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What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R = 8.314Imol−1K−1)
- a)34.7kJmol−1
- b)15.1kJmol−1
- c)342kJmol−1
- d)296kJmol−1
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
What is the activation energy for a reaction if its rate doubles when...
k2 = 2k1
T1 = 20 + 273 = 293K or T2 = 35 + 273 = 308K
R = 8.314Jmol−1K−1
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What is the activation energy for a reaction if its rate doubles when...
Calculating Activation Energy:
To calculate the activation energy for a reaction, we can use the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 Jmol^−1K^−1)
T = temperature in Kelvin
Given Information:
- Initial temperature (T1) = 20°C = 293K
- Final temperature (T2) = 35°C = 308K
- Rate doubles when temperature is raised
Using Rate Constant:
According to the Arrhenius equation, if the rate doubles, then the ratio of rate constants at two different temperatures is:
k2/k1 = 2
Substitute the values of T1, T2, and R into the Arrhenius equation:
A * e^(-Ea/RT2) / A * e^(-Ea/RT1) = 2
e^(-Ea/R * (1/T2 - 1/T1)) = 2
Solving for Ea:
Ea = -R * ln(2) / (1/T2 - 1/T1)
Ea = -8.314 * ln(2) / (1/308 - 1/293)
Ea = 34.7 kJmol^−1
Therefore, the activation energy for the reaction is 34.7 kJmol^−1, which corresponds to option 'A'.
To calculate the activation energy for a reaction, we can use the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 Jmol^−1K^−1)
T = temperature in Kelvin
Given Information:
- Initial temperature (T1) = 20°C = 293K
- Final temperature (T2) = 35°C = 308K
- Rate doubles when temperature is raised
Using Rate Constant:
According to the Arrhenius equation, if the rate doubles, then the ratio of rate constants at two different temperatures is:
k2/k1 = 2
Substitute the values of T1, T2, and R into the Arrhenius equation:
A * e^(-Ea/RT2) / A * e^(-Ea/RT1) = 2
e^(-Ea/R * (1/T2 - 1/T1)) = 2
Solving for Ea:
Ea = -R * ln(2) / (1/T2 - 1/T1)
Ea = -8.314 * ln(2) / (1/308 - 1/293)
Ea = 34.7 kJmol^−1
Therefore, the activation energy for the reaction is 34.7 kJmol^−1, which corresponds to option 'A'.
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What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R = 8.314Imol−1K−1)a)34.7kJmol−1b)15.1kJmol−1c)342kJmol−1d)296kJmol−1Correct answer is option 'A'. Can you explain this answer?
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