Answer:
The factor by which the rate constant increases is [tex]\frac{k_2}{k_1} = 6.71 *10^{14}[/tex]
Explanation:
From the question we are told that
The activation energy of oil before addition of Oil Spill Easter
[tex]E_a_1 = 137.kJ/mol[/tex]
The activation energy of oil after addition of Oil Spill Easter
[tex]E_a_2 = 52.0 kJ/mol[/tex]
The temperature is [tex]T = 27^o C[/tex]
The objective of this solution is to obtain the factor by which the rate constant increase at 27°C
This can be obtained using the Arrhenius equation
This equation is mathematically represented as
[tex]k = A exp (\frac{-E_a}{R T} )[/tex]
Where A is the frequency
k is the rate constant
R is the gas constant
This Arrhenius equation show a relationship between rate of reaction and
Activation energy
The decomposition of oil without catalyst is expressed as
[tex]k_1 = A exp ( \frac{-E_a_1}{RT} )[/tex]
The decomposition of oil with catalyst is expressed as
[tex]k_1 = A exp ( \frac{-E_a_2}{RT} )[/tex]
Now dividing [tex]k_2 \ by \ k_1[/tex]
[tex]\frac{k_2}{k_1} = \frac{A exp (\frac{-E_a_2}{RT} )}{A exp (\frac{-E_a}{RT} ) }[/tex]
applying rule of exponent
[tex]\frac{k_2}{k_1} = exp (\frac{-E_a_2}{RT} + \frac{E_a_1}{RT} )[/tex]
[tex]\frac{k_2}{k_1} = exp(\frac{1}{RT} (E_a_1 - E_a_2) )[/tex]
Substituting value
[tex]\frac{k_2}{k_1} = exp (\frac{1}{(8.3 * 294)} (137 * 10^3 - 52 * 10^{3}) )[/tex]
[tex]\frac{k_2}{k_1} = exp(34.14)[/tex]
[tex]\frac{k_2}{k_1} = 6.71 *10^{14}[/tex]
