Ch.15 - Chemical Kinetics

Tro - Chemistry: A Molecular Approach 6th Edition

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Tro 6th Edition

Ch.15 - Chemical Kinetics

Problem 69

Chapter 15, Problem 69

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 -7012 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 of the plot, which is -7012 K, and set it equal to \( -\frac{E_a}{R} \).

Solve for the activation energy \( E_a \) by rearranging the equation: \( E_a = -\text{slope} \times R \).

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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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.

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, as it is directly related to the slope in the Arrhenius equation.

Linear Regression in Kinetics

In chemical kinetics, linear regression is often used to analyze the relationship between the natural logarithm of the rate constant (ln k) and the inverse of temperature (1/T). A linear plot indicates a consistent relationship, allowing for the determination of activation energy from the slope. The slope of the line in this context is equal to -Ea/R, where R is the gas constant, enabling the calculation of activation energy when the slope is known.

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Related Practice

Textbook Question

The diagram shows the energy of a reaction as the reaction progresses. Label each blank box in the diagram.

a. reactants b. products c. activation energy (E_a) d. enthalpy of reaction (ΔH_rxn)

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Textbook Question

The half-life for the radioactive decay of C-14 is 5715 years and is independent of the initial concentration. If a sample of C-14 initially contains 1.5 mmol of C-14, how many millimoles are left after 2725 years?

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

1575

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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

2534

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Textbook Question

The activation energy of a reaction is 44.2 kJ/mol and the frequency factor is 1.9⨉10¹¹/ s. Calculate the rate constant of the reaction at 25 °C.

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