## timo86m Group Title for nin And using calc to derive this formula one year ago one year ago

1. timo86m Group Title

2. timo86m Group Title

@nincompoop

3. nincompoop Group Title

4. timo86m Group Title

Yes i remember integrating with time.

5. timo86m Group Title

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

6. timo86m Group Title

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

7. timo86m Group Title

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

8. abb0t Group Title

Is this for thermodynamics? D:

9. abb0t Group Title

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: $[A] = [A]_0 e^{(-k_{1^{st}}t)}$

10. nincompoop Group Title

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

11. abb0t Group Title