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Chapter 27: Problem 9

Consider the reaction mechanism $$A+B \stackrel{k_{1}}{\rightleftharpoons} C$$ $$\mathrm{C} \stackrel{k_{2}}{\Rightarrow} \mathrm{P}$$ Write the expression for \(d[\mathrm{P}] / d t,\) the rate of product formation. Assume equilibrium is established in the first reaction before any appreciable amount of product is formed, and thereby show that $$\frac{d[\mathrm{P}]}{d t}=k_{2} K_{c}[\mathrm{A}][\mathrm{B}]$$ where \(K_{c}\) is the equilibrium constant for step ( 1 ) of the reaction mechanism. This assumption is called the fast- equilibrium approximation.

Short Answer

Expert verified
The rate of product formation is \(\frac{d[P]}{dt} = k_2 K_c [A][B]\).

Step by step solution

01

Identify the Fast Equilibrium

The first step is to identify that the reaction \(A + B \rightleftharpoons C\) is in fast equilibrium. This means that the forward and reverse reaction rates are equal at equilibrium. Thus, the rate of formation of C is equal to the rate of its decomposition at equilibrium.
02

Write the Equilibrium Expression

At equilibrium, the rate of formation of \(C\) (forward reaction) is \(k_1 [A][B]\) and the rate of its decomposition (reverse reaction) is \(k_{-1} [C]\). Since the reaction is at equilibrium, \(k_1 [A][B] = k_{-1} [C]\). Thus, the equilibrium constant \(K_c\) is given by: \[K_c = \frac{[C]}{[A][B]}\]
03

Express \([C]\) in Terms of \([A]\) and \([B]\)

Rearrange the equilibrium expression to solve for \([C]\): \[[C] = K_c [A][B]\]
04

Relate \(d[P]/dt\) to \([C]\)

The rate of product formation \(\frac{d[P]}{dt}\) depends on the concentration of \(C\). From the reaction \(C \stackrel{k_{2}}{\Rightarrow} P\), we know the rate \(\frac{d[P]}{dt} = k_2 [C]\).
05

Substitute \([C]\) in the Rate Expression

Replace \([C]\) in the rate expression with the expression from step 3: \[\frac{d[P]}{dt} = k_2 (K_c [A][B])\] which simplifies to: \[\frac{d[P]}{dt} = k_2 K_c [A][B]\]
06

Conclusion

We have shown that, under the assumption of rapid equilibrium in the first step, the rate of product formation is given by \(\frac{d[P]}{dt} = k_2 K_c [A][B]\).

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Key Concepts

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

Reaction Mechanism
In chemical reactions, the reaction mechanism is a step-by-step sequence of elementary reactions by which overall chemical change occurs. It provides insight into the path taken by the reactants to transform into products.
In the given problem, the reaction mechanism involves two steps: the reversible formation of an intermediate \(C\) from reactants \(A\) and \(B\), and the subsequent transformation of \(C\) into a product \(P\).
This mechanism helps us understand how the reaction proceeds and allows for the calculation of rates using detailed biochemical kinetics.
Equilibrium Constant
The equilibrium constant, denoted \(K_c\), is a crucial concept in understanding how reactions behave at equilibrium. It is defined by the concentrations of products raised to the power of their stoichiometric coefficients, divided by the concentrations of reactants raised to their coefficients, for the reversible reaction at equilibrium.
For the fast-equilibrium step in the exercise \(A + B \rightleftharpoons C\), the equilibrium constant expression is: \[K_c = \frac{[C]}{[A][B]}\]
This expression allows us to relate the concentration of the intermediate \(C\) with the initial reactants \(A\) and \(B\), reflecting the dependency balance between these at equilibrium.
Rate of Reaction
The rate of reaction refers to the speed at which reactants are converted into products in a chemical reaction. It is determined by factors such as concentration, temperature, and the presence of a catalyst.
In our problem, the focus is on the rate at which product \(P\) forms. The fast-equilibrium approximation lets us derive the rate from the intermediate and reactant concentrations. Once \(C\) is formed, its conversion to \(P\) happens via the rate constant \(k_2\), defined by:\[\frac{d[P]}{dt} = k_2 [C]\]
This formula signifies how quickly products are formed based on the presence of \(C\) and its conversion rate.
Product Formation Rate
The product formation rate in this context implies the speed at which \(P\) appears in the reaction mixture. This rate is crucial for understanding the overall kinetics of the reaction and verifying the proposed mechanism.
Through the fast-equilibrium approximation, we simplify the expression for product formation due to the complexity brought by intermediate steps. The transformed equation for the product formation, using the relationship from equilibrium, gives:\[\frac{d[P]}{dt} = k_2 K_c [A][B]\]
With this expression, we predict the rate based on known values \(k_2\), \(K_c\), and reactant concentrations \([A]\), \([B]\), emphasizing how connected equilibria and kinetics are in multistep reactions.

