(a) Using a graphing calculator or standard graphing software, such as that on the Web site for this book, make an appropriate Arrhenius plot of the data shown here for the conversion of cyclopropane into propene and calculate the activation energy for the reaction. (b) What is the value of the rate constant at \(600^{\circ} \mathrm{C}\) ? $$ \begin{array}{lcccc} T(\mathrm{~K}) & 750 . & 800 . & 850 . & 900 . \\ k\left(\mathrm{~s}^{-1}\right) & 1.8 \times 10^{-4} & 2.7 \times 10^{-3} & 3.0 \times 10^{-2} & 0.26 \end{array} $$

Short Answer

Expert verified
The activation energy (Ea) can be calculated using the slope of the Arrhenius plot and the gas constant. The rate constant at 600°C can be extrapolated from the Arrhenius equation.

Step by step solution

01

Graph the Arrhenius Plot

An Arrhenius plot graphs the natural logarithm of the rate constant, ln(k), against the inverse of the temperature in kelvins, 1/T. Using the data provided, calculate ln(k) for each rate constant and plot these values against 1/T (in K⁻¹). Use a graphing calculator or standard graphing software to create this plot.
02

Determine the Slope of the Line

The Arrhenius equation in logarithmic form is ln(k) = -Ea/R * (1/T) + ln(A), where Ea is the activation energy, R is the gas constant, and A is the pre-exponential factor. The slope of the line in the Arrhenius plot corresponds to -Ea/R. Determine the slope of the line from the graph.
03

Calculate Activation Energy (Ea)

After finding the slope from the graph, use the gas constant R (8.314 J/mol·K) to calculate the activation energy, Ea, using the relationship Ea = -slope * R.
04

Extrapolate the Rate Constant at 600°C

Using the equation of the line obtained from the Arrhenius plot, extrapolate to find the value of ln(k) at the inverse temperature corresponding to 600°C (873.15 K). Convert this value back from ln(k) to k by using the exponential function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Activation Energy
Activation energy, denoted by the symbol Ea, refers to the minimum quantity of energy required for a chemical reaction to proceed. It acts as an energy barrier that reactants must overcome to transform into products. In the context of an Arrhenius plot, Ea can be graphically determined from the slope of a line when plotting the natural logarithm of the rate constant, \( \ln(k) \), against the inverse of temperature, \( 1/T \). The relationship between them is given by the Arrhenius equation in its logarithmic form: \[ \ln(k) = -\frac{Ea}{R} \cdot \left(\frac{1}{T}\right) + \ln(A) \] Here, R stands for the universal gas constant, and A is the pre-exponential factor that relates to the frequency of collisions and the orientation of reactant molecules. The higher the activation energy, the slower the reaction proceeds at a given temperature, since fewer reactant molecules will have the necessary energy to reach the transition state.
Rate Constant
The rate constant, represented by 'k', is a crucial component in the rate law of a chemical reaction, expressing the speed at which a reaction takes place under certain conditions. It is influenced by various factors, including temperature and activation energy. In the Arrhenius equation, the rate constant is dependent on the temperature and the nature of the reactants. A higher rate constant indicates a faster reaction. As temperature increases, more molecules have the kinetic energy needed to surpass the activation energy barrier, leading to an increase in the rate constant. The dependence of the rate constant on temperature can be studied and visualized through an Arrhenius plot, where variations in \( \ln(k) \) are observed in response to changes in the inverse temperature \( 1/T \). This plot helps chemists understand the temperature sensitivity of a reaction and predict reaction rates under different temperature conditions.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a mathematical function that is essential when working with exponential relationships in chemistry, such as the Arrhenius equation. It is the inverse operation of raising the base of the natural logarithm, represented by 'e', to a power. This function is particularly used to linearize exponential data so that they can be easily analyzed. For instance, the Arrhenius equation in exponential form is complex and difficult to work with directly. However, by taking the natural logarithm of the rate constant, \( \ln(k) \), this non-linear relationship becomes a straight line when plotted against the inverse of temperature, simplifying the process of calculating the activation energy and the pre-exponential factor. When interpreting an Arrhenius plot, understanding the properties of the natural logarithm, like its continuity and the fact that it turns multiplication into addition, helps students analyze the reaction kinetics more intuitively.
Temperature Inverse
In chemistry, the relationship between reaction rates and temperature is often explored using the inverse of temperature, expressed in \( 1/T \), where T is the temperature in kelvins (K). Using the inverse of temperature instead of the temperature itself allows for a linear relationship when coupled with other logarithmic values like the natural logarithm of the rate constant. In an Arrhenius plot, plotting \( \ln(k) \) versus \( 1/T \) reveals the effect of temperature on the reaction rate. As the temperature increases, the value of \( 1/T \) decreases, typically resulting in an increasing rate constant, and thus a faster reaction. This inverse relationship is helpful for chemists to predict how a reaction will behave at different temperatures and is instrumental in determining the optimal conditions for a reaction to occur efficiently.

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Most popular questions from this chapter

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) For a reaction with a very large equilibrium constant, the rate constant of the forward reaction is much larger than the rate constant of the reverse reaction. (b) At equilibrium, the rate constants of the forward and reverse reactions are equal. (c) Increasing the concentration of a reactant increases the rate of a reaction by increasing the rate constant in the forward direction.

All radioactive decay processes follow first-order kinetics. The half-life of the radioactive isotope tritium \(\left({ }^{3} \mathrm{H}\right.\), or \(\left.\mathrm{T}\right)\) is \(12.3\) years. How much of a \(25.0\)-mg sample of tritium would remain after \(10.9\) years?

The mechanism proposed for the oxidation of iodide ion by the hypochlorite ion in aqueous solution is as follows: Step \(1 \mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HClO}+\mathrm{OH}^{-}\)and its reverse (both fast, equilibrium) Step \(2 \mathrm{I}^{-}+\mathrm{HClO} \longrightarrow \mathrm{HIO}+\mathrm{Cl}^{-}\)(slow) Step \(3 \mathrm{HIO}+\mathrm{OH}^{-} \longrightarrow \mathrm{IO}^{-}+\mathrm{H}_{2} \mathrm{O}\) (fast) Write the rate law for the formation of HIO implied by this mechanism.

Because partial pressures are proportional to concentrations, rate laws for gas-phase reactions can also be expressed in terms of partial pressures, for instance, as Rate \(=\) \(k P_{\mathrm{X}}\) for a first-order reaction of a gas \(\mathrm{X}\). What are the units for the rate constants when partial pressures are expressed in torr and time is expressed in seconds for (a) zero-order reactions; (b) first-order reactions; (c) second-order reactions?

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes by first- order kinetics with a rate constant of \(3.7 \times 10^{-5} \mathrm{~s}^{-1}\) at \(298 \mathrm{~K}\). (a) What is the half-life (in hours) for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(298 \mathrm{~K}\) ? (b) If \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}=0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), what will be the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) after \(3.5 \mathrm{~h}\) ? (c) How much time (in minutes) will elapse before the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration decreases from \(0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.0135 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) ?

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