Respuesta :
Answer:
The activation energy is lowered by +85.35 KJ/mol
Explanation:
Rate constant is related to the activation energy and the temperature of the reaction through
K = Ae⁻ᴱᵃ/ᴿᵀ
In K = (In A) - (Ea/RT)
In K = (-Ea/RT) + In A
where k = the activity constant
Ea = activation energy
R = molar gas constant
T = absolute temperature in Kelvin
₁₂
At point 1, without the catalyst
In K₁ = (-Ea₁/RT) + In A (eqn 1)
At point 2 with the catalyst
In K₂ = (-Ea₂/RT) + In A (eqn 2)
Note that the molar gas constant, the absolute temperature in Kelvin and the activity constant are all the same for both points.
Subtract (eqn 1) from (eqn 2)
In K₂ - In K₁ = (-Ea₂/RT) - (-Ea₁/RT) + In A - In A
In (K₂/K₁) = (1/RT) (Ea₁ - Ea₂)
We were told that K₂/K₁ = 2.3 × 10¹⁴, then find the difference in Ea
R = 8.314 J/mol.K, T = 37°C = 310.15 K
In (2.3 × 10¹⁴) = (1/(8.314×310.15) (Ea₁ - Ea₂)
33.1 = (1/2578.5871) (Ea₁ - Ea₂)
(Ea₁ - Ea₂) = 33.1 × 2578.5871 = 85351.23 J/mol
Therefore, the activation energy is lowered by +85.35 KJ/mol
The activation energy is lowered by +85.35 KJ/mol
Relation between Rate constant and Activation Energy:
[tex]K = Ae^{\frac{-E_a}{RT} }\\\\In K = (In A) -( \frac{-E_a}{RT})\\\\In K = (\frac{-E_a}{RT}) + In A[/tex]
where
- k = the activity constant
- Ea = activation energy
- R = molar gas constant
- T = absolute temperature in Kelvin
1. At point 1, without the catalyst
[tex]In K_1 = (-Ea_1/RT) + In A[/tex]............(i)
2. At point 2 with the catalyst
[tex]In K_ = (-Ea_2/RT) + In A[/tex]...........(ii)
On subtracting equation (i) from (ii)
[tex]In K_2 - In K_1 = (-Ea_2/RT) - (-Ea_1/RT) + In A - In A\\\\In (K_2/K_2) = (1/RT) (Ea_1 - Ea_2)[/tex]
Given:
- K₂/K₁ = 2.3 × 10¹⁴
- R = 8.314 J/mol.K, T = 37°C = 310.15 K
On solving:
[tex]In (2.3 * 10^{14}) = (1/(8.314*310.15) (Ea_1 - Ea_2)\\\\33.1 = (1/2578.5871) (Ea_1 - Ea_2)\\\\(Ea_1 - Ea_2) = 33.1 * 2578.5871\\\\ (Ea_1 - Ea_2)= 85351.23 J/mol[/tex]
Thus, the activation energy is lowered by +85.35 KJ/mol.
Find more information about Activation energy here:
brainly.com/question/1380484
