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When substrates are present at relatively high concentration and are catalyzed by enzymes, the effect on reaction rate of changing substrate concentration can be described by zeroth-order kinetics. Calculate by what factor the rate of an enzyme-catalyzed reaction changes when the substrate concentration is changed from \(1.5 \times 10^{-2} \mathrm{M}\) to \(4.5 \times 10^{-2} \mathrm{M}\)

Short Answer

Expert verified
The reaction rate does not change; it remains constant due to zeroth-order kinetics.

Step by step solution

01

Understand Zeroth-Order Kinetics

In zeroth-order kinetics, the reaction rate is independent of the concentration of the reactant. This means that the rate of reaction remains constant, regardless of changes in substrate concentration.
02

Identify Initial and Final Conditions

We are given an initial substrate concentration of \(1.5 \times 10^{-2} \mathrm{M}\) and a final substrate concentration of \(4.5 \times 10^{-2} \mathrm{M}\). Since the reaction follows zeroth-order kinetics, the rate of reaction at both these concentrations is the same.
03

Calculate the Factor of Rate Change

Because zeroth-order kinetics implies that the rate does not change with substrate concentration, the factor by which the rate changes, when moving from an initial to a final substrate concentration, is 1.

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Key Concepts

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

Enzyme-Catalyzed Reactions
Enzymes play a crucial role in facilitating biochemical reactions by acting as catalysts. They speed up reactions without being consumed in the process.
Enzyme-catalyzed reactions are prevalent in biological systems due to their efficiency and specificity. These reactions occur when enzymes lower the activation energy needed for reactions to proceed, which increases the reaction rate. Key characteristics of enzyme-catalyzed reactions include:
  • Specificity: Enzymes are selective, usually catalyzing only a single or closely related set of reactions.
  • Reusability: Enzymes are not consumed or permanently altered during the reaction process.
  • Substrate Binding: Enzymes bind to substrates—reactants in the enzyme-catalyzed reaction—at their active sites.
This interaction is critical because it allows enzymes to transform substrates into products effectively. When substrate concentration is extremely high, saturation of enzyme active sites occurs, which is why in some cases the reaction can display zeroth-order kinetics, where the rate remains constant regardless of further increases in substrate concentration.
Reaction Rate
The reaction rate refers to the speed at which a chemical reaction proceeds. For enzyme-catalyzed reactions, understanding the reaction rate is essential for determining how quickly substrates are converted to products.
Reaction rates can be influenced by several factors:
  • Concentration: Typically, an increase in substrate concentration speeds up the reaction up to a point of saturation.
  • Temperature: Higher temperatures can increase reaction rates until the enzyme becomes denatured.
  • pH: Each enzyme has an optimal pH at which its activity is maximized.
However, in zeroth-order kinetics, the reaction rate is notably independent of substrate concentration. As a result, the concentration of reactants does not affect the reaction rate, leading to a constant rate under these conditions. This happens because enzyme sites are saturated with substrate, preventing any further increase in reaction rate.
Substrate Concentration
Substrate concentration is one of the critical factors affecting enzyme-catalyzed reactions. In general, as substrate concentration increases, the reaction rate also increases, until a saturation point is reached.
At this saturation point, all active sites of the enzyme molecules are occupied, and the reaction rate becomes constant, exhibiting zeroth-order kinetics. To understand the impact of substrate concentration on enzymatic reactions, consider the following:
  • Low substrate concentration leads to a proportional increase in reaction rate due to more frequent collisions between enzyme and substrate.
  • High substrate concentrations can lead to saturation where the reaction rate levels off.
  • In extreme cases of substrate concentration, enzyme active sites are continuously occupied, leading to a reaction rate that is constant and independent of further increases in substrate concentration.
This concept is crucial when investigating enzyme activity because it helps scientists and researchers determine conditions under which maximum enzyme efficiency and effectiveness are achieved.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the factors influencing them. It offers insights into the speed at which reactions occur and allows predictions of reaction behavior under varying conditions.
Some core principles of chemical kinetics include:
  • Rate Laws: Which express the rate of a reaction as a function of the concentration of reactants.
  • Order of Reaction: The sum of the powers of concentration terms in the rate law, which can be zero, first, second, or higher.
  • Activation Energy: The minimum energy needed for a reaction to occur, often lowered by the presence of a catalyst such as an enzyme.
In the context of zeroth-order kinetics, chemical kinetics dictates that the reaction rate is independent of the concentration of substrates. This is a special case seen in enzyme-catalyzed reactions where substrate saturation leads to a constant reaction rate. Understanding these kinetics is vital for fields such as drug design and metabolic engineering, where controlling reaction rates can be crucial.

