the specific rate constant for a reaction increases by a factor of 4 if the tempreature is changed from 27℃to 47℃ find the activation energy for the reaction Related: Collision theory of Chemical Reactions - Chemical Kinetics, CBSE, Class 12, Chemistry

Question

the specific rate constant for a reaction increases by a factor of 4 if the tempreature is changed from 27℃to 47℃ find the activation energy for the reaction Related:  Collision theory of Chemical Reactions - Chemical Kinetics, CBSE, Class 12, Chemistry

Answer

53.33 kj mol-1

Answer

Activation Energy Calculation
The specific rate constant (k) for a reaction is given by the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
- k is the specific rate constant
- A is the pre-exponential factor
- Ea is the activation energy
- R is the gas constant
- T is the temperature in Kelvin

Given Data
- Initial temperature (T1) = 27℃ = 300K
- Final temperature (T2) = 47℃ = 320K
- Rate constant ratio (k2/k1) = 4

Calculations
1. Using the Arrhenius equation, we can write:
k2/k1 = (A * e^(-Ea/RT2)) / (A * e^(-Ea/RT1))
4 = e^(-Ea/R * (1/T2 - 1/T1))
2. Substitute the values of T1, T2, and R:
4 = e^(-Ea/8.314 * (1/320 - 1/300))
3. Solve for Ea:
ln(4) = -Ea/8.314 * (1/320 - 1/300)
Ea = -8.314 * (1/320 - 1/300) * ln(4)

Results
By solving the above equation, we can determine the activation energy (Ea) for the reaction when the temperature is changed from 27℃ to 47℃. This calculation is essential for understanding the effect of temperature on reaction rates based on the Collision theory of Chemical Reactions.

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This section contains 2 multiple choice questions. Each question has four choices (

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