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?

Question

asked Oct 6, 2021 135k views

1 Answer

NKijak · Answer 1 · 2021-10-12T05:38:31+0000

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

= 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.