Question

A reaction rate increases by a factor of 655 in the presence of a catalyst at 37°C. The activation energy of the original pathway is 106 kJ/mol. What is the activation energy of the new pathway, all other factors being equal?

Answer 1

Answer : The activation energy of the new pathway is, 89.2869 kJ/mol

Solution :

According to the Arrhenius equation,

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

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_1$ = rate of reaction without catalyst  = x

$K_2$ = rate of reaction with catalyst  = 655 x

$Ea_2$ = activation energy with catalyst  = ?

$Ea_1$ = activation energy without catalyst  = 106 kJ/mole = 106000 J/mole

R = gas constant = 8.314 J/mole.K

T = temperature = $37^oC=273+37=310K$

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

$\frac{655x}{x}=e^{\frac{106000-Ea_2}{(8.314)\times (310)}}$

$Ea_2=89286.9J/mol=89.2869kJ/mol$

Therefore, the activation energy of the new pathway is, 89.2869 kJ/mol

Answer 2

The activation energy of the new pathway is mathematically given as

Ea2  = 89.2869 kJ/mol

What is the activation energy of the new pathway, all other factors being equal?

Question Parameter(s):

A reaction rate increases by a factor of 655 in the presence of a catalyst at 37°C.

The activation energy of the original pathway is 106 kJ/mol.

Generally, the equation for the  Arrhenius equation is mathematically given as

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

Therefore expression for catalyst

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

$\frac{655x}{x}=e^{\frac{106000-Ea_2}{(8.314)\times (310)}}$

In conclusion, the activation energy of the new pathway is,

$Ea_2=89286.9J/mol$

Ea2  = 89.2869 kJ/mol

Read more about Energy

https://brainly.com/question/13439286