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The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is: (Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol−1 K−1)
- a)107.2 kJ mol−1
- b)53.6 kJ mol−1
- c)26.8 kJ mol−1
- d)214.4 kJ mol−1
Correct answer is option 'D'. Can you explain this answer?
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The rate of a reaction quadruples when the temperature changes from 30...
To determine the activation energy of the reaction, we can use the Arrhenius equation:
k = A * e^(-Ea / (R * T))
Where:
k is the rate constant
A is the preexponential factor
Ea is the activation energy
R is the gas constant (8.314 J mol^-1 K^-1)
T is the temperature in Kelvin
Given that the rate of the reaction quadruples when the temperature changes from 300 to 310 K, we can write:
k2 = 4 * k1
Using the Arrhenius equation for both temperatures, we have:
k1 = A * e^(-Ea / (R * 300))
k2 = A * e^(-Ea / (R * 310))
Substituting k2 = 4 * k1, we get:
4 * k1 = A * e^(-Ea / (R * 310))
Dividing both sides by k1 and simplifying, we have:
4 = e^(-Ea / (R * 310))
Taking the natural logarithm of both sides, we get:
ln(4) = -Ea / (R * 310)
Simplifying further, we have:
Ea = -ln(4) * R * 310
Substituting the values of ln(4) = 0.693 and R = 8.314 J mol^-1 K^-1, we have:
Ea = -0.693 * 8.314 * 310
Ea ≈ -17372.31 J/mol
Therefore, the activation energy of the reaction is approximately -17372.31 J/mol.
k = A * e^(-Ea / (R * T))
Where:
k is the rate constant
A is the preexponential factor
Ea is the activation energy
R is the gas constant (8.314 J mol^-1 K^-1)
T is the temperature in Kelvin
Given that the rate of the reaction quadruples when the temperature changes from 300 to 310 K, we can write:
k2 = 4 * k1
Using the Arrhenius equation for both temperatures, we have:
k1 = A * e^(-Ea / (R * 300))
k2 = A * e^(-Ea / (R * 310))
Substituting k2 = 4 * k1, we get:
4 * k1 = A * e^(-Ea / (R * 310))
Dividing both sides by k1 and simplifying, we have:
4 = e^(-Ea / (R * 310))
Taking the natural logarithm of both sides, we get:
ln(4) = -Ea / (R * 310)
Simplifying further, we have:
Ea = -ln(4) * R * 310
Substituting the values of ln(4) = 0.693 and R = 8.314 J mol^-1 K^-1, we have:
Ea = -0.693 * 8.314 * 310
Ea ≈ -17372.31 J/mol
Therefore, the activation energy of the reaction is approximately -17372.31 J/mol.
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The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:(Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol−1 K−1)a)107.2 kJ mol−1b)53.6 kJ mol−1c)26.8 kJ mol−1d)214.4 kJ mol−1Correct answer is option 'D'. Can you explain this answer?
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The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:(Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol−1 K−1)a)107.2 kJ mol−1b)53.6 kJ mol−1c)26.8 kJ mol−1d)214.4 kJ 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 quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:(Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol−1 K−1)a)107.2 kJ mol−1b)53.6 kJ mol−1c)26.8 kJ mol−1d)214.4 kJ 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 quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:(Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol−1 K−1)a)107.2 kJ mol−1b)53.6 kJ mol−1c)26.8 kJ mol−1d)214.4 kJ mol−1Correct answer is option 'D'. Can you explain this answer?.
The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:(Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol−1 K−1)a)107.2 kJ mol−1b)53.6 kJ mol−1c)26.8 kJ mol−1d)214.4 kJ 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 quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:(Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol−1 K−1)a)107.2 kJ mol−1b)53.6 kJ mol−1c)26.8 kJ mol−1d)214.4 kJ 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 quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:(Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol−1 K−1)a)107.2 kJ mol−1b)53.6 kJ mol−1c)26.8 kJ mol−1d)214.4 kJ mol−1Correct answer is option 'D'. Can you explain this answer?.
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