# What is the temperature that would produce a rate constant of 6×10^5 sec−1 if the activation energy is 40 kJ/mol and the Arrhenius factor is 6.2×10^12?

*print*Print*list*Cite

Expert Answers

lemjay | Certified Educator

Apply the Arrhenius equation which is:

`ln k = -E_a/RT + lnA`

where `k` - rate constant of reaction

`E_a` - energy of activation

`R` - gas constant, 8.314 J/(mol K)

`T` - temperature in Kelvin and

`A` - pre-exponential factor or frequency factor or Arrhenius factor.

So, substitute `k = 6xx10^5 s^(-1)` , `E_a = 40 000` `J/(mol)` , `A = 6.2xx10^12` and `R = 8.314` `J/(mol K)` .

`ln (6xx10^5) = -40000/(8.314T) + ln (6.2xx10^12)`

Isolate T.

`T = 40000/(8.314( ln(6.2xx10^12) - ln(6xx10^5)))`

`T = 297.89`

**Hence, the temperature is 297.89 K.**