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For a reaction, the rate constant at 25 °C is doubled when the temperature is raised to 45 °C. The activation energy (in kJ mol–1) of the reaction is _______ [Given: ln2 = 0.693]
Correct answer is '27'. Can you explain this answer?
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For a reaction, the rate constant at 25 °C is doubled when the te...
Solution:
Given data:
Temperature T1 = 25 °C = 298 K
Temperature T2 = 45 °C = 318 K
Rate constant k1 at temperature T1 = k
Rate constant k2 at temperature T2 = 2k
Activation energy Ea = ?
We know that the Arrhenius equation is given by:
k = A e^(-Ea/RT)
Where,
k = rate constant
A = frequency factor
Ea = activation energy
R = gas constant = 8.314 J mol^-1 K^-1
T = temperature in Kelvin
Doubling the rate constant k to 2k on increasing the temperature from T1 to T2 implies that the activation energy Ea is constant.
Therefore, we can write:
k1 = A e^(-Ea/RT1)
k2 = A e^(-Ea/RT2)
Dividing k2 by k1, we get:
k2/k1 = e^(Ea/R * (1/T1 - 1/T2))
Substituting the given values, we get:
2 = e^(Ea/8.314 * (1/298 - 1/318))
Taking natural logarithm on both sides, we get:
ln2 = Ea/8.314 * (1/298 - 1/318)
Solving for Ea, we get:
Ea = ln2 * 8.314 / (1/298 - 1/318)
Ea = 27.09 kJ mol^-1 (approx.)
Therefore, the activation energy of the reaction is 27 kJ mol^-1.
Given data:
Temperature T1 = 25 °C = 298 K
Temperature T2 = 45 °C = 318 K
Rate constant k1 at temperature T1 = k
Rate constant k2 at temperature T2 = 2k
Activation energy Ea = ?
We know that the Arrhenius equation is given by:
k = A e^(-Ea/RT)
Where,
k = rate constant
A = frequency factor
Ea = activation energy
R = gas constant = 8.314 J mol^-1 K^-1
T = temperature in Kelvin
Doubling the rate constant k to 2k on increasing the temperature from T1 to T2 implies that the activation energy Ea is constant.
Therefore, we can write:
k1 = A e^(-Ea/RT1)
k2 = A e^(-Ea/RT2)
Dividing k2 by k1, we get:
k2/k1 = e^(Ea/R * (1/T1 - 1/T2))
Substituting the given values, we get:
2 = e^(Ea/8.314 * (1/298 - 1/318))
Taking natural logarithm on both sides, we get:
ln2 = Ea/8.314 * (1/298 - 1/318)
Solving for Ea, we get:
Ea = ln2 * 8.314 / (1/298 - 1/318)
Ea = 27.09 kJ mol^-1 (approx.)
Therefore, the activation energy of the reaction is 27 kJ mol^-1.
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For a reaction, the rate constant at 25 °C is doubled when the temperature is raised to 45 °C. The activation energy (in kJ mol–1) of the reaction is _______ [Given: ln2 = 0.693]Correct answer is '27'. Can you explain this answer?
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