Question

In Arrhenius equation for a certain reaction , the values of A and `E_(a)` (activation energy are `4xx10^(13) "sec"^(-1) and 98.6 kJ "mol"^(-1)` respectively. At what temeperatue, the reaction will have specific constant `1.1xx10^(-3) "sec"^(-1)`?

Answer 1

Accroding to Arrhenius equation, `k=Ae^(-E_(a)//RT)`
or `"log"_(e)k="log"_(e)A-(E_(a))/(RT)"log"_(e)e`
`or 2.303 "log"k=2.303"log"_(10)A-(E_(a))/(RT)`
or `2.303"log"(1.1xx10^(-3))=2.303"log"K(4xx10)^(13)-(98xx10^(3))/(8.314xxT)`
`T=(98.6xx10^(3))/(8.314xx2.303xx16.56)K=310.96K`