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When enzymes are present at very low concentration, their effect on reaction rate can be described by firstorder kinetics. Calculate by what factor the rate of an enzyme-catalyzed reaction changes when the enzyme concentration is changed from \(1.5 \times 10^{-7} \mathrm{M}\) to \(4.5 \times 10^{-6} \mathrm{M}\)

Short Answer

Expert verified
The reaction rate increases by a factor of 30.

Step by step solution

01

Understanding First-Order Kinetics

In first-order kinetics, the rate of reaction is directly proportional to the concentration of the reactant, in this case, the enzyme. Formula for first-order kinetics is given by:\[ \text{Rate} = k[E] \]Where \( k \) is the rate constant and \([E]\) is the enzyme concentration.
02

Identify Initial and Final Rates

Using the first-order kinetics formula, the initial rate \( R_1 \) and final rate \( R_2 \) can be expressed as:\[ R_1 = k(1.5 \times 10^{-7} \text{ M}) \]\[ R_2 = k(4.5 \times 10^{-6} \text{ M}) \]
03

Calculate the Rate Change Factor

The factor by which the rate changes can be calculated by taking the ratio of the final rate \( R_2 \) to the initial rate \( R_1 \):\[ \text{Factor} = \frac{R_2}{R_1} = \frac{k(4.5 \times 10^{-6})}{k(1.5 \times 10^{-7})} \]
04

Simplify the Expression

Cancel out the common terms, the rate constants \( k \), and simplify the expression:\[ \text{Factor} = \frac{4.5 \times 10^{-6}}{1.5 \times 10^{-7}} \]
05

Calculate the Factor

Compute the numerical factor:\[ \text{Factor} = \frac{4.5}{1.5} \times 10^{(-6 + 7)} = 3 \times 10^{1} = 30 \]

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Key Concepts

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

First-Order Kinetics
First-order kinetics is a simple yet fundamental concept in chemistry and biology. It describes how the rate of a reaction is directly proportional to the concentration of a specific reactant. In this case, that reactant is an enzyme. This type of reaction suggests that if you double the concentration of the enzyme, the reaction rate also doubles.
Understanding first-order kinetics helps us predict how changes in enzyme concentration affect the overall speed of a reaction. For enzyme-catalyzed reactions at low concentrations, as mentioned in the exercise, this relationship can be remarkably linear.
Thus, the formula for first-order kinetics, \( \text{Rate} = k[E] \), becomes essential, where \( k \) is the rate constant and \( [E] \) represents enzyme concentration. This formula is a core principle used to determine how fast a reaction will proceed when the concentration of enzymes changes.
Enzyme Concentration
Enzyme concentration refers to the amount of enzyme present in a reaction mixture. Enzymes act as biological catalysts, speeding up reactions without being consumed. Their concentration can have a huge impact on the reaction rate, especially when operating under first-order kinetics.
As demonstrated in the exercise, changing the enzyme concentration from \(1.5 \times 10^{-7} \text{ M} \) to \(4.5 \times 10^{-6} \text{ M} \) significantly alters the reaction rate. This change directly affects how quickly a reaction progresses, making it faster or slower.
It’s important to consider that while increasing enzyme concentration generally increases the rate, there is often a plateau. At extremely high concentrations, other factors like substrate availability might become limiting, switching the system from first-order to a zero-order kinetics as the enzyme saturates with substrate.
Reaction Rate Change
In enzyme-catalyzed reactions, measuring how the reaction rate changes is crucial. The reaction rate is the speed at which the reactants are converted into products.
When we increase enzyme concentration in a reaction following first-order kinetics, the reaction rate changes proportionally. This means the rate increases by a factor equal to the change in enzyme concentration.
In the provided problem, the rate change can be calculated by the ratio of final enzyme concentration to the initial one, showing the impact of enzyme concentration on reaction speed. Such calculations help in understanding the efficiency and capacity of the enzyme to facilitate the reaction.
Rate Constant
The rate constant, denoted as \( k \), is a specific coefficient that makes the relationship between reaction rate and concentration accurate and quantifiable. In first-order kinetics, the rate constant remains unchanged regardless of changes in enzyme concentration.
This constant is an intrinsic property of a reaction, influenced by factors such as temperature and pH, but independent of the concentration of enzymes or other reactants.
While calculating reaction rate changes, as demonstrated in the exercise, the rate constant \( k \) cancels out, emphasizing that the proportional change in rate is solely due to changes in enzyme concentration. This feature of \( k \) allows us to understand and predict reaction behaviors in varying conditions.

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Most popular questions from this chapter

For the reaction $$ 2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) $$ make qualitatively correct plots of the concentrations of \(\mathrm{NO}_{2}(\mathrm{~g}), \mathrm{NO}(\mathrm{g}),\) and \(\mathrm{O}_{2}(\mathrm{~g})\) versus time. Draw all three graphs on the same axes; assume that you start with \(\mathrm{NO}_{2}(\mathrm{~g})\) at a concentration of \(1.0 \mathrm{~mol} / \mathrm{L}\). Explain how you would determine, from these plots, (a) the initial rate of the reaction. (b) the final rate (that is, the rate as time approaches infinity).

Draw a reaction energy diagram for an endothermic process. Mark the positions of reactants, products, and activated complex. Indicate the activation energies of the forward and reverse processes and explain how \(\Delta_{r} E\) for the reaction can be calculated from the diagram.

Suppose a chemical reaction has an activation energy of \(76 \mathrm{~kJ} / \mathrm{mol}\), as in the example in Figure \(11.12 .\) Calculate by what factor the rate of the reaction at \(50 .{ }^{\circ} \mathrm{C}\) is increased over its rate at \(25^{\circ} \mathrm{C}\).

For the reaction of iodine atoms with hydrogen molecules in the gas phase, these rate constants were obtained experimentally. \(2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) 2 \begin{tabular}{lc} \hline\(T(\mathrm{~K})\) & \(10^{-5} \mathrm{k}\left(\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}\right)\) \\ \hline 417.9 & 1.12 \\ 480.7 & 2.60 \\ 520.1 & 3.96 \\ 633.2 & 9.38 \\ 666.8 & 11.50 \\ 710.3 & 16.10 \\ 737.9 & 18.54 \\ \hline \end{tabular} (a) Calculate the activation energy and frequency factor for this reaction. (b) Estimate the rate constant of the reaction at \(400.0 \mathrm{~K}\).

Consider dissolving sugar as a simple process in which kinetics is important. Suppose that you dissolve an equal mass of each kind of sugar listed. Which dissolves the fastest? Which dissolves the slowest? Explain why in terms of rates of heterogeneous reactions. (If you are not sure which is fastest or slowest, try them all.) (a) Rock candy sugar (large sugar crystals) (b) Sugar cubes (c) Granular sugar (d) Powdered sugar

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