Understanding Laws of Logs: How to Derive Chemistry Kinetics Equation

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In summary, the chemistry kinetics equation [A] = [A]_0*e^{-kt} is derived through the application of the exponential to both sides of the equation ln[A] - ln[A]_0 = -kt, resulting in [A] = [A]_0*e^{-kt} instead of [A] = [A]_0 - e^{-kt}.
  • #1
mycotheology
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I'm reading about how the chemistry kinetics equations are derived and here's something I don't get. How does this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-6.gif
get turned into this:
[itex][A] = [A]_{0}*e^{-kt}[/itex]?
When I try to derive it, I first get this:
[itex]ln[A] - ln[A]_{0} = -kt[/itex].
Then I isolate ln[A] and get:
[itex]ln[A] = ln[A]_{0} - kt[/itex]
then I reverse the ln on both sides of the equation and get:
[itex][A] = [A]_{0} - e^{-kt}[/itex].
I don't understand how the two terms end up multiplied rather than subtracted.
 
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  • #2
mycotheology said:
I'm reading about how the chemistry kinetics equations are derived and here's something I don't get. How does this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-6.gif
get turned into this:
[itex][A] = [A]_{0}*e^{-kt}[/itex]?
When I try to derive it, I first get this:
[itex]ln[A] - ln[A]_{0} = -kt[/itex].
Then I isolate ln[A] and get:
[itex]ln[A] = ln[A]_{0} - kt[/itex]
then I reverse the ln on both sides of the equation and get:
[itex][A] = [A]_{0} - e^{-kt}[/itex].

I don't understand how the two terms end up multiplied rather than subtracted.

The red part is the wrong part.
By reverse you use e as a index for exponentiation, so let's say :
ln a = ln b - c
e^(ln a) = e^(ln b - c)
u got your exponentialtion wrong.
 
  • #3
mycotheology said:
I'm reading about how the chemistry kinetics equations are derived and here's something I don't get. How does this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-6.gif
get turned into this:
[itex][A] = [A]_{0}*e^{-kt}[/itex]?
When I try to derive it, I first get this:
[itex]ln[A] - ln[A]_{0} = -kt[/itex].
Then I isolate ln[A] and get:
[itex]ln[A] = ln[A]_{0} - kt[/itex]
then I reverse the ln on both sides of the equation and get:
[itex][A] = [A]_{0} - e^{-kt}[/itex].
I don't understand how the two terms end up multiplied rather than subtracted.

The easiest method to see this is to apply the exponential to each side.

##
\begin{eqnarray*}
\displaystyle ln\frac{[A]}{[A]_0} &=& -kt\\
\displaystyle e^{ln\frac{[A]}{[A]_0}}&=& e^{-kt}\\
\displaystyle \frac{[A]}{[A]_0} &=& e^{-kt}\\
\displaystyle [A] &=& [A]_0 e^{-kt}\\
\end{eqnarray*}
##
 

1. What are the basic laws of logs?

The basic laws of logs include the product rule, quotient rule, power rule, and change of base rule. These laws govern how logarithms can be manipulated and simplified.

2. How can logarithms be used in deriving chemistry kinetics equations?

Logarithms can be used to transform exponential equations into linear equations, which can then be graphed and analyzed to determine the rate of a chemical reaction and its dependence on certain factors such as concentration or temperature.

3. What is the process of deriving a chemistry kinetics equation using logarithms?

The process involves taking the natural logarithm of both sides of the rate law equation, which allows for the use of the laws of logs to simplify and rearrange the equation into a linear form. This linear form can then be used to determine the rate constant and reaction order.

4. How do the laws of logs relate to the principles of chemistry?

The laws of logs are based on mathematical principles and can be applied to various fields, including chemistry. In the context of chemistry kinetics, they allow for the manipulation and simplification of complex equations to determine important factors in chemical reactions.

5. What are some common mistakes when using logarithms to derive chemistry kinetics equations?

Some common mistakes include not taking the natural logarithm of both sides of the rate law equation, not properly applying the laws of logs, and not considering the proper units and dimensions when solving for the rate constant.

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