A reaction has a rate constant of 0.000122 s⁻¹ at 27 °C and 0.228 s⁻¹ at 77 °C. a. Determine the activation barrier for the reaction.

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Identify the given data: rate constants k1 = 0.000122 \text{ s}^{-1} at T1 = 27 \degree C and k2 = 0.228 \text{ s}^{-1} at T2 = 77 \degree C.

Convert the temperatures from Celsius to Kelvin: T1 = 27 + 273.15 \text{ K} and T2 = 77 + 273.15 \text{ K}.

Use the Arrhenius equation: k = A e^{-\frac{E_a}{RT}}, where k is the rate constant, A is the pre-exponential factor, E_a is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.

Apply the Arrhenius equation in its logarithmic form to find the activation energy: \ln\left(\frac{k2}{k1}\right) = \frac{E_a}{R}\left(\frac{1}{T1} - \frac{1}{T2}\right).

Solve for E_a (activation energy) by rearranging the equation: E_a = \frac{R \cdot \ln\left(\frac{k2}{k1}\right)}{\left(\frac{1}{T1} - \frac{1}{T2}\right)}.

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