Question

for nin

And using calc to derive this formula

Answer 1

@nincompoop

Answer 2

Yes i remember integrating with time.

Answer 3

The central science book is not that advanced then since it just gives you formulas but doesn't explain how to reach that conclusion

Answer 4

calc 2 in this case. There is also some advanced wave function kinda stuff.

Answer 5

in the electronic structure of matter chapter
luckily it dont go too far. It just gives you equations you will use.

Answer 6

Is this for thermodynamics? D:

Answer 7

For a reaction of the form:
A --> product
a first order rate law takes the form:
\[r = k_{1^{st}}[A]^1\]
writing it in differentials, you get:
\[\frac{ d[A] }{ dt } = -k_{1^{st}}[A]\]
The minus sign as A is a reactant and so its concentration decreases with time (whici is the slope of a graph of [A] against time will be negativ when you graph it)
The variables [A] and t can be separated to give:
\[\frac{ 1 }{ [A] }d[A] = -k_{1^{st}}dt\]
which CAN be integrated easily!
\[\int\limits \frac{ 1 }{ [A] }d[A] = \int\limits -k_{1^{st}}dt\] this gives: \[\ln[A] = -k_{1^{st}}t + C\]
The constant can be removed by supposing that at t = 0 since
\[[A] = [A]_0\]
which is the initial concentration of A giving us: \[\ln[A] = -k_{1^{st}}t + \ln[A]_0\]
to simplify it a bit more to a form that might be more familiar to you is: \[[A] = [A]_0 e^{(-k_{1^{st}}t)} \]

Answer 8

@abb0t I think what you did is the same to what I attached

Answer 9

Oh. Lol. My bad. haha.