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The rate equation for the hydrolysis of sucrose to fructose and glucose $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ is " \(-\Delta[\text { sucrose }] / \Delta t=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] .\) After \(2.57 \mathrm{h}\) at \(27^{\circ} \mathrm{C}\) the sucrose concentration decreased from \(0.0146 \mathrm{M}\) to \(0.0132 \mathrm{M} .\) Find the rate constant, \(k.\)

Short Answer

Expert verified
The rate constant \( k \) is approximately \( 0.037 \text{ h}^{-1} \).

Step by step solution

01

Understand the Rate Equation

The given rate equation is \(-\Delta[\text{sucrose}]/\Delta t=k[\text{C}_{12}\text{H}_{22}\text{O}_{11}]\), indicating a first-order reaction. This means the rate of reaction is directly proportional to the concentration of sucrose.
02

Determine Change in Concentration

The initial concentration of sucrose \( [\text{sucrose}]_0 \) is 0.0146 M, and the final concentration \([\text{sucrose}]\) is 0.0132 M after 2.57 hours. The change in concentration \( \Delta[\text{sucrose}] = 0.0146 \text{ M} - 0.0132 \text{ M} = 0.0014 \text{ M} \).
03

Calculate Average Reaction Rate

The average reaction rate is given by \(-\Delta[\text{sucrose}] / \Delta t \). Therefore, \(-\Delta[\text{sucrose}] / \Delta t = 0.0014 \text{ M} / 2.57 \text{ h} \approx 0.000544 \text{ M/h} \).
04

Solve for Rate Constant

Use the rate equation \(-\Delta[\text{sucrose}] / \Delta t = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}] \) to solve for \(k\). Substitute in the average reaction rate and the initial concentration to find \( k = 0.000544 \text{ M/h} / 0.0146 \text{ M} \approx 0.037 \text{ h}^{-1} \).

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Key Concepts

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

Rate Equation
In chemical kinetics, the rate equation is a fundamental expression that relates the rate of a chemical reaction to the concentration of reactants. The rate equation signifies how the concentration of a substance changes over time, providing insight into the dynamics of a reaction.
  • The general form is: \[ -\frac{d[ ext{A}]}{dt} = k [ ext{A}]^n [ ext{B}]^m \ldots \] where \([\text{A}]\) and \([\text{B}]\) are concentrations, \(k\) is the rate constant, and \(n, m\) are the reaction orders.
  • In this specific exercise, we are given the rate equation for sucrose hydrolysis as \[ -\frac{\Delta[\text{sucrose}]}{\Delta t} = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}]. \]
This indicates a first-order reaction, where the rate is directly proportional to the concentration of sucrose. Understanding the rate equation is crucial for predicting how fast a reaction proceeds and is essential for controlling chemical processes.
First-Order Reaction
A first-order reaction is characterized by its direct proportionality between the reaction rate and the concentration of a single reactant. This straightforward relationship simplifies many analyses in chemistry.
  • The rate law for a first-order reaction is: \[ -\frac{d[ ext{A}]}{dt} = k [\text{A}] \]This tells us that if the concentration of \([\text{A}]\) doubles, the rate of reaction also doubles.
  • In the case of sucrose hydrolysis, the rate equation given is of first-order, meaning the concentration of sucrose is the sole factor influencing how fast the hydrolysis occurs.
  • This type of reaction has a unique feature: it follows an exponential decay model, often described with a half-life, which is the time taken for half of the reactant to undergo reaction.
Understanding a first-order reaction helps achieve better control over reaction rates, which is vital in many industrial and laboratory processes.
Rate Constant
The rate constant \(k\) is a pivotal factor in the rate equation, representing the intrinsic speed of a reaction under specific conditions.
  • It has units that depend on the reaction order, with the first-order reaction having units of \(\text{time}^{-1}\), such as \(\text{h}^{-1}\).
  • In our exercise, we calculated the rate constant \(k\) for sucrose hydrolysis to be approximately 0.037 \(\text{h}^{-1}\), indicative of the reaction's speed at the given temperature and concentration.
  • The rate constant incorporates the conditions at which the reaction is conducted (e.g., temperature, pressure, and presence of a catalyst), making it a crucial parameter for predicting and controlling reaction behavior.
Knowing the rate constant allows chemists to compare different reactions, predict reaction times, and manipulate conditions to direct the course of a reaction.

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Most popular questions from this chapter

Experimental data are listed here for the reaction \(\mathrm{A} \longrightarrow 2 \mathrm{B}.\) $$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\\\(\mathrm{s})\end{array} & \begin{array}{l}{[\mathrm{B}]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data, connect the points with a smooth line, and calculate the rate of change of [B] for each \(10-\) s interval from 0.0 to 40.0 s. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of \([\mathrm{A}]\) related to the rate of change of \([\mathrm{B}]\) in each time interval? Calculate the rate of change of \([\mathrm{A}]\) for the time interval from 10.0 to \(20.0 \mathrm{s}\) (c) What is the instantaneous rate when \([\mathrm{B}]=0.750\) \(\mathrm{mol} / \mathrm{L} ?\)

The reaction between ozone and nitrogen dioxide at \(231 \mathrm{K}\) is first order in both \(\left[\mathrm{NO}_{2}\right]\) and \(\left[\mathrm{O}_{3}\right]\) $$2 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})$$ (a) Write the rate equation for the reaction. (b) If the concentration of \(\mathrm{NO}_{2}\) is tripled, what is the change in the reaction rate? (c) What is the effect on reaction rate if the concentration of \(\mathrm{O}_{3}\) is halved?

Formic acid decomposes at \(550^{\circ} \mathrm{C}\) according to the equation $$\mathrm{HCO}_{2} \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ The reaction follows first-order kinetics. In an experiment, it is determined that \(75 \%\) of a sample of \(\mathrm{HCO}_{2} \mathrm{H}\) has decomposed in 72 seconds. Determine \(t_{1 / 2}\) for this reaction.

Isotopes are often used as "tracers" to follow an atom through a chemical reaction, and the following is an example. Acetic acid reacts with methanol by eliminating a molecule of water and forming methyl acetate (See Chapter 11 ). $$\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{CH}_{3} \mathrm{OH} \quad \longrightarrow \quad \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O}$$ Explain how you could use the isotope \(^{18} \mathrm{O}\) to show whether the oxygen atom in the water comes from the - OH of the acid or the - OH of the alcohol.

A reaction has the following experimental rate equation: Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}] .\) If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

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