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What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20ºC to 35ºC ? (R = 8.314 J mol–1 K–1)
- a)342 kJ mol–1
- b)269 kJ mol–1
- c)34.7 kJ mol–1
- d)15.1 kJ mol–1
Correct answer is option 'C'. Can you explain this answer?
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The activation energy for a reaction cannot be determined solely from the information provided. To calculate the activation energy, you need to know the rate constant at two different temperatures. The rate constant is related to the rate of reaction and can be determined experimentally.
If you have the rate constant at two different temperatures, you can use the Arrhenius equation to calculate the activation energy:
k1 = A * e^(-Ea / (R * T1))
k2 = A * e^(-Ea / (R * T2))
where k1 and k2 are the rate constants at temperatures T1 and T2 respectively, Ea is the activation energy, A is the pre-exponential factor, and R is the gas constant.
By taking the ratio of the rate constants, you can eliminate A and solve for Ea:
k2 / k1 = (e^(-Ea / (R * T2))) / (e^(-Ea / (R * T1)))
Take the natural logarithm of both sides:
ln(k2 / k1) = (-Ea / (R * T2)) + (Ea / (R * T1))
Rearrange the equation to solve for Ea:
Ea = -R * ((ln(k2 / k1)) / ((1 / T2) - (1 / T1)))
Given that the rate doubles when the temperature is raised, it implies that k2 is twice k1. However, since the specific values of k1, k2, T1, and T2 are not provided, it is not possible to calculate the activation energy using this information alone.
If you have the rate constant at two different temperatures, you can use the Arrhenius equation to calculate the activation energy:
k1 = A * e^(-Ea / (R * T1))
k2 = A * e^(-Ea / (R * T2))
where k1 and k2 are the rate constants at temperatures T1 and T2 respectively, Ea is the activation energy, A is the pre-exponential factor, and R is the gas constant.
By taking the ratio of the rate constants, you can eliminate A and solve for Ea:
k2 / k1 = (e^(-Ea / (R * T2))) / (e^(-Ea / (R * T1)))
Take the natural logarithm of both sides:
ln(k2 / k1) = (-Ea / (R * T2)) + (Ea / (R * T1))
Rearrange the equation to solve for Ea:
Ea = -R * ((ln(k2 / k1)) / ((1 / T2) - (1 / T1)))
Given that the rate doubles when the temperature is raised, it implies that k2 is twice k1. However, since the specific values of k1, k2, T1, and T2 are not provided, it is not possible to calculate the activation energy using this information alone.
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What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20ºC to35ºC ? (R = 8.314 J mol–1 K–1)a)342 kJ mol–1b)269 kJ mol–1c)34.7 kJ mol–1d)15.1 kJ mol–1Correct answer is option 'C'. Can you explain this answer?
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What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20ºC to35ºC ? (R = 8.314 J mol–1 K–1)a)342 kJ mol–1b)269 kJ mol–1c)34.7 kJ mol–1d)15.1 kJ mol–1Correct answer is option 'C'. 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 What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20ºC to35ºC ? (R = 8.314 J mol–1 K–1)a)342 kJ mol–1b)269 kJ mol–1c)34.7 kJ mol–1d)15.1 kJ mol–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20ºC to35ºC ? (R = 8.314 J mol–1 K–1)a)342 kJ mol–1b)269 kJ mol–1c)34.7 kJ mol–1d)15.1 kJ mol–1Correct answer is option 'C'. Can you explain this answer?.
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20ºC to35ºC ? (R = 8.314 J mol–1 K–1)a)342 kJ mol–1b)269 kJ mol–1c)34.7 kJ mol–1d)15.1 kJ mol–1Correct answer is option 'C'. 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 What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20ºC to35ºC ? (R = 8.314 J mol–1 K–1)a)342 kJ mol–1b)269 kJ mol–1c)34.7 kJ mol–1d)15.1 kJ mol–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20ºC to35ºC ? (R = 8.314 J mol–1 K–1)a)342 kJ mol–1b)269 kJ mol–1c)34.7 kJ mol–1d)15.1 kJ mol–1Correct answer is option 'C'. Can you explain this answer?.
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