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timo86m Group TitleBest ResponseYou've already chosen the best response.1
@nincompoop
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
Yes i remember integrating with time.
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
The central science book is not that advanced then since it just gives you formulas but doesn't explain how to reach that conclusion
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
calc 2 in this case. There is also some advanced wave function kinda stuff.
 one year ago

timo86m Group TitleBest ResponseYou've already chosen the best response.1
in the electronic structure of matter chapter luckily it dont go too far. It just gives you equations you will use.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
Is this for thermodynamics? D:
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
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: \[[A] = [A]_0 e^{(k_{1^{st}}t)} \]
 one year ago

nincompoop Group TitleBest ResponseYou've already chosen the best response.1
@abb0t I think what you did is the same to what I attached
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
Oh. Lol. My bad. haha.
 one year ago
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