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Chapter 14: Problem 94

Benzene diazonium chloride, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NNCl}\), decomposes by a first-order rate law. $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NNCl} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{N}_{2}(g) $$ If the rate constant at \(20^{\circ} \mathrm{C}\) is \(4.3 \times 10^{-5} / \mathrm{s}\), how long will it take for \(75 \%\) of the compound to decompose?

Short Answer

Expert verified
It takes approximately 8.96 hours for 75% of the compound to decompose.

Step by step solution

01

Identify Parameters

We are given that the reaction follows a first-order rate law and is described by the following reaction: \( \text{C}_6\text{H}_5\text{NNCl} \rightarrow \text{C}_6\text{H}_5\text{Cl} + \text{N}_2(g) \). The rate constant \( k \) is \( 4.3 \times 10^{-5} \text{ s}^{-1} \), and we need to find the time for \( 75\% \) of the compound to decompose.
02

Determine Fraction Remaining

If \( 75\% \) of benzene diazonium chloride decomposes, \( 25\% \) remains. This means the fraction remaining \( \left( \frac{[A]}{[A]_0} \right) \) is \( 0.25 \).
03

Use First-Order Kinetics Formula

For a first-order reaction, the formula for the concentration at a time \( t \) is given by the equation: \( \ln \left( \frac{[A]}{[A]_0} \right) = -kt \).
04

Substitute Values into Equation

Substitute \( \ln(0.25) \) for \( \ln \left( \frac{[A]}{[A]_0} \right) \) and \( 4.3 \times 10^{-5} \text{ s}^{-1} \) for \( k \) into the equation: \[ \ln(0.25) = -(4.3 \times 10^{-5})t \].
05

Solve for Time

Calculate \( t \) by solving the equation: \[ t = \frac{\ln(0.25)}{-(4.3 \times 10^{-5})} \]. Calculate \( \ln(0.25) \approx -1.3863 \), so: \[ t = \frac{-1.3863}{-(4.3 \times 10^{-5})} \approx 32228.84 \text{ s} \approx 8.96 \text{ hours}. \]

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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, often represented as \( k \), is a crucial component in determining the speed of a chemical reaction. In first-order kinetics, the rate of reaction is directly proportional to the concentration of one reactant.

A rate constant's value is pivotal because it helps predict how fast the reactants will be transformed into products under certain conditions. It typically varies with temperature. In our exercise, the rate constant is given as \( 4.3 \times 10^{-5} \text{ s}^{-1} \), indicating it takes about the same amount of time for the concentration of the reactant to decrease by a certain proportion.

Remember, for first-order reactions, if you change the concentration of the reactant, the rate changes accordingly. However, \( k \) remains constant at a constant temperature until new conditions are applied.
Half-life
The half-life of a reaction is the time it takes for half of the original quantity of a reactant to be consumed in a chemical reaction. In the context of first-order reactions, the half-life is particularly interesting because it is constant and does not depend on the initial concentration.

This characteristic simplicity makes first-order reactions relatively straightforward to study and predict. If you know the rate constant \( k \), you can calculate the half-life using the formula: \[ t_{1/2} = \frac{0.693}{k} \] With this formula, you can see how the rate constant influences the half-life period. So in our case of benzene diazonium chloride, you could use the given rate constant to figure out the time for half to decompose if needed.
Chemical Reaction Rates
Chemical reaction rates define how fast a reaction occurs. This "speed" can be affected by several factors:
  • Concentration of reactants
  • Temperature
  • Presence of a catalyst
  • Surface area
In a first-order reaction like the decomposition of benzene diazonium chloride, the rate is proportional to the concentration of one single reactant. This implies that if you double the concentration of \( \text{C}_6\text{H}_5\text{NNCl} \), the rate of decomposition is also doubled at a constant temperature.

Understanding reaction rates is essential because it allows chemists to control and manipulate reactions to obtain desired outcomes efficiently.
Benzene Diazonium Chloride Decomposition
Benzene diazonium chloride is a compound with significant interest in organic chemistry, particularly for its decomposition process. Its decomposition follows first-order kinetics as shown in the problem.

