Question

if the rate of a particular reaction is 4 ties faster at 373K that it was at 323K what is the activation energy for the reaction

Answer 1

27.77 kJ/mol is the activation energy for the reaction.

Step-by-step explanation:

Rate of the reaction at 323 K =
`k_1=k`

Rate of the reaction at 373 K =
`k_2=4k`

Activation energy for the reaction is calculated by formula:

`\log (k_2)/(k_1)=(E_a)/(2.303* R)[(T_2-T_1)/(T_2* T_1)]`

`E_a` = Activation energy

`T_1` = Temperature when rate of the reaction was
`k_1`

`T_2` = Temperature when rate of the reaction was
`k_2`

Substituting the values:

`\log (4k)/(k)=(E_a)/(2.303* 8.314 J /mol K)[(373 K-323K)/(373 K* 323 K)]`

`E_a=27,776.98 J/mol=27.77 kJ/mol`

27.77 kJ/mol is the activation energy for the reaction.