Need Help with natural log problem

[Trey803 5,000+ posts]

k=e^-Ea/RT

Ea=activation energy

R=8.314

T=310...............therefore RT=2577.34

k1=1

k2=4*10^5

so I need to solve these two equations in order to find Ea(activation energy) of both

1=e^-Ea/2577.34 Ea=?

and

4*10^5=e^-Ea/2577.34 Ea=?

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[Crzy 10+ year member]

such easy problems, last eq set ea = to 2577.34(4e5) to get ea, then plug into first EQ to get e if you need to for that problem also

 

[Trey803 5,000+ posts]

Ea is not a constant....I am ultimately trying to find (Ea1-Ea2)

 

[jmanpc 5,000+ posts]

The only natural logs i deal with are turds.

 

[DDSC 10+ year member]

They should have plenty stuff from EA at game stop in the sports section =) I hate math, what the hell would you actually use this for on the job or shall I say what jobs would you use this with while working

 

[Trey803 5,000+ posts]

The only natural logs i deal with are turds.

since you have a ****** this actually makes sense

 

[Trey803 5,000+ posts]

They should have plenty stuff from EA at game stop in the sports section =) I hate math, what the hell would you actually use this for on the job or shall I say what jobs would you use this with while working

one that pays a lot of money

 

[Trey803 5,000+ posts]

bump......maybe I am approaching the problem wrong but this is a chemistry problem involving two different activation energies with the second increasing the reaction rate by 4*10^5 of the first reaction rate

k1=x Ea1=?

k2=(4*10^5)(x) Ea2=?

 

[cajun_duck 10+ year member]

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