Question

The rate constants of some reactions double with every 10 degree rise in temperature. Assume that a reaction takes place at 295 K and 305 K. What must the activation energy be for the rate constant to double as described?

Answer 1

Iwc have 305 a9s 0.300 x 19.1471ニEas( 10 9 9935 0 30 10x 19-141 x 8911335.0 Jouits Ca 51 335 k

Answer 2

Here, Let the rate constant at 295K be k1=Ae^-E/RT=Ae^{-E/(8.314*295)}= Ae^-E/2452.63

So, reaction constant at 305K be k2= Ae^-E/RT =Ae^{-E/(8.314*305)}= Ae^-E/2535.77

According to question,

 k2/k1=2

or,=2

or e^-E/2535.77=2*e^-E/2452.63

or ln(e^-E/2535.77)=ln (2*e^-E/2452.63)

or -E/2535.77=ln 2 +ln(e^-E/2452.63)

or -E/2535.77=0.693 -E/2452.63

or 1.337*10^-5E=0.693

E=51840 Joul/mole