Question

A certain reaction has an activation energy of 28.90 kJ / mol. At what Kelvin temperature will the reaction proceed 5.00 times faster than it did at 313 K?

Answer 1

Answer: 365 K

Step-by-step explanation:

According to the Arrhenius equation,

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

or,

`\log ((K_2)/(K_1))=(Ea)/(2.303* R)[(1)/(T_1)-(1)/(T_2)]`

where,

`K_1` = rate constant at
`T_1` = 1.00

`K_2` = rate constant at
`T_2` = 5.00

`Ea` = activation energy for the reaction = 28.90 kJ/mol= 28900 j/mol

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature = 313 K

`T_2` = final temperature = ?

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

`\log ((5.00)/(1.00))=(28900)/(2.303* 8.314J/mole.K)[(1)/(313K)-(1)/(T_2K)]`

`0.69=(28900)/(2.303* 8.314J/mole.K)[(1)/(313K)-(1)/(T_2K)]`

`T_2=365K`

Therefore, 365 K is required to increase the reaction rate by 5.00 times.