# The rate constant of a rxn is 1.2×10^3 sec−1 @ 400 C and 2.5×10^5 s−1 @ 410 C.Calculate the energy of activation. Answer in units of kJ/mol

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### 1 Answer

To solve for the energy of activation, use the Arrhenius formula which is:

`ln (k_2/k_1)=-E_a/R(1/T_2-1/T_1)`

where `E_a ` - activation energy

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

`k_1` and `k_2` - rate constant of reaction at `T_1` and `T_2` , respectively and,

`T_1` and `T_2` - temperatures in Kelvin

So, let's express the given temperatures in Kelvin.

`T_1 = 400 + 273.15=673.15` K

`T_2= 410 + 273.15=683.15` K

Substitute `k_1= 1.2 xx 10^3 s(-1)` ,` k_2=2.5xx10^5 s^(-1)` , `T_1=673.15` , `T_2=683.15` and `R=8.314` to the formula.

`ln ((2.5xx10^5)/(1.2xx 10^3)) = -E_a/8.314(1/683.15-1/673.15)`

Isolate `E_a` .

`E_a = -(8.314*ln ((2.5xx10^5)/(1.2xx10^3)))/(1/683.15-1/673.15)`

`E_a = 2041311` `J/(mol)` `= 2041.311` `(kJ)/(mol)`

**Hence, the energy of activation is `2041.311` `(kJ)/(mol)` .**