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Chapter 15: Problem 74

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: Fast, reversible: \(\quad \mathrm{HA} \rightleftarrows \mathrm{H}^{+}+\mathrm{A}^{-}\) Step 2: Fast, reversible: \(\quad \mathrm{X}+\mathrm{H}^{+} \rightleftarrows \mathrm{XH}^{+}\) Step 3: Slow \(\quad \mathrm{XH}^{+} \rightarrow\) 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 \( \text{Rate} = k_{eff} [\mathrm{X}][\mathrm{HA}] \). The reaction is first order in HA. Doubling \( [\mathrm{HA}] \) doubles the rate.

Step by step solution

01

Identify Rate-Determining Step

From the provided mechanism steps, we identify that Step 3 is the slow step: \( \mathrm{XH}^{+} \rightarrow \text{products} \). This is the rate-determining step and dictates the rate of the overall reaction.
02

Write the Rate Law Expression for the Rate-Determining Step

Since Step 3 is the slow step, its rate law is given by: \( \text{Rate} = k_3 [\mathrm{XH}^{+}] \), where \( k_3 \) is the rate constant for Step 3.
03

Determine the Equilibrium Expressions for Fast Steps

For Step 1: \( \mathrm{HA} \rightleftarrows \mathrm{H}^{+} + \mathrm{A}^{-} \), the equilibrium expression is \( K_1 = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]} \). For Step 2: \( \mathrm{X} + \mathrm{H}^{+} \rightleftarrows \mathrm{XH}^{+} \), the equilibrium expression is \( K_2 = \frac{[\mathrm{XH}^{+}]}{[\mathrm{X}][\mathrm{H}^{+}]} \).
04

Express [XH+] in Terms of HA, X, and Equilibrium Constants

Using the equilibrium expressions:1. Solve for \([\mathrm{H}^{+}]\) from Step 1: \([\mathrm{H}^{+}] = K_1 \frac{[\mathrm{HA}]}{[\mathrm{A}^{-}]}\)2. Substitute \([\mathrm{H}^{+}]\) in Step 2 equilibrium: \([\mathrm{XH}^{+}] = K_2 [\mathrm{X}][\mathrm{H}^{+}]\)3. Substitute \([\mathrm{H}^{+}]\) to get: \([\mathrm{XH}^{+}] = K_2 K_1 \frac{[\mathrm{X}][\mathrm{HA}]}{[\mathrm{A}^{-}]}\)
05

Substitute [XH+] into the Rate Law and Simplify

Substituting \([\mathrm{XH}^{+}]\) from the previous step into the rate law: \[ \text{Rate} = k_3 K_2 K_1 \frac{[\mathrm{X}][\mathrm{HA}]}{[\mathrm{A}^{-}]} \] Simplifying, if \([\mathrm{A}^{-}]\) remains relatively constant due to it being a product of a fast equilibrium, the rate law simplifies to: \[ \text{Rate} = k_{eff} [\mathrm{X}][\mathrm{HA}] \]where \( k_{eff} = k_3 K_2 K_1 \frac{1}{[\mathrm{A}^{-}]} \) is an effective rate constant.
06

Determine Reaction Order with Respect to HA

From the derived rate law \( \text{Rate} = k_{eff} [\mathrm{X}][\mathrm{HA}] \), it is evident that the reaction is first order with respect to \([\mathrm{HA}]\).
07

Predict Effect of Doubling HA

Since the reaction is first order with respect to \([\mathrm{HA}]\), doubling \([\mathrm{HA}]\) would double the rate of the reaction.

