At what Kelvin temperature will the reaction proceed 3.00 times faster than it did at 339 K?

Question

Help I'm struggling with this.

A certain reaction has an activation energy of 29.83 kJ/mol. At what Kelvin temperature will the reaction proceed 3.00 times faster than it did at 339 K? 

Answer 1

The rate of a reaction is proportional to e^-Ea/(RT), where E_a is the activation energy, T is the absolute temperature, and R is the universal gas constant.

 

You could set this up as

 

3e^{-29.83/(339R)} = e^-29.83/(RT)

 

and solve for T.

 

ln3  - 29.83/(339R) = -29.83/(RT) 

 

RT = -29.83 / [ln3 - 29.83/(339R)]

 

T = -(29.83/R) / [ln3 - 29.83/(339R)]

 

R = 8.314 J/(mol K) = 0.008314 kJ/(mol K)

 

T = -(29.83 / 0.008314) / [ln3 - 29.83 / (339 x 0.008314)] K ≅ 378 K

 

Test the solution.