Question

For a given reaction, the rate constant doubles with the temperature increases from 20.6 oC to 38.8 oC. What is the activation energy (Ea) for the reaction? Ea = ___ kJ/mo

Answer 1

The activation energy (Ea) for the reaction is approximately 62.177 kJ/mol.

Step-by-step explanation:

The activation energy (Ea) can be determined using the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and the activation energy. The Arrhenius equation is given by:

k = Ae-Ea/RT

By comparing the rate constants at two different temperatures, we can determine the activation energy. If the rate constant doubles when the temperature increases from 20.6 oC to 38.8 oC, it means that the exponential term e-Ea/RT is equal to 2. Taking the natural logarithm of both sides, we can solve for Ea:

ln(k2/k1) = (-Ea/R) * (1/T2 - 1/T1)

Plugging in the values, we can solve for Ea:

Ea ≈ 62.177 kJ/mol

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