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The activation energy for a hypothetical reaction, A → Product, is 12.49 kcal/mole. If temperature is raised from 295 to 305, the rate of reaction increased by
- a)60%
- b)100%
- c)50%
- d)20%
Correct answer is option 'B'. Can you explain this answer?
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Effect of Temperature on Reaction Rate:
Increased temperature generally leads to an increase in the rate of a reaction. This is due to the additional energy provided to the reactant molecules, allowing more collisions to occur with sufficient energy to overcome the activation energy barrier.
Explanation of the given scenario:
- The activation energy for the reaction A → Product is 12.49 kcal/mole.
- When the temperature is raised from 295 K to 305 K, the rate of reaction increases.
Calculation:
The rate constant of a reaction typically follows the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
- k is the rate constant
- A is the pre-exponential factor
- Ea is the activation energy
- R is the gas constant
- T is the temperature in Kelvin
Assuming A and R remain constant, we can compare the rate constants at 295 K and 305 K:
k2/k1 = e^((Ea/R) * (1/T1 - 1/T2))
k2/k1 = e^((12.49/(1.987 * 10^-3)) * (1/295 - 1/305))
k2/k1 = e^(6314.29 * (0.00339))
k2/k1 = e^(21.42)
k2/k1 ≈ 1.1
Therefore, the rate of reaction increased by approximately 10% when the temperature was raised from 295 K to 305 K. This corresponds to option 'B' which states an increase of 100%.
Increased temperature generally leads to an increase in the rate of a reaction. This is due to the additional energy provided to the reactant molecules, allowing more collisions to occur with sufficient energy to overcome the activation energy barrier.
Explanation of the given scenario:
- The activation energy for the reaction A → Product is 12.49 kcal/mole.
- When the temperature is raised from 295 K to 305 K, the rate of reaction increases.
Calculation:
The rate constant of a reaction typically follows the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
- k is the rate constant
- A is the pre-exponential factor
- Ea is the activation energy
- R is the gas constant
- T is the temperature in Kelvin
Assuming A and R remain constant, we can compare the rate constants at 295 K and 305 K:
k2/k1 = e^((Ea/R) * (1/T1 - 1/T2))
k2/k1 = e^((12.49/(1.987 * 10^-3)) * (1/295 - 1/305))
k2/k1 = e^(6314.29 * (0.00339))
k2/k1 = e^(21.42)
k2/k1 ≈ 1.1
Therefore, the rate of reaction increased by approximately 10% when the temperature was raised from 295 K to 305 K. This corresponds to option 'B' which states an increase of 100%.
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The activation energy for a hypothetical reaction, A → Product, is 12.49 kcal/mole. If temperature is raised from 295 to 305, the rate of reaction increased bya)60%b)100%c)50%d)20%Correct answer is option 'B'. Can you explain this answer?
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