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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 process where an acid serves as a catalyst to speed up a biochemical reaction. In the given mechanism, an acid (HA) dissociates to provide protons (\( \mathrm{H}^{+} \)) that facilitate the conversion of reactants to products. The presence of \( \mathrm{H}^{+} \) lowers the activation energy, thereby speeding up the reaction.
This process is important in biochemical pathways, such as those involving proteins and nucleic acids, where acids or bases modify the speed of reaction.
  • Acids donate protons to the reactants, thereby stabilizing transition states.
  • Enhanced reaction speed can make certain metabolic processes feasible under physiological conditions.
In summary, the ability of acids to stabilize charged intermediates or transition states is key to acid-catalyzed reactions.
Reaction Mechanisms
A reaction mechanism describes the step-by-step sequence of elementary reactions by which a chemical change occurs. In the provided example, three distinct steps depict how reactants transform into products.
Each step has a characteristic speed, denoted as fast or slow, affecting the overall reaction pace.
  • Step 1 and Step 2 are fast, reversible steps where equilibrium is quickly attained.
  • Step 3 is the slow step, known as the rate-determining step, crucial for understanding the reaction speed.
This combination of steps forms an intricate mechanism, detailing interactions and atomic rearrangements that result in the final products.
Rate Law
The rate law provides a mathematical description of the speed of a reaction in terms of the concentration of reactants. From the mechanism, the rate law is derived from the slowest step, known as the rate-determining step.
The given mechanism suggests that the rate law for the reaction is expressed as:
\[ \text{Rate} = k_{eff} [\mathrm{X}][\mathrm{HA}] \]
  • Here, \( k_{eff} \) is the effective rate constant, aggregating several parameters like equilibrium constants for earlier steps.
  • The rate law indicates how changes in concentration of the reactants affect the rate of reaction.
Understanding the rate law is crucial in predicting how the reaction rate will change when reactant concentrations are altered.
Reaction Order
The reaction order provides insight into the relationship between the concentration of reactants and the rate of reaction. In the context of the mechanism given, the reaction is first order with respect to HA.
The overall reaction order is determined by summing the exponents of the concentrations in the rate law:
\[ \text{Rate} = k_{eff} \cdot [\mathrm{X}]^1 [\mathrm{HA}]^1 \]
  • This indicates a first-order dependency on both \( \mathrm{X} \) and \( \mathrm{HA} \) individually.
  • Doubling the concentration of \( \mathrm{HA} \) results in doubling the reaction rate due to its first-order nature.
Understanding the order helps in controlling and optimizing reaction conditions.
Equilibrium Expressions
Equilibrium expressions arise from fast, reversible steps in a mechanism. They define the balance of products and reactants at equilibrium for each elementary step.
In the given exercise, two equilibrium expressions are formulated from the first two steps:
  • Step 1: \( K_1 = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]} \)
  • Step 2: \( K_2 = \frac{[\mathrm{XH}^{+}]}{[\mathrm{X}][\mathrm{H}^{+}]} \)
These expressions play a pivotal role in deriving the concentration of crucial intermediates like \( \mathrm{XH}^{+} \).
With these expressions, you can relate concentrations at equilibrium to help understand the dynamics of the overall reaction system.

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Most popular questions from this chapter

The decomposition of HOF occurs at \(25^{\circ} \mathrm{C}\) $$\mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ Using the data in the table below, determine the rate law, and then calculate the rate constant. $$\begin{array}{cc}{[\mathrm{HOF}](\mathrm{mol} / \mathrm{L})} & \mathrm{Time}(\mathrm{min}) \\ \hline 0.850 & 0 \\\0.810 & 2.00 \\\0.754 & 5.00 \\\0.526 & 20.0 \\\0.243 & 50.0 \\\\\hline\end{array}$$

The following statements relate to the reaction for the formation of HI: $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HI}(\mathrm{g}) \quad \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]$$ Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is a first-order reaction: $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ The rate constant for the reaction is \(2.8 \times 10^{-3} \mathrm{min}^{-1}\) at \(600 \mathrm{K}\). If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.24 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) how long will it take for the concentration to drop to \(0.31 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

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. Step 1: Fast, endothermic $$\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}$$ Step 2: Slow $$\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}$$ (a) Write an equation for the overall reaction. (b) Draw a reaction coordinate diagram for this reaction. (c) Show that the rate law for this reaction is Rate \(=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right]\)

At temperatures below \(500 \mathrm{K},\) the reaction between carbon monoxide and nitrogen dioxide $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g})$$ has the following rate equation: Rate \(=k\left[\mathrm{NO}_{2}\right]^{2} .\) Which of the three mechanisms suggested here best agrees with the experimentally observed rate equation? Mechanism \(1 \quad\) single, elementary step $$\mathrm{NO}_{2}+\mathrm{CO} \rightarrow \mathrm{CO}_{2}+\mathrm{NO}$$ Mechanism \(2 \quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2}+\mathrm{NO}_{2} \rightarrow\mathrm{NO}_{3}+\mathrm{NO}\\\&\text { Fast } \quad \mathrm{NO}_{3}+\mathrm{CO} \rightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}\end{aligned}$$ Mechanism \(3 \quad\) Two steps $$\begin{array}{ll}\text { Slow } & \mathrm{NO}_{2} \rightarrow \mathrm{NO}+\mathrm{O} \\\\\text { Fast } & \mathrm{CO}+\mathrm{O} \rightarrow \mathrm{CO}_{2}\end{array}$$
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