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Ch.14 - Chemical Kinetics
Chapter 14, Problem 64

The rate constant (k) for a reaction was measured as a function of temperature. A plot of ln k versus 1/T (in K) is linear and has a slope of -1.01 * 10^4 K. Calculate the activation energy for the reaction.

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1
Identify the relationship between the slope of the plot and the activation energy using the Arrhenius equation: \( \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \).
Recognize that the slope of the plot \( \ln k \) versus \( \frac{1}{T} \) is equal to \( -\frac{E_a}{R} \), where \( E_a \) is the activation energy and \( R \) is the universal gas constant.
Use the given slope \(-1.01 \times 10^4 \text{ K}\) to set up the equation: \(-\frac{E_a}{R} = -1.01 \times 10^4 \text{ K}\).
Solve for the activation energy \( E_a \) by rearranging the equation: \( E_a = 1.01 \times 10^4 \text{ K} \times R \).
Substitute the value of the universal gas constant \( R = 8.314 \text{ J/mol K} \) into the equation to find \( E_a \).
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