Chemistry- two point Arrhenius Equation (miscalculation?)

Click For Summary
SUMMARY

The discussion centers on calculating the rate constant (K2) at 75 degrees Celsius using the two-point Arrhenius equation. Given the first-order rate constant (K1) of 2.0 x 103 s-1 and an activation energy (Ea) of 15.0 kJ/mol, the correct formula to use is K2 = K1(e[(Ea/R)(1/T1-1/T2)]). The user attempts to compute K2 but struggles with the calculations, leading to incorrect results. The expected value for K2 is approximately 4.8 x 103 s-1.

PREREQUISITES
  • Understanding of the Arrhenius equation
  • Basic knowledge of thermodynamics, specifically activation energy
  • Familiarity with unit conversions, particularly kJ to J
  • Proficiency in using exponential functions in calculations
NEXT STEPS
  • Review the Arrhenius equation and its applications in chemical kinetics
  • Practice calculations involving activation energy and rate constants
  • Learn about temperature conversions in Kelvin for chemical equations
  • Explore LaTeX formatting for clearer mathematical presentations
USEFUL FOR

Chemistry students, educators, and anyone involved in chemical kinetics or thermodynamics who seeks to understand the Arrhenius equation and its practical applications.

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
2K