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The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such
a reaction will be:
a reaction will be:
(R = 8.314 JK–1 mol–1 and log 2 = 0.301)
- a)48.6 kJ mol–1
- b)58.5 kJ mol–1
- c)60.5 kJ mol–1
- d)53.6 kJ mol–1
Correct answer is option 'D'. Can you explain this answer?
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The rate of a reaction doubles when its temperature changes from 300K ...
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The rate of a reaction doubles when its temperature changes from 300K ...
Mol^-1 K^-1)
To determine the activation energy (Ea), we can use the Arrhenius equation:
k = Ae^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J mol^-1 K^-1)
T = temperature in Kelvin
Given that the rate of the reaction doubles when the temperature changes from 300K to 310K, we can write:
k2 = 2k1
Substituting the values into the Arrhenius equation for each temperature:
k1 = A*e^(-Ea/(R*300))
k2 = A*e^(-Ea/(R*310))
Since k2 = 2k1, we can write:
2k1 = A*e^(-Ea/(R*310))
Dividing the two equations:
(2k1)/(k1) = (A*e^(-Ea/(R*310)))/(A*e^(-Ea/(R*300)))
2 = e^((Ea/(R*300)) - (Ea/(R*310)))
Taking the natural logarithm of both sides:
ln(2) = (Ea/(R*300)) - (Ea/(R*310))
Rearranging the equation to solve for Ea:
ln(2) = Ea/(R*300) - Ea/(R*310)
ln(2) = Ea(R*310 - R*300)/(R*300*R*310)
ln(2) = Ea(10)/(8.314*300*310)
Simplifying the equation:
Ea = (8.314*300*310*ln(2))/(10)
Calculating Ea:
Ea ≈ 50.9 kJ/mol
To determine the activation energy (Ea), we can use the Arrhenius equation:
k = Ae^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J mol^-1 K^-1)
T = temperature in Kelvin
Given that the rate of the reaction doubles when the temperature changes from 300K to 310K, we can write:
k2 = 2k1
Substituting the values into the Arrhenius equation for each temperature:
k1 = A*e^(-Ea/(R*300))
k2 = A*e^(-Ea/(R*310))
Since k2 = 2k1, we can write:
2k1 = A*e^(-Ea/(R*310))
Dividing the two equations:
(2k1)/(k1) = (A*e^(-Ea/(R*310)))/(A*e^(-Ea/(R*300)))
2 = e^((Ea/(R*300)) - (Ea/(R*310)))
Taking the natural logarithm of both sides:
ln(2) = (Ea/(R*300)) - (Ea/(R*310))
Rearranging the equation to solve for Ea:
ln(2) = Ea/(R*300) - Ea/(R*310)
ln(2) = Ea(R*310 - R*300)/(R*300*R*310)
ln(2) = Ea(10)/(8.314*300*310)
Simplifying the equation:
Ea = (8.314*300*310*ln(2))/(10)
Calculating Ea:
Ea ≈ 50.9 kJ/mol
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The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of sucha reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)a)48.6 kJ mol–1b)58.5kJ mol–1c)60.5kJ mol–1d)53.6kJ mol–1Correct answer is option 'D'. Can you explain this answer?
Question Description
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of sucha reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)a)48.6 kJ mol–1b)58.5kJ mol–1c)60.5kJ mol–1d)53.6kJ mol–1Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of sucha reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)a)48.6 kJ mol–1b)58.5kJ mol–1c)60.5kJ mol–1d)53.6kJ mol–1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of sucha reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)a)48.6 kJ mol–1b)58.5kJ mol–1c)60.5kJ mol–1d)53.6kJ mol–1Correct answer is option 'D'. Can you explain this answer?.
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of sucha reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)a)48.6 kJ mol–1b)58.5kJ mol–1c)60.5kJ mol–1d)53.6kJ mol–1Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of sucha reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)a)48.6 kJ mol–1b)58.5kJ mol–1c)60.5kJ mol–1d)53.6kJ mol–1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of sucha reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)a)48.6 kJ mol–1b)58.5kJ mol–1c)60.5kJ mol–1d)53.6kJ mol–1Correct answer is option 'D'. Can you explain this answer?.
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