Question

Pleaseeee help if you can!!
using the differential equation dN/dt=kN, where N is absorbance and t is time, derive an equation relating absorbance to time:

time: 14.00 absorbance:0.20
time:16.30 absorbance:0.50

Answer 1

@sloths

Answer 2

Are there options or no?

Answer 3

@sloths its to measure absorbance of a microorganism at different time periods. the k is a constant but that is the only equation and numbers ive been given

Answer 4

Ok that's fine, but is it multiple choice?

Answer 5

nope no multiple choice

Answer 6

im in a bit over my head here.. what is d for?

Answer 7

same, i think d means change in, so change in time or change in absorbance @sloths

Answer 8

ok so i'll explain it as best i tell

Answer 9

please if you can!

Answer 10

it looks like you have to find the difference between first and second test, like take the first time and subtract it from the second one, 16.3-14 and put that on bottom
so far we have
\[\frac{ dn }{ 2.3}=kn\]
then you'd take the difference of the absorbance, 0.5-0.2.
then you put it on top and get this:
\[\frac{ 0.3 }{ 2.3 }=kn\]
thats really all i understand. did that help at all??

Answer 11

yes i think ive understood it finally, thank you so much!! @sloths

Answer 12

sure any time! medal me maybe?

Answer 13

how do you do it? I'm new here haha @sloths

Answer 14

haha same i've been here two days.. you click best response on the right side of any of my answers

Answer 15

\[\frac{dN}{dt}=kN\]
This equation is separable:
\[\frac{dN}{N}=k~dt\]
Integrate both sides:
\[\begin{align*}\int\frac{dN}{N}&=\int k~dt\\
\ln|N|&=kt+C\\
e^{\ln|N|}&=e^{kt+C}\\
N&=e^{kt+C}&\text{assuming absorbance is always positive}\\
N&=e^Ce^{kt}\\
N&=Ce^{kt}&\text{since }e^{C}\text{ is itself a constant}
\end{align*}\]
At time \(t=14\), the absorbance is \(N=0.2\); at \(t=16.3\), \(N=0.5\). So, you have two equations:
\[\begin{cases}0.2=Ce^{14k}\\
0.5=Ce^{16.3k}\end{cases}\]
Solve for \(C\) and \(k\).