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For the reversible, one-step reaction \(\mathrm{A}+\mathrm{B} \rightleftarrows \mathrm{C}+\mathrm{D}\), the forward rate constant is \(52.4 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\) and the rate constant for the reverse reaction is \(32.1 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\). The activation energy was found to be \(35.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the forward reaction and \(44.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the reverse reaction. (a) What is the equilibrium constant for the reaction? (b) Is the reaction exothermic or endothermic? (c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

Short Answer

Expert verified
The equilibrium constant, K, is 1.63. The reaction is exothermic. Raising the temperature will increase both rate constants, but K will decrease as the reaction shifts to favor the reverse reaction.

Step by step solution

01

Calculate the Equilibrium Constant

The equilibrium constant, K, can be calculated from the rate constants of the forward and reverse reactions using the relation: \(K = \frac{k_{forward}}{k_{reverse}}\). Substitute the given rate constants to find K.
02

Determine the Thermodynamic Nature of the Reaction

Since activation energy for the forward reaction is less than the activation energy for the reverse reaction, the forward reaction requires less energy to proceed. This indicates that the overall reaction is exothermic.
03

Predict the Effect of Temperature on Rate Constants and Equilibrium Constant

If the temperature is raised, according to the Arrhenius equation, both the forward and reverse rate constants will increase. However, because the reaction is exothermic, the equilibrium constant K will decrease with an increase in temperature as per Le Chatelier's principle.

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Key Concepts

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

Rate Constants

Rate constants are vital in the study of reaction kinetics, as they quantify the speed of a chemical reaction. For the reaction \(\mathrm{A}+\mathrm{B} \rightleftarrows \mathrm{C}+\mathrm{D}\), the forward rate constant and reverse rate constant are symbolic of how swiftly reactants convert to products and vice versa, respectively. Specifically, the greater the value of a rate constant, the faster the reaction proceeds in that direction. The units of rate constants (\(\mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\)) also suggest the reaction order, implicating a second-order reaction influenced by the concentration of two reactants.

Understanding this concept is fundamental when studying dynamic chemical systems and dictates the feasibility and extent to which reactions can occur, ultimately manifesting as an equilibrium constant under specific conditions.

Thermodynamic Nature of Reactions

The thermodynamic nature of a chemical reaction informs us about the energy changes during the reaction. It is most commonly described as either endothermic or exothermic. An exothermic reaction, like the given reaction where the activation energy for the forward process is lower, releases heat into the surrounding, signifying that the products have less energy than the reactants. Conversely, in an endothermic reaction, heat is absorbed from the environment, meaning that the products store more energy than the reactants.

The indicators of thermodynamic nature provide insights into the energy profile of a reaction and influence the reaction's spontaneity and how conditions such as temperature affect reaction progress.

Arrhenius Equation

The Arrhenius equation is a mathematical expression that links the rate constant (\( k \)) of a chemical reaction to temperature and activation energy (\( E_a \) ). It is generally presented as \( k = A \exp\left(-\frac{E_a}{RT}\right) \), where \( A \) is the frequency factor, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. This equation elucidates that increasing the temperature or having a lower activation energy will accelerate the reaction rate because it amplifies the number of particles with sufficient energy to overcome the energy barrier of the activation energy.

In the context of our exercise, higher temperatures will enhance both the forward and reverse rate constants, yet they won't increase uniformly due to different activation energies, an aspect that profoundly affects the equilibrium state.

Le Chatelier's Principle

Le Chatelier's principle elegantly predicts how a system at equilibrium reacts to disturbances. It posits that if a dynamic equilibrium is subjected to change in conditions, such as temperature, pressure, or concentration, the system adjusts to counter that change and restore a new equilibrium. When the temperature is heightened for an exothermic reaction, the principle suggests that the equilibrium will shift to favor the endothermic reverse reaction to absorb excess heat, thereby reducing the equilibrium constant (\( K \)).

This principle is crucial for understanding and controlling chemical reactions, especially in industrial applications where yield optimization is essential. It enforces the concept that equilibrium is a balance, not a static state, sensitive to external influences.

