Question

What is the activation energy (in kJ/mol) of a reaction whose rate constant increases by a factor of 89 upon increasing the temperature from 307 K to 343 K? R = 8.314 J/(mol • K). Only enter the numerical value as an integer in the answer box below. Do NOT type in the unit (kJ/mol).

Answer 1

Answer:

15.9 KJ/mol

Explanation:

Given data:

Temperature = T1 =  307 K

Temperature = T2 = 343 K

Gas constant  R= 8.314 J/(mol • K)

rate constant = k2/K1 = 89

To find:

Activation energy (in kJ/mol) = Ea = ?

Formula:

The Arrhenius equation gives the relation between temperature and reaction rates:

$ln\frac{K2}{K1} = \frac{Ea}{R} (\frac{1}{T1} - \frac{1}{T2})$

here, in this equation

k = the rate constant

Ea = the activation energy

R = the Universal Gas Constant

T= the temperature

Solution:

$ln\frac{89 K1}{K1} = \frac{Ea}{8.314 J/mol . k\\} (\frac{1}{307 K} - \frac{1}{343 K})$

ln 89 = Ea / 8.314 J/mol.K  x (0.0325 - 0.00291)

ln 89 = Ea / 8.314 J/mol.K  x  (2.95 x 10^2 )

4.488 = Ea  / 8.314 J/mol.K  x  (2.95 x 10^2)

Ea = 4.488  x  (2.95 x 10^2) /  8.314 J/mol.K

   = 0.1324 / 8.314

Ea = 0.0159

Ea = 1.59 x 10^2 J/mol  

    = 15.9 KJ/mol