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

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

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

$\log (\frac{5.00}{1.00})=\frac{28900}{2.303\times 8.314J/mole.K}[\frac{1}{313K}-\frac{1}{T_2K}]$

$0.69=\frac{28900}{2.303\times 8.314J/mole.K}[\frac{1}{313K}-\frac{1}{T_2K}]$

$T_2=365K$

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