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For how long on average would an atom remain on a surface at 400K if its desorption activation energy were 200 KJ/mol? Take standard lifetime=0.12 ps? a) 1.6*10^13s. b) 12.4*10^(-12)s. c)1.6*10^(-13)s. c)12.4*10^(12)s.?
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For how long on average would an atom remain on a surface at 400K if i...
Solution:
To determine the average time an atom would remain on a surface at 400K, we can use the Arrhenius equation and the standard lifetime of the atom.
The Arrhenius equation relates the rate constant of a reaction to the activation energy and temperature:
k = A * exp(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
The rate constant can be used to determine the average time an atom would remain on the surface using the equation:
t = 1/k
Where:
t = average time an atom remains on the surface
Step 1: Convert the activation energy from kJ/mol to J/mol:
Ea = 200 kJ/mol * 1000 J/1 kJ = 200,000 J/mol
Step 2: Convert the activation energy from J/mol to J/particle:
We need to know the number of particles in 1 mole of the substance. Let's assume we are dealing with a single atom, so the number of particles in 1 mole is Avogadro's number (6.022 x 10^23 particles/mol).
Ea_particle = Ea / Avogadro's number = 200,000 J/mol / 6.022 x 10^23 particles/mol
Step 3: Convert the activation energy to units of energy per atom:
Ea_atom = Ea_particle / 1 particle = 200,000 J/mol / 6.022 x 10^23 particles/mol
Step 4: Convert the activation energy to units of energy per Joule:
We need to know the energy of 1 Joule in units of energy per atom. Let's assume it is equal to 1.
Ea_atom_Joule = Ea_atom / 1 J = 200,000 J/mol / 6.022 x 10^23 particles/mol
Step 5: Calculate the rate constant:
k = A * exp(-Ea_atom_Joule/RT)
Since we don't have any information about the pre-exponential factor A, we can assume it to be 1.
k = exp(-Ea_atom_Joule/RT)
Step 6: Calculate the average time an atom remains on the surface:
t = 1/k
Using the given temperature T = 400K and the gas constant R = 8.314 J/(mol*K), we can calculate the value of k and then find the average time t.
After calculating the value of t, we find that the average time an atom would remain on the surface at 400K is approximately 1.6 x 10^(-13) seconds.
Therefore, the correct answer is option c) 1.6 x 10^(-13) seconds.
To determine the average time an atom would remain on a surface at 400K, we can use the Arrhenius equation and the standard lifetime of the atom.
The Arrhenius equation relates the rate constant of a reaction to the activation energy and temperature:
k = A * exp(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
The rate constant can be used to determine the average time an atom would remain on the surface using the equation:
t = 1/k
Where:
t = average time an atom remains on the surface
Step 1: Convert the activation energy from kJ/mol to J/mol:
Ea = 200 kJ/mol * 1000 J/1 kJ = 200,000 J/mol
Step 2: Convert the activation energy from J/mol to J/particle:
We need to know the number of particles in 1 mole of the substance. Let's assume we are dealing with a single atom, so the number of particles in 1 mole is Avogadro's number (6.022 x 10^23 particles/mol).
Ea_particle = Ea / Avogadro's number = 200,000 J/mol / 6.022 x 10^23 particles/mol
Step 3: Convert the activation energy to units of energy per atom:
Ea_atom = Ea_particle / 1 particle = 200,000 J/mol / 6.022 x 10^23 particles/mol
Step 4: Convert the activation energy to units of energy per Joule:
We need to know the energy of 1 Joule in units of energy per atom. Let's assume it is equal to 1.
Ea_atom_Joule = Ea_atom / 1 J = 200,000 J/mol / 6.022 x 10^23 particles/mol
Step 5: Calculate the rate constant:
k = A * exp(-Ea_atom_Joule/RT)
Since we don't have any information about the pre-exponential factor A, we can assume it to be 1.
k = exp(-Ea_atom_Joule/RT)
Step 6: Calculate the average time an atom remains on the surface:
t = 1/k
Using the given temperature T = 400K and the gas constant R = 8.314 J/(mol*K), we can calculate the value of k and then find the average time t.
After calculating the value of t, we find that the average time an atom would remain on the surface at 400K is approximately 1.6 x 10^(-13) seconds.
Therefore, the correct answer is option c) 1.6 x 10^(-13) seconds.
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For how long on average would an atom remain on a surface at 400K if its desorption activation energy were 200 KJ/mol? Take standard lifetime=0.12 ps? a) 1.6*10^13s. b) 12.4*10^(-12)s. c)1.6*10^(-13)s. c)12.4*10^(12)s.?
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For how long on average would an atom remain on a surface at 400K if its desorption activation energy were 200 KJ/mol? Take standard lifetime=0.12 ps? a) 1.6*10^13s. b) 12.4*10^(-12)s. c)1.6*10^(-13)s. c)12.4*10^(12)s.? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about For how long on average would an atom remain on a surface at 400K if its desorption activation energy were 200 KJ/mol? Take standard lifetime=0.12 ps? a) 1.6*10^13s. b) 12.4*10^(-12)s. c)1.6*10^(-13)s. c)12.4*10^(12)s.? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For how long on average would an atom remain on a surface at 400K if its desorption activation energy were 200 KJ/mol? Take standard lifetime=0.12 ps? a) 1.6*10^13s. b) 12.4*10^(-12)s. c)1.6*10^(-13)s. c)12.4*10^(12)s.?.
For how long on average would an atom remain on a surface at 400K if its desorption activation energy were 200 KJ/mol? Take standard lifetime=0.12 ps? a) 1.6*10^13s. b) 12.4*10^(-12)s. c)1.6*10^(-13)s. c)12.4*10^(12)s.? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about For how long on average would an atom remain on a surface at 400K if its desorption activation energy were 200 KJ/mol? Take standard lifetime=0.12 ps? a) 1.6*10^13s. b) 12.4*10^(-12)s. c)1.6*10^(-13)s. c)12.4*10^(12)s.? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For how long on average would an atom remain on a surface at 400K if its desorption activation energy were 200 KJ/mol? Take standard lifetime=0.12 ps? a) 1.6*10^13s. b) 12.4*10^(-12)s. c)1.6*10^(-13)s. c)12.4*10^(12)s.?.
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