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Most popular questions from this chapter

For \(T<500 \mathrm{K},\) the reaction $$\mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \stackrel{k_{\text {obs }}}{\longrightarrow} \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g})$$ has the rate law $$\frac{d\left[\mathrm{CO}_{2}\right]}{d t}=k_{\mathrm{obs}}\left[\mathrm{NO}_{2}\right]^{2}$$ Show that the following mechanism is consistent with the observed rate law $$\mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_{3}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \quad(\text { rate determining })$$ $$\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g})$$ Express \(k_{\text {obs }}\) in terms of \(k_{1}\) and \(k_{2}\).

The standard Gibbs energy change of reaction for $$2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ is \(-457.2 \mathrm{kJ}\) at \(298 \mathrm{K}\). At room temperature, however, this reaction does not occur and mixtures of gaseous hydrogen and oxygen are stable. Explain why this is so. Is such a mixture indefinitely stable?

Determine the rate law for the reaction $$ \mathrm{I}(\mathrm{g})+\mathrm{I}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k}{\Longrightarrow} \mathrm{I}_{2}(\mathrm{~g})+\mathrm{M}(\mathrm{g}) $$ where \(M\) is any molecule present in the reaction container. Give the units of \(k\). Determine the molecularity of this reaction. Is this reaction identical to $$ \mathrm{I}(\mathrm{g})+\mathrm{I}(\mathrm{g}) \stackrel{k}{\Longrightarrow} \mathrm{I}_{2}(\mathrm{~g}) $$ Explain.

The thermal decomposition of ethylene oxide occurs by the mechanism $$\begin{array}{l} \mathrm{H}_{2} \mathrm{COCH}_{2}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{H}_{2} \mathrm{COCH}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \\ \mathrm{H}_{2} \mathrm{COCH}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ \mathrm{CH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{COCH}_{2}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{COCH}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \\ \mathrm{CH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{COCH}(\mathrm{g}) \stackrel{k_{4}}{\longrightarrow} \text { products } \end{array}$$ Which of these reaction(s) are the initiation, propagation, and termination step(s) of the reaction mechanism? Show that if the intermediates \(\mathrm{CH}_{3}\) and \(\mathrm{H}_{2} \mathrm{COCH}\) are treated by the steady-state approximation, the rate law, \(d[\text { products }] / d t,\) is first order in ethylene oxide concentration.

The Lindemann reaction mechanism for the isomerization reaction $$\mathrm{CH}_{3} \mathrm{NC}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3} \mathrm{CN}(\mathrm{g})$$ is $$\begin{array}{c} \mathrm{CH}_{3} \mathrm{NC}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons} \mathrm{CH}_{3} \mathrm{NC}^{*}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \\ \mathrm{CH}_{3} \mathrm{NC}^{*}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CN}(\mathrm{g}) \end{array}$$ Under what conditions does the steady-state approximation apply to \(\mathrm{CH}_{3} \mathrm{NC}^{*} ?\)
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