/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 The gas-phase decomposition of \... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 41

The gas-phase decomposition of \(\mathrm{CH}_{3} \mathrm{CHO}(g)\) occurs according to the equation $$ \mathrm{CH}_{3} \mathrm{CHO}(g) \rightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) $$ and is second order. The value of the rate constant is \(0.105 \mathrm{M}^{-1} \cdot \mathrm{s}^{-1}\) at \(490^{\circ} \mathrm{C}\). If the concentration of \(\mathrm{CH}_{3} \mathrm{CHO}(g)\) is \(0.012 \mathrm{M}\) initially, what will be its concentration \(5.0\) minutes later?

Short Answer

Expert verified
After 5 minutes, the concentration of \( \mathrm{CH}_{3} \mathrm{CHO}(g) \) is approximately \( 0.00871 \, \text{M} \).

Step by step solution

01

Write the rate law expression for a second-order reaction.

The rate law for a second-order reaction in one reactant can be written as \( rac{1}{[ ext{A}]} = rac{1}{[ ext{A}]_0} + kt \), where \( [\text{A}] \) is the concentration at time \( t \), \( [\text{A}]_0 \) is the initial concentration, \( k \) is the rate constant, and \( t \) is the time.
02

Plug in the known values.

Given \( [\text{A}]_0 = 0.012 \, \text{M} \), \( k = 0.105 \, \text{M}^{-1}\cdot \text{s}^{-1} \), and \( t = 5.0 \times 60 = 300 \, \text{s} \). Substitute these into the rate law: \( \frac{1}{[\text{A}]} = \frac{1}{0.012} + 0.105 \times 300 \).
03

Calculate the concentration at 5.0 minutes later.

First, calculate \( \frac{1}{0.012} = 83.33 \). Multiply \( k \times t = 0.105 \times 300 = 31.5 \). Add these: \( 83.33 + 31.5 = 114.83 \). Then calculate \( [\text{A}] = \frac{1}{114.83} \approx 0.00871 \, \text{M} \).

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Key Concepts

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

Understanding Rate Law in Second-Order Reactions
In chemical kinetics, the rate law is an equation that relates the rate of a reaction to the concentration of the reactants. For second-order reactions, the rate depends on the concentration of one reactant squared or the concentration of two different reactants proportional to their respective orders. This makes the formula slightly more complex than for a first-order reaction.

The general rate law equation for a second-order reaction that involves only one reactant, like the decomposition of \(\mathrm{CH}_{3} \mathrm{CHO}(g)\), is:
  • \(\frac{1}{[\text{A}]} = \frac{1}{[\text{A}]_0} + kt\)
Here, \([\text{A}]\) is the concentration at any time \(t\), \([\text{A}]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time elapsed. This equation shows that the inverse of the concentration is directly proportional to time, indicating a decrease in concentration over time following a particular pattern.
Diving Into Reaction Kinetics
Reaction kinetics, or chemical kinetics, is the study of the rates of chemical processes and the factors affecting them. Every reaction proceeds at a certain rate, and understanding the kinetics helps predict how quickly a reaction will occur. Kinetics also provide insights into the reaction mechanism, which is the step-by-step sequence of elementary reactions by which a chemical change occurs.

Key factors influencing reaction rates include:
  • Concentration of reactants: Higher concentrations typically lead to higher reaction rates.
  • Temperature: Increasing temperature generally speeds up reactions as particles have more energy.
  • Catalysts: These substances increase the reaction rate without being consumed in the process.
In our example of \(\mathrm{CH}_{3} \mathrm{CHO}(g)\) decomposition, reaction kinetics helps us use the known rate constant \(k\) to describe how the reaction progresses over time.
The Nature of Decomposition Reactions
Decomposition reactions involve breaking down a compound into simpler substances. These types of reactions are critical in many scientific and industrial processes. In a decomposition reaction, one substance is converted into two or more different substances.

For our specific reaction, \(\mathrm{CH}_{3} \mathrm{CHO}(g)\) decomposes into \(\mathrm{CH}_{4}(g)\) and \(\mathrm{CO}(g)\). This transformation highlights the breakdown of an organic molecule into simpler gaseous products.

Typically, decomposition reactions can be triggered by various factors such as:
  • Heat: As in our exercise, heat is used to decompose ethyl aldehyde at high temperatures.
  • Light: Some reactions are sensitive to light, causing molecules to break down.
  • Electricity: Applying an electric current can also cause decomposition in certain reactions.
The understanding of decomposition reactions is crucial for predicting the outcomes of reactions and controlling them in real-world settings.

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Most popular questions from this chapter

Express the rate of the following reaction equation in terms of the rate of concentration change for each of the three species involved: $$ 2 \mathrm{NOCl}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$

Determine the rate law for the reaction described by $$ \mathrm{NO}(g)+\mathrm{H}_{2}(g) \rightarrow \text { products } $$ from the initial-rate data tabulated below: \begin{tabular}{ccc} \hline Initial pressure of \(P_{\mathrm{H}_{2}} /\) Torr & Initial pressure of \(P_{\mathrm{No}} /\) Torr & Initial rate of reaction/Torr \(\cdot \mathrm{s}^{-1}\) \\ \hline 400 & 159 & 34 \\ 400 & 300 & 125 \\ 289 & 400 & 160 \\ 205 & 400 & 110 \\ 147 & 400 & 79 \\ \hline \end{tabular} Calculate the value of the rate constant for this reaction.

Identify in each of the following cases the order of the reaction rate law with respect to the reactant \(\mathrm{A}\), where \(\mathrm{A} \rightarrow\) Products: (a) The half-life of \(\mathrm{A}\) is independent of the initial concentration of \(\mathrm{A}\). (b) The rate of decrease of \(\mathrm{A}\) is a constant. (c) A twofold increase in the initial concentration of A leads to a \(1.41\) -fold increase in the initial rate. (d) A twofold increase in the initial concentration of A leads to a fourfold increase in the initial rate. (e) The time required for \([\mathrm{A}]_{0}\) to decrease to \([\mathrm{A}]_{0} / 2\) is equal to the time required for \([\mathrm{A}]\) to decrease from \([\mathrm{A}]_{0} / 2\) to \([\mathrm{A}]_{0} / 4\)

The reaction described by $$ \mathrm{NO}_{2}(g) \rightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ is a second-order reaction with a rate constant \(k=\) \(0.54 \mathrm{M}^{-1} \cdot \mathrm{s}^{-1} .\) How long will it take for \(\left[\mathrm{NO}_{2}\right]\) to be \(10.0 \%\) of its initial value of \(2.00 \mathrm{M} ?\)

The rate law for the reaction described by the equation $$ 2 \mathrm{~N}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) $$ is second order in \(\left[\mathrm{N}_{2} \mathrm{O}\right]\). The reaction was carried out at \(900 \mathrm{~K}\) with an initial concentration of \(\mathrm{N}_{2} \mathrm{O}(g)\) of \(2.0 \times 10^{-2}\) M. It took 4500 seconds for \(\left[\mathrm{N}_{2} \mathrm{O}\right]\) to fall to half its initial value. Determine the value of the rate constant for this reaction.
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