Question

A certain reaction with an activation energy of 185 kJ/mol was run at 525 K and again at 545 K . What is the ratio of f at the higher temperature to f at the lower temperature?

Answer 1

Answer: The ratio of f at the higher temperature to f at the lower temperature is 4.736

Step-by-step explanation:

According to the Arrhenius equation,

`f=A* e^{(-Ea)/(RT)}`

or,

`\log ((f_2)/(f_1))=(Ea)/(2.303* R)[(1)/(T_1)-(1)/(T_2)]`

where,

`f_1` = rate constant at 525K

`K_2` = rate constant at 545K

`Ea` = activation energy for the reaction = 185kJ/mol= 185000J/mol (1kJ=1000J)

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature = 525 K

`T_2` = final temperature = 545 K

Now put all the given values in this formula, we get

`\log ((f_2)/(f_1))=(185000J/mol)/(2.303* 8.314J/mole.K)[(1)/(525K)-(1)/(545K)]`

`\log ((f_2)/(f_1))=0.6754`

`((f_2)/(f_1))=4.736`

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 4.736