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Ch.14 - Chemical Kinetics
All textbooksTro 4th EditionCh.14 - Chemical KineticsProblem 7
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Chapter 14, Problem 7

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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Textbook Question

Consider the reaction:

2 HBr (g) → H2 (g) + Br2 (g)

a. Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

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Textbook Question

Consider the reaction:

2 HBr (g) → H2 (g) + Br2 (g)

b. In the first 25.0 s of this reaction, the concentration of HBr dropped from 0.600 M to 0.512 M. Calculate the average rate of the reaction during this time interval.

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