k = Ae^-(Ea/RT) on a math term

[Mcat35]

Advertisement - Members don't see this ad

I know the meaning behind it.. increase temp = increase collison = increase rate,
dcrease activation energy = increase rate.. how does this work mathematically? I'm pretty bad at natural log approximations lol...

 

[libraryismyhome]

I know the meaning behind it.. increase temp = increase collison = increase rate,
dcrease activation energy = increase rate.. how does this work mathematically? I'm pretty bad at natural log approximations lol...

Sorry for being irrelevent, but God~~~ I hate memorizing that equation... 😡

 

[Mcat35]

Sorry for being irrelevent, but God~~~ I hate memorizing that equation... 😡

its not really one to memorize I don't think, you just need to know that the rate constant is affected by temperature and activation energy. Since everything else are constants, changing these would affect rate.

I just want to know the math behind it, check back to the thread and see if someone answers soon.

 

[ilovemcat]

I know the meaning behind it.. increase temp = increase collison = increase rate,
dcrease activation energy = increase rate.. how does this work mathematically? I'm pretty bad at natural log approximations lol...

You guys definitely don't won't be using this equation for any sort of calculation. Just focus on the concepts. The "A" term is basically a hidden constant that accounts for the frequency of collisions and their orientation in space. Certain orientations are more favorable than others for a given reaction. Some questions asking about the rate constant for a given reaction are common sense. Lowering the activation energy and increasing the temperature both increase the reaction rate by increasing the rate constant (k). The activation energy can be lowered by adding a catalyst (which decrease Ea of both the forward AND reverse directions). Also by increasing the temperature, you're increasing the overall kinetic energy of the reactants and so more reactants on average will collide more often (more frequency of collisions) and more reactants will collide with the proper orientation. These are all things that effect the rate constant. The actual rate of the reaction in addition to the above changes (which effect the rate constant only) could also increase by increasing the concentration of reactants. Note that the rate constant is independent of concentration. From these relationships you can workout that the rate constant must be inversely proportional to activation energy (lower Ea, increases rate) and directly proportional to temperature changes (increase temp, increases rate and vice versa).

 

[ilovemcat]

I just want to add that if they are mean enough to make you calculate the rate constant, changes are you'd be using the rate law (plus associated data) to solve for it, not this equation. This equation is only useful for understanding the concepts.

 

[Mcat35]

I just want to add that if they are mean enough to make you calculate the rate constant, changes are you'd be using the rate law (plus associated data) to solve for it, not this equation. This equation is only useful for understanding the concepts.

alright thank Ilovemcat. Some reassurance goes a long way.

 

[whiteshadodw]

I know the meaning behind it.. increase temp = increase collison = increase rate,
dcrease activation energy = increase rate.. how does this work mathematically? I'm pretty bad at natural log approximations lol...

plot out y=e^(-x) on your calculator

then...think of some situations. such as, if Ea increases, then x increases, so follow the plot to the right, if Ea decreases then go left.

as the Ea/RT term gets larger, then e^(-some really large number)=some super small positive number. as the Ea/RT term gets smaller, like much smaller than one, but still positive (Ea/RT <<<<<<< 1) then, e^(-some super small positive #)=~1

conversely. if T increases, then x gets small, and if you let T increase to infinity then x tends to 0. e^-0 = 1. if you know the graph for y=e^(-x) you can have an idea of what the arrhenius equation looks like.

 

[Mcat35]

plot out y=e^(-x) on your calculator

then...think of some situations. such as, if Ea increases, then x increases, so follow the plot to the right, if Ea decreases then go left.

as the Ea/RT term gets larger, then e^(-some really large number)=some super small positive number. as the Ea/RT term gets smaller, like much smaller than one, but still positive (Ea/RT <<<<<<< 1) then, e^(-some super small positive #)=~1

conversely. if T increases, then x gets small, and if you let T increase to infinity then x tends to 0. e^-0 = 1. if you know the graph for y=e^(-x) you can have an idea of what the arrhenius equation looks like.

ah ok, lol.. gotta go buy some batteries for my TI83, haven't used it in a while lol.

 

[whiteshadodw]

ah ok, lol.. gotta go buy some batteries for my TI83, haven't used it in a while lol.

http://url3.tk/graph.tk/

 

[Mcat35]

http://url3.tk/graph.tk/

thanks what a useful little tool.