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Chapter 7: Problem 4

When a catalyst is present, the hydrolysis of sucrose (see Problem 3 ) is much more rapid. If the initial concentration of sucrose is \(0.050 \mathrm{M}\), it takes \(6.9 \times 10^{-5} \mathrm{~s}\) for the concentration to decrease by half to \(0.025 \mathrm{M}\). What is the rate of disappearance of sucrose in the presence of a catalyst?

Short Answer

Expert verified
The rate of disappearance of sucrose is approximately \(362.32 \mathrm{M/s}\).

Step by step solution

01

Understand the Problem

We need to find the rate at which sucrose disappears. The formula to find the rate of disappearance for a reactant A is given by \( -\frac{\Delta [A]}{\Delta t} \), where \([A]\) is the concentration and \( \Delta t \) is the change in time.
02

Identify Given Values

From the problem, we have the initial concentration of sucrose \([A]_0 = 0.050 \mathrm{M}\), the final concentration \([A]_f = 0.025 \mathrm{M}\), and the time taken for this change, \( \Delta t = 6.9 \times 10^{-5} \mathrm{s} \).
03

Calculate Change in Concentration

The change in concentration \( \Delta [A] \) is given by \([A]_f - [A]_0 = 0.025 \mathrm{M} - 0.050 \mathrm{M} = -0.025 \mathrm{M}\).
04

Calculate Rate of Disappearance

Substitute the values into the rate formula: \( -\frac{-0.025 \mathrm{M}}{6.9 \times 10^{-5} \mathrm{s}} = \frac{0.025 \mathrm{M}}{6.9 \times 10^{-5} \mathrm{s}} \).
05

Compute the Rate

Calculate the rate of disappearance: \( \frac{0.025}{6.9 \times 10^{-5}} \approx 362.32 \mathrm{M/s} \).

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Key Concepts

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

Reaction Rate
The reaction rate is essentially about the speed at which a chemical reaction occurs. For students delving into chemistry, understanding reaction rates is key. In simple terms, it measures how fast reactants transform into products.
The reaction rate can be influenced by several factors such as:
  • Concentration: Higher concentrations typically lead to faster reaction rates.
  • Temperature: Increasing temperature generally increases the reaction rate.
  • Catalysts: These substances speed up reactions without being consumed in the process.
  • Surface area: More exposed surface area can lead to faster reactions.
For our exercise, we're focusing on how catalysts can impact the reaction rate by speeding up the hydrolysis of sucrose. This is important because it allows us to achieve desired chemical conversions more efficiently.
Sucrose Hydrolysis
Sucrose hydrolysis is the process where sucrose, a common sugar, breaks down into glucose and fructose. This reaction is important in biology and food processing as it's foundational to how organisms utilize sugar.
Without a catalyst, sucrose hydrolysis occurs slowly, taking a much longer time to complete. However, when a catalyst is introduced, the reaction accelerates significantly. In our exercise, we observe that the presence of a catalyst reduces the time it takes for the sucrose concentration to halve from 0.050 M to 0.025 M to just 6.9 x 10-5 seconds. This example demonstrates the pivotal role catalysts play in speeding up reactions.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that examines the rate of chemical reactions and the factors affecting them. Kinetics studies include exploring reaction speeds and the steps within reaction mechanisms.
Key components in chemical kinetics involve understanding rate laws, which describe how the reaction rate depends on reactant concentrations. In our exercise, the focus is on how the concentration of sucrose decreases over time in the presence of a catalyst.
The rate of disappearance of sucrose is calculated using the formula \(-\frac{\Delta [A]}{\Delta t}\), leading to a rate of approximately 362.32 M/s in our scenario. This calculation showcases how kinetics allows us to quantify reaction rates and understand the underlying processes influencing them, making it a central aspect of understanding chemical reactions.

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Most popular questions from this chapter

In the absence of allosteric effectors, the enzyme phosphofructokinase displays Michaelis-Menten kinetics (see Fig. 7.15). The \(v_{0} / V_{\max }\) ratio is \(0.9\) when the concentration of the substrate, fructose6-phosphate, is \(0.10 \mathrm{mM}\). Calculate the \(K_{\mathrm{M}}\) for phosphofructokinase under these conditions.

The rate of hydrolysis of trehalose to its constituent glucose monomers in the absence of a catalyst is even slower than hydrolysis of sucrose (see Problem 3) and may indicate that the two sugars are hydrolyzed by different mechanisms. If the initial concentration of trehalose is \(0.050 \mathrm{M}\), it takes \(6.6 \times 10^{6}\) years for the concentration to decrease by half. What is the rate of disappearance of trehalose in the absence of a catalyst?

Computer-modeling studies have shown that uncompetitive and noncompetitive inhibitors of enzymes are more effective than competitive inhibitors. These studies have important implications in drug design. Propose a hypothesis that explains these results.

Inhibitor A at a concentration of \(2 \mu \mathrm{M}\) doubles the apparent \(K_{\mathrm{M}}\) for an enzymatic reaction, whereas inhibitor \(B\) at a concentration of \(9 \mu \mathrm{M}\) quadruples the apparent \(K_{\mathrm{M}}\). What is the ratio of the \(K_{\mathrm{I}}\) for inhibitor B to the \(K_{\mathrm{I}}\) for inhibitor \(\mathrm{A}\) ?

A bacterial enzyme catalyzes hydrolysis of the disaccharide maltose to produce two glucose monosaccharides. During an interval of one minute, the concentration of maltose decreases by \(65 \mathrm{mM}\). What is the rate of disappearance of maltose in the enzyme-catalyzed reaction?
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