The activation energy of a reaction is 56.8 kJ/mol and the frequency factor is 1.5⨉1011/ s. Calculate the rate constant of the reaction at 25 °C.
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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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 \( E_a \).
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Arrhenius Equation
The Arrhenius equation describes how the rate constant (k) of a chemical reaction depends on temperature (T) and activation energy (Ea). It is expressed as k = A * e^(-Ea/RT), where A is the pre-exponential factor, R is the universal gas constant, and T is the temperature in Kelvin. This relationship indicates that as temperature increases, the rate constant typically increases, reflecting a higher likelihood of overcoming the activation energy barrier.
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Arrhenius Equation
Activation Energy (Ea)
Activation energy is the minimum energy required for a chemical reaction to occur. It represents the energy barrier that reactants must overcome to transform into products. A higher activation energy means that fewer molecules have sufficient energy to react at a given temperature, resulting in a slower reaction rate. The activation energy can be determined from the slope of a plot of ln k versus 1/T using the equation Ea = -slope * R.
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Activity Series Chart
Linear Relationship in ln k vs. 1/T
The linear relationship between ln k and 1/T indicates that the reaction follows the Arrhenius behavior, allowing for the determination of activation energy. The slope of this line is equal to -Ea/R, where R is the gas constant. This linearity suggests that the rate constant changes exponentially with temperature, providing a straightforward method to calculate Ea from experimental data by rearranging the equation to solve for activation energy.
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Relationship between ∆E°, ∆G°, and K
Related Practice
Textbook Question
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Open Question
The rate constant of a reaction at 32 °C is 0.055 s⁻¹. If the frequency factor is 1.2 × 10¹³ s⁻¹, what is the activation barrier?
Textbook Question
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 -7445 K. Calculate the activation energy for the reaction.
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Textbook Question
The data shown here were collected for the first-order reaction: N2O(g) → N2(g) + O(g) Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction.
Temperature (K) Rate Constant (1 , s)
800 3.24⨉10- 5
900 0.00214
1000 0.0614
1100 0.955
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Textbook Question
The tabulated data show the rate constant of a reaction measured at several different temperatures. Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction.
Temperature (K) Rate Constant (1 , s)
300 0.0134
310 0.0407
320 0.114
330 0.303
340 0.757
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Open Question
The tabulated data were collected for the second-order reaction: Cl(g) + H2(g) → HCl(g) + H(g). Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction. Temperature (K) and Rate Constant (L/mol # s) are as follows: 90 K, 0.00357; 100 K, 0.0773; 110 K, 0.956; 120 K, 7.781.