is a BS in earth science, so please. My speciality is the minerals that came out of the melt in volcanos and the sediment patterns and lithology of alluvial fans.
Here is how we can check if 0.0116 is the constant?
t half life = 0.693/0.0116 = 59.7 days = 60 days
ln (Nt/N0) = -kt= -(0.0116 days^-1) (t)
t=120 days after expiration
Here is my troubled step. I know the counts per minute after expiration, but I do not know at initial or at half life. I am guessing from here on.
t=120 days after expiration
N0=20000 counts per minute
What is Nt @ 120 days after expiration
ln(Nt/N0) = -kt= -(0.0116 days ^-1) * (120 days)
ln(Nt/N0) = -kt = -1.3920
Nt/N0=e^-1.3920=4.0229
N0=20000 counts per minute
Nt=?
Nt=4.0229 *N0=(4.0229)(20000 counts per minute)
Nt=80457.7558 counts per minute
Nt=80458 counts per minute
This does not seem correct, because we should have less and not more. This higher count than 20000 counts per minute indicates it was before the expiration date
We will have to reject this answer and keep the k=0.0116 days^-1
This is expected, because I am not in your class and I do not have your book or attended the lecture. I do not know who or where, typical. The quesions posted by high school, elementary school, and middle school are viewed more than this one. I am not a teacher. I am turtle earth scientists, telescope builder, camera man, microscope repair man, and lives near a corner.
Let us try again:
ln(Nt/N0) = -kt = -(0.0116 days^-1) (60 days)=-0.6960
(Nt/N0)=e^-0.6960=0.4986
N0=?
Nt=20000 counts per minute
20000 counts per minute/N0=0.4986
20000 counts per minute = 0.4986 * N0
20000 counts per minute/0.4986 = N0
40000 counts per minute = N0
When the 125-l insulin was new, it was 40000 counts per minute
Expiration date it was 20000 counts per minute as given from your message.
60 days +120 days, 4 months after the expiration date= 180 days
ln(Nt/N0) = -kt =-(0.0116 days^-1) (180 days)=-2.0880
(Nt/N0) = e^-2.0880=0.1239
N0=20000 counts per minute
Nt=? That is your answer to this question. What is the counts per minute 120 days or 4 months after the expiration date?
(Nt/20000 counts per minute at the expiration date)=0.1239
Nt=01239 * 20000 counts per minute
Nt=2478.6952 counts per minute
Nt=2479 counts per minute
This answer seems correct, but I am not sure. I am not responsible, if you failed school. I am a turtle. Leave me to crawl back into my refuge area. 2479 counts per minute is less than 20000 counts per minute. The time is beyond half life, expiration and initial. It might be right. This is a hard problem. I still do not know what to do with the 50 micro liter fact.
You know what to do with that?
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