/*! 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 22 The dimerization of butadiene, \... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 22

The dimerization of butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) to form 1,5-cyclooctadiene is a second-order process that occurs when the diene is heated. In an experiment, a sample of 0.0087 mol of \(\mathrm{C}_{4} \mathrm{H}_{6}\) was heated in a \(1.0-\mathrm{L}\) flask. After 600 seconds, \(21 \%\) of the butadiene had dimerized. Calculate the rate constant for this reaction.

Short Answer

Expert verified
The rate constant \( k \) is approximately \( 0.051 \; \text{M}^{-1}\text{s}^{-1} \).

Step by step solution

01

Understand the Reaction Order

The problem states that the dimerization of butadiene is a second-order reaction. The rate law for a second-order reaction is given as \( rate = k[A]^2 \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.
02

Determine the Initial Concentration

Initially, we have 0.0087 mol of \( \text{C}_4\text{H}_6 \) in a 1.0 L flask. The initial concentration \( [A]_0 \) is \( 0.0087 \; \text{M} \).
03

Calculate Remaining Concentration After Reaction

21% of the butadiene has reacted, meaning 79% remains. So, the concentration after 600 seconds, \( [A] \), is \( 0.79 \times 0.0087 = 0.006873 \; \text{M} \).
04

Use Second-Order Integrated Rate Law

The integrated rate law for a second-order reaction is \( \frac{1}{[A]} = kt + \frac{1}{[A]_0} \). Insert the known values: \([A] = 0.006873 \; \text{M}\), \([A]_0 = 0.0087 \; \text{M}\), and \( t = 600 \; \text{s} \).
05

Calculate the Rate Constant \( k \)

Rearrange the rate law equation: \( k = \frac{1/[A] - 1/[A]_0}{t} \). Substitute \([A]_0 = 0.0087\), \([A] = 0.006873\), and \(t = 600\) to find \( k \):\[ k = \frac{1/0.006873 - 1/0.0087}{600} \k = \frac{145.526 - 114.943}{600} \k \approx 0.05097 \; \text{M}^{-1}\text{s}^{-1}\]
06

Conclusion

Therefore, the rate constant \( k \) for the dimerization of butadiene is approximately \( 0.051 \; \text{M}^{-1}\text{s}^{-1} \).

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Key Concepts

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

Rate Constant
The rate constant, denoted by the symbol \( k \), is a crucial factor in determining the speed of a chemical reaction. It is specific to a particular reaction at a given temperature. The rate constant provides insight into how fast a reaction proceeds. For a second-order reaction like the dimerization of butadiene, the rate constant has units of \( ext{M}^{-1} ext{s}^{-1} \).
The value of \( k \) can be affected by several factors:
  • Temperature: Higher temperatures typically increase the rate constant, thus speeding up the reaction.
  • Catalysts: Presence of catalysts can alter \( k \) by providing a different pathway for the reaction.
  • Nature of reactants: Different substances have different intrinsic reaction rates.
Understanding the rate constant helps chemists predict how reactant concentrations change over time, which is critical in designing reactors and industrial processes.
Dimerization
Dimerization is a chemical reaction where two identical molecules join to form a single, larger molecule. In this exercise, dimerization involves the reaction of two butadiene molecules to form 1,5-cyclooctadiene.
The process has specific characteristics:
  • It involves the formation of a bond between two identical entities.
  • The reaction typically proceeds via a mechanism that involves an intermediate state.
  • Dimerization can be reversible, depending on the conditions and the strength of the bonds formed.
In organic chemistry, dimerization is used to synthesize new compounds and materials with desirable properties. In the context of this reaction, understanding the process can help in predicting and controlling the synthesis of desired cyclized products.
Butadiene
Butadiene, with the chemical formula \( ext{C}_4 ext{H}_6 \), is an important starting material in the chemical industry. It is a conjugated diene, meaning it has two double bonds separated by a single carbon-carbon bond.
Here are some essential properties of butadiene:
  • It is a colorless gas at room temperature and has a mild, aromatic odor.
  • Butadiene is highly reactive due to its conjugated system, making it an excellent candidate for polymerization and dimerization reactions.
  • It is used in the manufacture of synthetic rubber and plastics, such as styrene-butadiene rubber (SBR).
In this dimerization reaction, butadiene acts as the reactant, forming larger molecules through the bonding of two butadiene units. Understanding its behavior aids in the manipulation of reaction conditions for optimal product yield.
Integrated Rate Law
The integrated rate law for a chemical reaction provides a equation that relates the concentrations of reactants over time. For second-order reactions like the dimerization of butadiene, the integrated rate law is expressed as:\[ \frac{1}{[A]} = kt + \frac{1}{[A]_0} \]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \( t \), and \( k \) is the rate constant.
Here鈥檚 why integrated rate laws are useful:
  • They allow calculation of concentrations at any given time, aiding in tracking the progress of the reaction.
  • They enable determination of the order of reaction and computation of the rate constant.
  • They facilitate predictions about how long a reaction will take to reach a certain extent.
Using the integrated rate law, we can transform experimental data into actionable insights, such as calculating rate constants and half-life of the reaction.

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Most popular questions from this chapter

The decomposition of phosphine, \(\mathrm{PH}_{3},\) proceeds according to the equation $$\mathrm{PH}_{3}(\mathrm{g}) \rightarrow 1 / 4 \mathrm{P}_{4}(\mathrm{g})+3 / 2 \mathrm{H}_{2}(\mathrm{g})$$ It is found that the reaction has the following rate equation: Rate \(=k\left[\mathrm{PH}_{3}\right] .\) The half-life of \(\mathrm{PH}_{3}\) is 37.9 seconds at \(120^{\circ} \mathrm{C}.\) (a) How much time is required for three fourths of the \(\mathrm{PH}_{3}\) to decompose? (b) What fraction of the original sample of \(\mathrm{PH}_{3}\) remains after 1.00 minute?

The following statements relate to the reaction for the formation of HI: $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HI}(\mathrm{g}) \quad \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]$$ Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

If the rate constant for a reaction triples when the temperature rises from \(3.00 \times 10^{2} \mathrm{K}\) to \(3.10 \times 10^{2} \mathrm{K},\) what is the activation energy of the reaction?

Draw a reaction coordinate diagram for an exothermic reaction that occurs in a single step. Identify the activation energy and the net energy change for the reaction on this diagram. Draw a second diagram that represents the same reaction in the presence of a catalyst, assuming a single step reaction is involved here also. Identify the activation energy of this reaction and the energy change. Is the activation energy in the two drawings different? Does the energy evolved in the two reactions differ?

Nitrosyl bromide, NOBr, is formed from \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\) : $$2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})$$ Experiments show that this reaction is second order in NO and first order in \(\mathrm{Br}_{2}\). (a) Write the rate equation for the reaction. (b) How does the initial reaction rate change if the concentration of \(\mathrm{Br}_{2}\) is changed from \(0.0022 \mathrm{mol} / \mathrm{L}\) to \(0.0066 \mathrm{mol} / \mathrm{L} ?\) (c) What is the change in the initial rate if the concentration of NO is changed from \(0.0024 \mathrm{mol} / \mathrm{L}\) to \(0.0012 \mathrm{mol} / \mathrm{L} ?\)
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