Given the following data for the reaction A $\rightarrow$ B, determine the activation energy, E_a, of the reaction. $$\begin{array} { c c } k ( M / s ) & T ( \mathbf { K } ) \ 2.04 \times 10 ^ { - 4 } & 250 \ 6.78 \times 10 ^ { - 3 } & 400 \end{array}$$

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The Arrhenius equation is:$$k = Ae^{-Ea/RT}$$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 temperatureWe can take the natural logarithm of both sides of the equation to get:$$ln(k) = ln(A) - \frac{Ea}{RT}$$This equation is linear, with a slope of -Ea/R. We can plot ln(k) vs. 1/T and find the slope of the line to get Ea.Using the data given, we get:$$ln(2.04 	imes 10^{-4}) = ln(A) - \frac{Ea}{8.314 J/mol K 	imes 250 K}$$$$ln(6.78 	imes 10^{-3}) = ln(A) - \frac{Ea}{8.314 J/mol K 	imes 400 K}$$Subtracting the first equation from the second, we get:$$ln(6.78 	imes 10^{-3}) - ln(2.04 	imes 10^{-4}) = \frac{Ea}{8.314 J/mol K} 	imes (400 K - 250 K)$$$$ln(3.32) = \frac{Ea}{8.314 J/mol K} 	imes 150 K$$$$Ea = 19.4 kJ/mol$$Therefore, the activation energy of the reaction is 19.4 kJ/mol.

Given the Following Data for the Reaction A →\rightarrow→ B, Determine the Activation Energy, Ea, of the Reaction

Given the Following Data for the Reaction A $\rightarrow$ B, Determine the Activation Energy, E_a, of the Reaction