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Chapter 11: Problem 84

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: reversible HA(aq) \(\rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq}) \quad\) fast Step 2: reversible \(\mathrm{X}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{XH}^{+}(\mathrm{aq}) \quad\) fast Step 3: \(\mathrm{XH}^{+}(\mathrm{aq}) \longrightarrow\) products slow (a) Derive the rate law from this mechanism. (b) Determine the order of reaction with respect to HA. (c) Determine how doubling the concentration of HA would affect the rate of the reaction.

Short Answer

Expert verified
(a) Rate law: \( \text{rate} = k[X][HA] \). (b) First order with respect to HA. (c) Doubling [HA] doubles the rate.

Step by step solution

01

Understand the Mechanism

The reaction mechanism provided includes three steps: a fast equilibrium dissociation of the acid (HA) into H鈦� and A鈦�, followed by a fast equilibrium reaction between X and H鈦� to form XH鈦�, and finally a slow conversion of XH鈦� into products. The slow step will determine the rate law.
02

Identify the Rate-Determining Step

The rate of the overall reaction is determined by the slowest step, which is Step 3: \( \mathrm{XH}^{+} \rightarrow \text{products} \). The rate of this step can be expressed as: \( \text{rate} = k_{3}[XH^+] \).
03

Express Intermediates in Terms of Reactants

Since [XH鈦篯 is involved in the rate-determining step, we express [XH鈦篯 using the fast pre-equilibria: - From Step 2, fast equilibrium \( \mathrm{X} + \mathrm{H}^+ \rightleftharpoons \mathrm{XH}^+ \) gives \( K_2 = \frac{[XH^+]}{[X][H^+]} \) or \( [XH^+] = K_2[X][H^+] \). - From Step 1, fast equilibrium \( \mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^- \) gives \( K_1 = \frac{[H^+][A^-]}{[HA]} \). Therefore, \( [H^+] = K_1 \frac{[HA]}{[A^-]} \). Substitute into the equation for [XH鈦篯.
04

Substitute and Derive the Rate Law

Substituting the expression for [H鈦篯 from Step 1 into the expression for [XH鈦篯 from Step 3 gives: \([XH^+] = K_2[X]K_1\frac{[HA]}{[A^-]}\). The rate can then be expressed as: \(\text{rate} = k_3K_2K_1\frac{[HA][X]}{[A^-]}\). If assuming [A鈦籡\ remains constant, the rate simplifies to \(\text{rate} = k[X][HA]\) because in a fast equilibrium, [A鈦籡 is usually constant.
05

Determine Reaction Order with Respect to HA

Since the derived rate law \(\text{rate} = k[X][HA]\) shows direct proportionality to [HA], HA is first order.
06

Effect of Doubling [HA]

If the concentration of HA is doubled, the rate of reaction will also double because the rate law is first order with respect to [HA]. This means \([HA]\) is directly proportional to the rate.

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Key Concepts

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

Reaction Rate Laws
In biochemical reaction mechanisms, understanding how reactants turn into products is crucial. Reaction rate laws help describe this process mathematically and show the relationship between the rate of a chemical reaction and the concentration of its reactants. For the mechanism provided, the reaction rate law is determined by the slowest step, known as the rate-determining step.
  • In our example, Step 3 is the slowest, and it determines the overall rate law.
  • The rate can be expressed as: \( \text{rate} = k_3[XH^+] \).
To use this, we must express the concentration of the intermediate \([XH^+]\) in terms of the concentrations of the reactants. This involves substituting what we know from the fast, reversible reactions in the mechanism. Ultimately, we find a rate law that describes how the concentration of reactants, like HA and X, affect the speed of the reaction. With these relationships understood, we can make predictions about how changes in concentrations will affect the reaction rate.
Acid Catalysis
Acid catalysis occurs when an acid accelerates the rate of a biochemical reaction. In this process, the acid donates a proton (H鈦�) to another substance, enhancing the reaction speed without being consumed by the reaction.
  • The mechanism involves the acid (HA) dissociating to contribute H鈦�.
  • This H鈦� then combines with the reactant \(X\), forming \(XH^+\).
The catalyst (in this case, the acid) is crucial because it provides an alternative pathway with a lower activation energy for the reaction. This means that more molecules have enough energy to react when they collide. As a result, reactions that might otherwise happen slowly or not at all occur more quickly, enhancing the efficiency of the biochemical processes.
Reaction Order
Understanding the order of a reaction with respect to each reactant helps in assessing how concentration changes affect the overall reaction rate. For the given mechanism, we determine the reaction order from the derived rate law, which in this case is first order with respect to HA.
  • The order of the reaction provides insight into how sensitive the rate is to concentration changes.
  • If doubling the concentration of HA doubles the reaction rate, that indicates a first-order relationship.
Interestingly, reaction orders do not always match stoichiometry but instead reflect how the concentration of a substance affects the rate. By understanding reaction orders, we can make informed predictions about the reaction kinetics, enabling better control over chemical processes in experiments and industrial applications.

