Chemistry- two point Arrhenius Equation (miscalculation?)

Click For Summary
The discussion revolves around calculating the rate constant at 75 degrees Celsius using the two-point Arrhenius equation. The provided values include a first-order rate constant of 2.0 x 10^3 s^-1 at 25 degrees Celsius and an activation energy of 15.0 kJ/mol. The formula K2 = K1(e^[(Ea/R)(1/T1-1/T2)]) is applied, but the user struggles with achieving the correct answer, suspecting it should be around 4.8 x 10^3 s^-1. There are also comments about the clarity of the calculations, suggesting the use of proper formatting for better readability. The thread highlights common challenges in applying the Arrhenius equation for temperature-dependent rate constants.
Flip
Messages
8
Reaction score
0
Chemistry- two point Arrhenius Equation (miscalculation??)

Homework Statement



At 25 degrees celsius, the first-order rate constant for a reaction is 2.0 x 10^3 s^-1. The activation energy is 15.0 kJ/mol. what is the value of the rate constant at 75 degrees celsius.

T1= 298.15 K
T2= 348.15 K
K1= 2.0 x 10^3 s^-1
Ea= 15.0 kJ/mol
K2= ??


Homework Equations



K2=K1(e^[(Ea/R)(1/T1-1/T2)])


The Attempt at a Solution



K2= 2.0 x 10^3 s^-1(e^[(15.0 kJ/mol / 8.3145 J/molK)(10^3 J / 1 kJ)(0.0036 K - 0.0029 K)])

I can't remember if the solution is supposed to be 4.8 x 10^3 s^-1. This is taken from my notes that I have on my course guide. I am basically trying to replicate the entire process, but I keep coming up with over sized answers that are to the 9th power or something... -.-
 
Physics news on Phys.org


lol i can't understand your work...try using the super and subscripts and latex...then maybe people can read what you're trying to say
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
93K
Replies
4
Views
22K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K