Respuesta :
Answer:
Explanation:
Given data
K1 = 2.57 M-1s-1
T1 = 701 K
K2= ?
T2 = 525 K
Ea = 150 kJ/ mol * 1000 J */ 1 kJ = 150000 J / mol
Formula to calculate the rate constant using the activation energy is as follows
ln[K2/K1] = (Ea/ R )*[(1/T1)-(1/T2)]
where R= 8.314 J per mol. K
lets put the values in the formula and calculate the activation energy
ln [K2/2.57] = 150000 J per mol / 8.314 J per K . mol * [(1/701)-(1/525)]
ln [K2/2.57] = -8.628
K2 / 2.57 = anti ln [-8.628]
K2/2.57 = 0.000179
K2 = 0.000179 *2.57
K2 = 0.00046
Therefore rate constant at 525 K is 4.6*10^-4 L/mol. s
The rate constant at 525 K is [tex]4.6*10^{-4} L/mol. s[/tex]
Given:
k₁= 2.57 M−1⋅s−1
T₁= 701K
T₂=525K
Ea=15000J/mol
To find:
k₂=?
What is Activation Energy?
The minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation or physical transport. It is given by:
[tex]k=Ae^\frac{-E_a}{RT}[/tex]
Since, two rate constants are given so for that case, the formula to be used is:
[tex]ln [\frac{k_2}{k_1} ]= (Ea/ R )*[\frac{1}{T_1}-\frac{1}{T_2}]\\\\ln [\frac{k_2}{2.57} ] = \frac{150000 J / mol}{8.314 J / K . mol} * [\frac{1}{701}-\frac{1}{525}] \\\\ln [\frac{k_2}{2.57} ] = -8.628\\\\\frac{k_2}{2.57} = \text{ anti ln} [-8.628]\\\\\frac{k_2}{2.57} = 0.000179\\\\k_2 = 0.000179 *2.57\\\\k_2 = 0.00046[/tex]
Therefore, the rate constant at 525 K is [tex]4.6*10^{-4}L/mol. s[/tex]
Find more information about Activation Energy here: brainly.com/question/1380484