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Most popular questions from this chapter

The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. $$ \text { Step 1: } \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {fast, endothermic } $$ $$ \text { Step } 2: \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} $$ slow (a) Write an equation for the overall reaction. (b) Draw a reaction energy diagram for this reaction. (c) Show that the rate law for this reaction is $$ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right] $$

For the reaction of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) at \(660 \mathrm{~K}\), \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\) \begin{tabular}{lcc} \hline \multicolumn{2}{c} { Concentration (mol/L) } & \\ \hline [NO] & {\(\left[\mathrm{O}_{2}\right]\)} & Rate of Disappearance of \(\mathrm{NO}\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 0.010 & 0.010 & \(2.5 \times 10^{-5}\) \\ 0.020 & 0.010 & \(1.0 \times 10^{-4}\) \\ 0.010 & 0.020 & \(5.0 \times 10^{-5}\) \\ \hline \end{tabular} (a) Determine the order with respect to each reactant. (b) Write the rate equation for the reaction. (c) Calculate the rate constant. (d) Calculate the rate when [NO] \(=0.025 \mathrm{~mol} / \mathrm{L}\) and \(\left[\mathrm{O}_{2}\right]=0.050 \mathrm{~mol} / \mathrm{L}\) (e) If \(\mathrm{O}_{2}\) disappears at a rate of \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}, \mathrm{cal}-\) culate the rate at which NO is consumed. Calculate the rate at which \(\mathrm{NO}_{2}\) is formed.

Measurements of the initial rate of hydrolysis of benzenesulfonyl chloride in aqueous solution at \(15^{\circ} \mathrm{C}\) in the presence of fluoride ion yielded the results in the table for a fixed concentration of benzenesulfonyl chloride of \(2 \times\) \(10^{-4} \mathrm{M}\). The reaction rate is known to be proportional to the concentration of benzenesulfonyl chloride. \begin{tabular}{cc} \hline & Initial Rate \(\times 10^{7}\) \\ {\(\left[\mathrm{~F}^{-}\right] \times 10^{2}(\mathrm{~mol} / \mathrm{L})\)} & \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 0 & 2.4 \\ 0.5 & 5.4 \\ 1.0 & 7.9 \\ 2.0 & 13.9 \\ 3.0 & 20.2 \\ 4.0 & 25.2 \\ 5.0 & 32.0 \\ \hline \end{tabular} Note that some reaction must be occurring in the absence of any fluoride ion, because at zero concentration of fluoride the rate is not zero. This residual rate should be subtracted from each observed rate to give the rate of the reaction being studied. (a) Derive the complete rate law for the reaction. (b) Calculate the rate constant \(k\) and express it in appropriate units.

Define the terms "activation energy" and "frequency factor." Write an equation that relates activation energy and frequency factor to reaction rate

For each of these rate laws, state the reaction order with respect to the hypothetical substances \(\mathrm{A}\) and \(\mathrm{B}\), and give the overall order. (a) Rate \(=k[\mathrm{~A}][\mathrm{B}]^{3}\) (b) Rate \(=k[\mathrm{~A}][\mathrm{B}]\) (c) Rate \(=k[\mathrm{~A}]\) (d) Rate \(=k[\mathrm{~A}]^{3}[\mathrm{~B}]\)
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