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Most popular questions from this chapter

(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 decomposition of iodoethane into ethene and hydrogen iodide, \(\mathrm{C}_{2} \mathrm{H} \mathrm{H}_{5} \mathrm{I}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{HI}(\mathrm{g})\), and determine the activation energy for the reaction. (b) What is the value of the rate constant at \(400^{\circ} \mathrm{C}\) ? $$ \begin{array}{lcccc} T(\mathbf{K}) & 660 & 680 & 720 & 760 \\ k\left(\mathbf{s}^{-1}\right) & 7.2 \times 10^{-4} & 2.2 \times 10^{-3} & 1.7 \times 10^{-2} & 0.11 \end{array} $$

When the rate of the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow\) \(2 \mathrm{NO}_{2}(\mathrm{~g})\) was studied, the rate was found to double when the \(\mathrm{O}_{2}\) concentration alone was doubled but to quadruple when the NO concentration alone was doubled. Which of the following mechanisms accounts for these observations? Explain your reasoning. (a) Step \(1 \mathrm{NO}+\mathrm{O}_{2} \longrightarrow \mathrm{NO}_{3}\) and its reverse (both fast, equilibrium) Step \(2 \mathrm{NO}+\mathrm{NO}_{3} \rightarrow \mathrm{NO}_{2}+\mathrm{NO}_{2}\) (slow) (b) Step \(1 \mathrm{NO}+\mathrm{NO} \rightarrow \mathrm{N}_{2} \mathrm{O}_{2}\) (slow) Step \(2 \mathrm{O}_{2}+\mathrm{N}_{2} \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}\) (fast) Step \(3 \mathrm{~N}_{2} \mathrm{O}_{4} \rightarrow \mathrm{NO}_{2}+\mathrm{NO}_{2}\) (fast)

(a) Using a graphing calculator or graphing software, such as that on the Web site for this book, calculate the activation energy for the acid hydrolysis of sucrose to give glucose and fructose from an Arrhenius plot of the data shown here. (b) Calculate the rate constant at \(37^{\circ} \mathrm{C}\) (body temperature). (c) From data in Appendix 2A, calculate the enthalpy change for this reaction, assuming that the solvation enthalpies of the sugars are negligible. Draw an energy profile for the overall process. $$ \begin{array}{cc} \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k\left(\mathrm{~s}^{-1}\right) \\ \hline 24 & 4.8 \times 10^{-3} \\ 28 & 7.8 \times 10^{-3} \\ 32 & 13 \times 10^{-3} \\ 36 & 20 . \times 10^{-3} \\ 40 . & 32 \times 10^{-3} \\ \hline \end{array} $$

The decomposition of gaseous hydrogen iodide, \(2 \mathrm{HI}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\), gives the data shown here for \(700 . \mathrm{K}\). $$ \begin{array}{lcccccc} \text { Time }(\mathrm{s}) & 0 . & 1000 . & 2000 . & 3000 . & 4000 . & 5000 . \\\ {[\mathrm{HI}]\left(\mathrm{mmol} \cdot \mathrm{L}^{-1}\right)} & 10.0 & 4.4 & 2.8 & 2.1 & 1.6 & 1.3 \end{array} $$ (a) Use a graphing calculator or standard graphing software, such as that on the Web site for this book, to plot the concentration of HI as a function of time. (b) Estimate the rate of decomposition of HI at each time. (c) Plot the concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) as a function of time on the same graph.

Manganate ions, \(\mathrm{MnO}_{4}^{2-}\), react at \(2.0 \mathrm{~mol} \cdot \mathrm{L}^{-1} \cdot \mathrm{min}^{-1}\) in acidic solution to form permanganate ions and manganese(IV) oxide: \(3 \mathrm{MnO}_{4}{ }^{2-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{MnO}_{4}{ }^{-}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{~s})+\) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\). (a) What is the rate of formation of permanganate ions? (b) What is the rate of reaction of \(\mathrm{H}^{+}(\mathrm{aq})\) ? (c) What is the unique rate of the reaction?

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