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Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is Step \(1: \quad\) Fast, reversible: \(\quad \mathrm{HA} \rightleftarrows \mathrm{H}^{+}+\mathrm{A}^{-}\) Step \(2: \quad\) Fast, reversible: \(\quad \mathrm{X}+\mathrm{H}^{+} \rightleftharpoons \mathrm{XH}^{+}\) Step 3: Slow \(\mathrm{XH}^{+} \rightarrow\) products What rate law is derived from this mechanism? What is the order of the reaction with respect to HA? How would doubling the concentration of HA affect the reaction?

Short Answer

Expert verified
The rate law is \( \text{Rate} = k'[X][HA] \), first order in HA; doubling \([HA]\) doubles the rate.

Step by step solution

01

Identify Steps and Reactions

Examine the given steps in the reaction mechanism. The first step is the ionization of the acid: \( \mathrm{HA} \rightleftarrows \mathrm{H}^{+} + \mathrm{A}^{-} \) which is fast and reversible. The second step involves \( \mathrm{X} \) and \( \mathrm{H}^{+} \) to form \( \mathrm{XH}^{+} \): \( \mathrm{X} + \mathrm{H}^{+} \rightleftharpoons \mathrm{XH}^{+} \), also fast and reversible. The third and slow step is the conversion of \( \mathrm{XH}^{+} \) to products.
02

Determine the Rate-Determining Step

In a mechanism with a single slow step, the overall rate of reaction is determined by this slow step. Therefore, Step 3 is the rate-determining step: \( \mathrm{XH}^{+} \rightarrow \text{products} \). The rate law will be based on this step.
03

Express Intermediate Concentrations

Since \( \mathrm{XH}^{+} \) appears in the slow step, express its concentration using the equilibrium established by the faster steps. From Step 2, assume equilibrium: \( \mathrm{[XH^{+}]} = K_2 [X][H^{+}] \). From Step 1 equilibrium, \( [H^{+}] = K_1 [HA] \). Thus, \( [XH^{+}] = K_2 K_1 [X][HA] \).
04

Write the Rate Law

The rate of the reaction is proportional to the concentration of \( \mathrm{XH}^{+} \), so: \( \text{Rate} = k \mathrm{[XH^{+}]} = k K_2 K_1 [X][HA] \). The rate law is \( \text{Rate} = k' [X][HA] \), where \( k' = k K_2 K_1 \).
05

Determine Reaction Order with Respect to HA

In the rate law \( \text{Rate} = k' [X][HA] \), the reaction is first order with respect to \( \mathrm{HA} \), as the concentration of \( \mathrm{HA} \) is raised to the power of one.
06

Effect of Doubling HA Concentration

Doubling the concentration of \( \mathrm{HA} \) will double the reaction rate, since the reaction is first order with respect to \( \mathrm{HA} \). The rate law shows a direct proportionality between the rate and \( \mathrm{HA} \) concentration.

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Key Concepts

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

Acid Catalysis
Acid catalysis is a common process in biochemical reactions where an acid facilitates the reaction. In our given mechanism, the acid involved is HA. The first step shows HA ionizing into hydrogen ions (\(\mathrm{H}^{+}\)) and its conjugate base (\(\mathrm{A}^{-}\)). Here, the acid donates a proton, which can then interact with the reactant X in subsequent steps. This proton donation increases the reactivity of X by converting it into a more reactive species, such as \(\mathrm{XH}^{+}\). This process is typical in acid-catalyzed reactions and plays a crucial role in facilitating the formation of products.

Moreover, acid catalysis often improves the reaction rate by lowering the activation energy needed for the reaction. The presence of \(\mathrm{H}^{+}\) can stabilize transition states or create favorable structures that facilitate the reaction. This can crucially influence how quickly a reaction takes place.
Rate-Determining Step
The rate-determining step is pivotal in understanding the overall reaction rate. In our mechanism, Step 3 is identified as the rate-determining step (\(\mathrm{XH}^{+} \rightarrow \text{products}\)). This step is much slower than the preceding steps, making it the bottleneck for the entire reaction process.

This means that the speed of the overall reaction is limited by this slow step. Therefore, the concentration of \(\mathrm{XH}^{+}\) directly influences the reaction rate. No matter how fast the other steps are, the reaction cannot progress faster than this slowest step.
Reaction Order
The reaction order tells us how the rate of reaction depends on the concentration of reactants. In the given rate law \( \text{Rate} = k' [X][HA] \), the term explains that the reaction is first order with respect to both \(\mathrm{X}\) and \(\mathrm{HA}\). A first order with respect to \(\mathrm{HA}\) means if you double the concentration of \(\mathrm{HA}\), the reaction rate would also double.
  • The exponent 1 in the rate law \([HA]^1\) indicates a direct proportionality.
  • "First order with respect to X" describes a similar relationship where a change in \(\mathrm{X}\) concentration linearly affects the rate.
This concept helps to predict how shifts in concentrations will alter the speed of the reaction, which is crucial for controlling reaction conditions in experiments and industrial processes.
Rate Law Derivation
Deriving the rate law from the reaction mechanism involves identifying rate-limiting factors. Following the slowest step (Step 3: \(\mathrm{XH}^{+} \rightarrow \text{products}\)), we derive that rate is proportional to \([\mathrm{XH}^{+}]\).

To express \([\mathrm{XH}^{+}]\), we use the equilibrium conditions derived from the first two fast steps. The second step, \(\mathrm{X} + \mathrm{H}^{+} \rightleftharpoons \mathrm{XH}^{+}\), allows us to express \([\mathrm{XH}^{+}] = K_2 [X][H^{+}]\). Similarly, the first step, \(\mathrm{HA} \rightleftharrows \mathrm{H}^{+} + \mathrm{A}^{-}\), gives us \([H^{+}] = K_1 [HA]\). Substituting these values, \([\mathrm{XH}^{+}] = K_2 K_1 [X][HA]\).The overall rate law becomes \(\text{Rate} = k K_2 K_1 [X][HA]\), simplified to \(\text{Rate} = k' [X][HA]\) where \(k'\) encompasses all different constants. This accurate rate law allows predictions on how various concentrations affect the reaction rate, and clearly shows how the initial steps influence the formation of intermediates crucial for the slowest step.

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Most popular questions from this chapter

Using the rate equation Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}],\) define the order of the reaction with respect to A and B. What is the total order of the reaction?

You want to study the hydrolysis of the beautiful green, cobalt-based complex called transdichlorobis-(ethylenediamine)cobalt(III) ion, In this hydrolysis reaction, the green complex ion trans- \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en \(=\) \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) trans-\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co-Cl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water. Slow: \(\left.\quad \text { trans-ICo(en) }_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ Fast: \(\quad\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq}) $$ (a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first-order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting In \(k\) versus \(1 / T .\) However, here we do not need to measure \(k\) directly. Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\) the time needed to achieve the gray color is a measure of \(k\). Use the data below to find the activation energy. green $$ \left.\underset{\text { red }}{\operatorname{Co}(\mathrm{en})_{2}}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray. The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker.

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