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Most popular questions from this chapter

In a time-resolved picosecond spectroscopy experiment, Sheps, Crowther, Carrier, and Crim (Journal of Physical Chemistry \(A,\) Vol. 110,\(2006 ;\) pp. \(3087-3092\) ) generated chlorine atoms in the presence of pentane. The pentane was dissolved in dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\). The chlorine atoms are free radicals and are very reactive. After a nanosecond the chlorine atoms have reacted with pentane molecules, removing a hydrogen atom to form \(\mathrm{HCl}\) and leaving behind a pentane radical with a single unpaired electron. The equation is \(\mathrm{Cl} \cdot(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{dcm}) \longrightarrow \mathrm{HCl}(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{11} \cdot(\mathrm{dcm})\) where \((\mathrm{dcm})\) indicates that a substance is dissolved in dichloromethane. Measurements of the concentration of chlorine atoms were made as a function of time at three different concentrations of pentane in the dichloromethane. These results are shown in the table. of Chlorine Atoms (M) \begin{tabular}{clcc} & \multicolumn{3}{c} { Concentration of Chlorine Atoms (m) } \\ Time (ps) & \multicolumn{4}{c} { for Different \(\mathrm{C}_{5} \mathrm{H}_{12}\) Concentrations } \\ \cline { 3 - 5 } & \(0.15-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.30-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.60-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) \\ \hline 100.0 & \(4.42 \times 10^{-5}\) & \(3.11 \times 10^{-5}\) & \(1.77 \times 10^{-5}\) \\ 140.0 & \(3.823 \times 10^{-5}\) & \(2.39 \times 10^{-5}\) & \(1.15 \times 10^{-5}\) \\\ 180.0 & \(3.38 \times 10^{-5}\) & \(1.94 \times 10^{-5}\) & \(8.48 \times 10^{-6}\) \\\ 220.0 & \(3.03 \times 10^{-5}\) & \(1.49 \times 10^{-5}\) & \(6.04 \times 10^{-6}\) \\\ 260.0 & \(2.68 \times 10^{-5}\) & \(1.19 \times 10^{-5}\) & \(4.12 \times 10^{-6}\) \\\ 300.0 & \(2.42 \times 10^{-5}\) & \(9.45 \times 10^{-6}\) & \(3.14 \times 10^{-6}\) \\\ 340.0 & \(2.08 \times 10^{-5}\) & \(7.75 \times 10^{-6}\) & \(2.38 \times 10^{-6}\) \\\ 380.0 & \(1.91 \times 10^{-5}\) & \(6.35 \times 10^{-6}\) & \(1.75 \times 10^{-6}\) \\\ 420.0 & \(1.71 \times 10^{-5}\) & \(4.58 \times 10^{-6}\) & \(1.61 \times 10^{-6}\) \\\ 460.0 & \(1.53 \times 10^{-5}\) & \(3.77 \times 10^{-6}\) & \(9.98 \times 10^{-7}\) \\\ \hline \end{tabular} (a) Determine the order of the reaction with respect to chlorine. (b) Determine whether the reaction rate depends on the concentration of pentane in dichloromethane. If so, determine the order of the reaction with respect to pentane. (c) Explain why the concentration of pentane in dichloromethane does not affect the data analysis that you performed in part (a). (d) Write the rate law for the reaction and calculate the rate of reaction for a concentration of chlorine atoms equal to \(1.0 \mu \mathrm{M}\) and a pentane concentration of \(0.23 \mathrm{M}\). (e) Sheps, Crowther, Carrier, and Crim found that the rate of formation of \(\mathrm{HCl}\) matched the rate of disappearance of Cl. From this they concluded that there were no intermediates and side reactions were not important. Explain the ic

The deep blue compound \(\mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}\) can be made from the chromate ion by using hydrogen peroxide in an acidic solution. \(\mathrm{HCrO}_{4}^{-}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The kinetics of this reaction have been studied, and the rate equation is Rate of disappearance of \(\mathrm{HCrO}_{4}^{-}=k\left[\mathrm{HCrO}_{4}^{-}\right]\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]\) One of the mechanisms suggested for the reaction is $$ \begin{array}{c} \mathrm{HCrO}_{4}^{-}+\mathrm{H}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CrO}_{4} \\ \mathrm{H}_{2} \mathrm{CrO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} $$ \(\mathrm{H}_{2} \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (a) Give the order of the reaction with respect to each reactant. (b) Show that the steps of the mechanism agree with the overall equation for the reaction. (c) Which step in the mechanism is rate-limiting? Explain your answer.

Nitrosyl bromide, NOBr, is formed from \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\). $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g}) $$ Experiment shows that the reaction is first-order in \(\mathrm{Br}_{2}\) and second-order in NO. (a) Write the rate law for the reaction. (b) If the concentration of \(\mathrm{Br}_{2}\) is tripled, determine how the reaction rate changes. (c) Determine what happens to the reaction rate when the concentration of \(\mathrm{NO}\) is doubled.

The ester methyl acetate, \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\), reacts with base to break one of the \(\mathrm{C}-\mathrm{O}\) bonds. CC(=O)OCC(O)C(C)O CC(=O)OCCOCCC(=O)O The rate law is rate \(=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{OH}^{-}\right]\) where $$ k=0.14 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1} \text {at } 25^{\circ} \mathrm{C} $$ (a) Calculate the initial rate at which the methyl acetate is converted to products when both reactants, \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\) and \(\mathrm{OH}^{-}\), have a concentration of \(0.025 \mathrm{M}\). (b) Calculate the rate at which methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), initially appears in the solution.

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 $$ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right] $$

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