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If the rate of reaction becomes double when temperature is increased from 25°C to 35°C, then activation energy of the reaction (in kJ) will be
- a)52.89 kJ
- b)65.2 kJ
- c)58.5 kJ
- d)45.3 kJ
Correct answer is option 'A'. Can you explain this answer?
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If the rate of reaction becomes double when temperature is increased ...
Given information:
- Initial temperature (T1) = 25°C
- Final temperature (T2) = 35°C
- Rate of reaction doubles when temperature is increased
Formula for calculating the activation energy (Ea):
- The Arrhenius equation is given by: k = A * e^(-Ea/RT)
- Where k = rate constant, A = pre-exponential factor, Ea = activation energy, R = gas constant, T = temperature in Kelvin
Using the given information:
- Let the initial rate constant be k1 and the final rate constant be k2
- According to the question, k2 = 2k1 (rate of reaction doubles)
- Taking the ratio of the rate constants: k2/k1 = e^(-Ea/R * (1/T2 - 1/T1))
Converting temperatures to Kelvin:
- Initial temperature (T1) = 25 + 273 = 298 K
- Final temperature (T2) = 35 + 273 = 308 K
Plugging in the values and solving for Ea:
- 2 = e^(-Ea/R * (1/308 - 1/298))
- Ln(2) = -Ea/R * (1/308 - 1/298)
- Solving for Ea gives Ea = 52.89 kJ
Therefore, the activation energy of the reaction is 52.89 kJ, which corresponds to option 'A'.
- Initial temperature (T1) = 25°C
- Final temperature (T2) = 35°C
- Rate of reaction doubles when temperature is increased
Formula for calculating the activation energy (Ea):
- The Arrhenius equation is given by: k = A * e^(-Ea/RT)
- Where k = rate constant, A = pre-exponential factor, Ea = activation energy, R = gas constant, T = temperature in Kelvin
Using the given information:
- Let the initial rate constant be k1 and the final rate constant be k2
- According to the question, k2 = 2k1 (rate of reaction doubles)
- Taking the ratio of the rate constants: k2/k1 = e^(-Ea/R * (1/T2 - 1/T1))
Converting temperatures to Kelvin:
- Initial temperature (T1) = 25 + 273 = 298 K
- Final temperature (T2) = 35 + 273 = 308 K
Plugging in the values and solving for Ea:
- 2 = e^(-Ea/R * (1/308 - 1/298))
- Ln(2) = -Ea/R * (1/308 - 1/298)
- Solving for Ea gives Ea = 52.89 kJ
Therefore, the activation energy of the reaction is 52.89 kJ, which corresponds to option 'A'.
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If the rate of reaction becomes double when temperature is increased ...
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If the rate of reaction becomes double when temperature is increased from 25°C to 35°C, then activation energy of the reaction (in kJ) will bea)52.89 kJb)65.2 kJc)58.5 kJd)45.3 kJCorrect answer is option 'A'. Can you explain this answer?
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If the rate of reaction becomes double when temperature is increased from 25°C to 35°C, then activation energy of the reaction (in kJ) will bea)52.89 kJb)65.2 kJc)58.5 kJd)45.3 kJCorrect answer is option 'A'. 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 If the rate of reaction becomes double when temperature is increased from 25°C to 35°C, then activation energy of the reaction (in kJ) will bea)52.89 kJb)65.2 kJc)58.5 kJd)45.3 kJCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the rate of reaction becomes double when temperature is increased from 25°C to 35°C, then activation energy of the reaction (in kJ) will bea)52.89 kJb)65.2 kJc)58.5 kJd)45.3 kJCorrect answer is option 'A'. Can you explain this answer?.
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