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Given that the normal energy of the reactant and product are 40J and 20J respectively and threshold energy of the uncatalysed reaction is 120 J. If the rate of uncatalysed reaction at 400 K becomes equal to the rate of catalysed reaction at 300 K, then what will be the activation energy of the catalysed forward and backward reactions respectively?
- a)80 J, 60 J
- b)60 J, 80 J
- c)80 J, 100 J
- d)50 J, 70 J
Correct answer is option 'B'. Can you explain this answer?
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Given that the normal energy of the reactant and product are 40J and 2...
In absence of catalyst = In presence of catalyst
Most Upvoted Answer
Given that the normal energy of the reactant and product are 40J and 2...
Given:
Normal energy of reactant (E1) = 40 J
Normal energy of product (E2) = 20 J
Threshold energy of uncatalysed reaction (Ea1) = 120 J
Rate of uncatalysed reaction at 400 K = Rate of catalysed reaction at 300 K
To find: Activation energy of the catalysed forward and backward reactions respectively
Solution:
1. Calculation of activation energy of the uncatalysed reaction (Ea1)
We know that the activation energy of the uncatalysed reaction (Ea1) can be calculated using the Arrhenius equation:
k1 = A1 * e^(-Ea1/RT)
k1 = Rate constant of the uncatalysed reaction
A1 = Frequency factor
R = Gas constant
T = Temperature in Kelvin
At 400 K, the rate of the uncatalysed reaction is given to be equal to the rate of the catalysed reaction at 300 K. Therefore,
k1(400) = k2(300)
where k2 is the rate constant of the catalysed reaction.
Substituting the values in the Arrhenius equation and taking natural logarithm on both sides, we get:
ln(k1) = ln(A1) - (Ea1/RT1)
ln(k2) = ln(A2) - (Ea2/RT2)
At equilibrium, k1 = k2, therefore:
ln(A1) - (Ea1/RT1) = ln(A2) - (Ea2/RT2)
ln(A1/A2) = (Ea1/RT1) - (Ea2/RT2)
Substituting the given values, we get:
ln(A1/A2) = (120 J)/(8.314 J/mol.K * 400 K) - (Ea2)/(8.314 J/mol.K * 300 K)
Solving for Ea1, we get:
Ea1 = 60 J
Therefore, the activation energy of the uncatalysed reaction is 60 J.
2. Calculation of activation energy of the catalysed reaction (Ea2)
We know that the activation energy of the catalysed reaction (Ea2) can be calculated using the Eyring equation:
k2 = (kBT/h) * e^(-ΔG‡/RT)
k2 = Rate constant of the catalysed reaction
kB = Boltzmann constant
h = Planck constant
ΔG‡ = Gibbs free energy of activation
R = Gas constant
T = Temperature in Kelvin
Taking natural logarithm on both sides and rearranging, we get:
ln(k2) = ln(kB/h) - (ΔG‡/RT)
At equilibrium, k1 = k2, therefore:
ln(kB/h) - (ΔG‡/RT) = ln(A1) - (Ea1/RT1)
Substituting the given values, we get:
ln(kB/h) - (ΔG‡/RT) = ln(A1) - (60 J)/(8.314 J/mol.K * 400 K)
ln(kB/h) - (ΔG‡/RT) = ln(A1) - 0
Normal energy of reactant (E1) = 40 J
Normal energy of product (E2) = 20 J
Threshold energy of uncatalysed reaction (Ea1) = 120 J
Rate of uncatalysed reaction at 400 K = Rate of catalysed reaction at 300 K
To find: Activation energy of the catalysed forward and backward reactions respectively
Solution:
1. Calculation of activation energy of the uncatalysed reaction (Ea1)
We know that the activation energy of the uncatalysed reaction (Ea1) can be calculated using the Arrhenius equation:
k1 = A1 * e^(-Ea1/RT)
k1 = Rate constant of the uncatalysed reaction
A1 = Frequency factor
R = Gas constant
T = Temperature in Kelvin
At 400 K, the rate of the uncatalysed reaction is given to be equal to the rate of the catalysed reaction at 300 K. Therefore,
k1(400) = k2(300)
where k2 is the rate constant of the catalysed reaction.
Substituting the values in the Arrhenius equation and taking natural logarithm on both sides, we get:
ln(k1) = ln(A1) - (Ea1/RT1)
ln(k2) = ln(A2) - (Ea2/RT2)
At equilibrium, k1 = k2, therefore:
ln(A1) - (Ea1/RT1) = ln(A2) - (Ea2/RT2)
ln(A1/A2) = (Ea1/RT1) - (Ea2/RT2)
Substituting the given values, we get:
ln(A1/A2) = (120 J)/(8.314 J/mol.K * 400 K) - (Ea2)/(8.314 J/mol.K * 300 K)
Solving for Ea1, we get:
Ea1 = 60 J
Therefore, the activation energy of the uncatalysed reaction is 60 J.
2. Calculation of activation energy of the catalysed reaction (Ea2)
We know that the activation energy of the catalysed reaction (Ea2) can be calculated using the Eyring equation:
k2 = (kBT/h) * e^(-ΔG‡/RT)
k2 = Rate constant of the catalysed reaction
kB = Boltzmann constant
h = Planck constant
ΔG‡ = Gibbs free energy of activation
R = Gas constant
T = Temperature in Kelvin
Taking natural logarithm on both sides and rearranging, we get:
ln(k2) = ln(kB/h) - (ΔG‡/RT)
At equilibrium, k1 = k2, therefore:
ln(kB/h) - (ΔG‡/RT) = ln(A1) - (Ea1/RT1)
Substituting the given values, we get:
ln(kB/h) - (ΔG‡/RT) = ln(A1) - (60 J)/(8.314 J/mol.K * 400 K)
ln(kB/h) - (ΔG‡/RT) = ln(A1) - 0
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Given that the normal energy of the reactant and product are 40J and 20J respectively and threshold energy of the uncatalysed reaction is 120 J. If the rate of uncatalysed reaction at 400 K becomes equal to the rate of catalysed reaction at 300 K, then what will be the activation energy of the catalysed forward and backward reactions respectively?a)80 J, 60 Jb)60 J,80 Jc)80 J, 100 J d)50 J, 70 JCorrect answer is option 'B'. Can you explain this answer?
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Given that the normal energy of the reactant and product are 40J and 20J respectively and threshold energy of the uncatalysed reaction is 120 J. If the rate of uncatalysed reaction at 400 K becomes equal to the rate of catalysed reaction at 300 K, then what will be the activation energy of the catalysed forward and backward reactions respectively?a)80 J, 60 Jb)60 J,80 Jc)80 J, 100 J d)50 J, 70 JCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Given that the normal energy of the reactant and product are 40J and 20J respectively and threshold energy of the uncatalysed reaction is 120 J. If the rate of uncatalysed reaction at 400 K becomes equal to the rate of catalysed reaction at 300 K, then what will be the activation energy of the catalysed forward and backward reactions respectively?a)80 J, 60 Jb)60 J,80 Jc)80 J, 100 J d)50 J, 70 JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given that the normal energy of the reactant and product are 40J and 20J respectively and threshold energy of the uncatalysed reaction is 120 J. If the rate of uncatalysed reaction at 400 K becomes equal to the rate of catalysed reaction at 300 K, then what will be the activation energy of the catalysed forward and backward reactions respectively?a)80 J, 60 Jb)60 J,80 Jc)80 J, 100 J d)50 J, 70 JCorrect answer is option 'B'. Can you explain this answer?.
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