When benzene diazonium chloride (\( \text{C}_6\text{H}_5\text{NNCl} \)) decomposes, it results in the formation of chlorobenzene (\( \text{C}_6\text{H}_5\text{Cl} \)) and nitrogen gas (\( \text{N}_2 \)), as depicted by the chemical equation: \[ \text{C}_6\text{H}_5\text{NNCl} \rightarrow \text{C}_6\text{H}_5\text{Cl} + \text{N}_2(g) \] This is a particularly useful reaction in the synthesis of aromatic compounds.

The decomposition is significant not only in lab-based applications but also in industrial processes due to the valuable products formed during the reaction.

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Most popular questions from this chapter

The reaction of thioacetamide with water is shown by the equation below: $$ \mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{CH}_{3} \mathrm{C}(\mathrm{O}) \mathrm{NH}_{2}(a q) $$ The rate of reaction is given by the rate law: $$ \text { Rate }=k\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}\right] $$ Consider one liter of solution that is \(0.20 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}\) and \(0.15 M\) in \(\mathrm{HCl}\) at \(25^{\circ} \mathrm{C}\). a. For each of the changes listed below, state whether the rate of reaction increases, decreases, or remains the same. Why? i. A 4 -g sample of \(\mathrm{NaOH}\) is added to the solution. ii. \(500 \mathrm{~mL}\) of water is added to the solution. b. For each of the changes listed below, state whether the value of \(k\) will increase, decrease, or remain the same. Why? i. A catalyst is added to the solution. ii. The reaction is carried out at \(15^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C}\).

Chlorine dioxide oxidizes iodide ion in aqueous solution to iodine; chlorine dioxide is reduced to chlorite ion. $$ 2 \mathrm{ClO}_{2}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+\mathrm{I}_{2}(a q) $$ The order of the reaction with respect to \(\mathrm{ClO}_{2}\) was determined by starting with a large excess of \(\mathrm{I}^{-}\), so that its concentration was essentially constant. Then $$ \text { Rate }=k\left[\mathrm{ClO}_{2}\right]^{m}\left[\mathrm{I}^{-}\right]^{n}=k^{\prime}\left[\mathrm{ClO}_{2}\right]^{m} $$ where \(k^{\prime}=k\left[\mathrm{I}^{-}\right]^{n} .\) Determine the order with respect to \(\mathrm{ClO}_{2}\) and the rate constant \(k^{\prime}\) by plotting the following data assuming first- and then second-order kinetics. [Data from \(\mathrm{H}\). Fukutomi and G. Gordon, J. Am. Chem. Soc., 89, 1362 \((1967) .]\) $$ \begin{array}{ll} \text { Time (s) } & {\left[\mathrm{ClO}_{2}\right](\mathrm{mol} / \mathrm{L})} \\ 0.00 & 4.77 \times 10^{-4} \\ 1.00 & 4.31 \times 10^{-4} \\ 2.00 & 3.91 \times 10^{-4} \\ 3.00 & 3.53 \times 10^{-4} \\ 5.00 & 2.89 \times 10^{-4} \\ 10.00 & 1.76 \times 10^{-4} \\ 30.00 & 2.4 \times 10^{-5} \\ 50.00 & 3.2 \times 10^{-6} \end{array} $$

The decomposition of nitrogen dioxide, $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ has a rate constant of \(0.498 \mathrm{M} / \mathrm{s}\) at \(319^{\circ} \mathrm{C}\) and a rate constant of \(1.81 \mathrm{M} / \mathrm{s}\) at \(354^{\circ} \mathrm{C}\). What are the values of the activation energy and the frequency factor for this reaction? What is the rate constant at \(420^{\circ} \mathrm{C}\) ?

A reaction of the form \(a \mathrm{~A} \longrightarrow\) Products is secondorder with a rate constant of \(0.225 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})\). If the initial concentration of A is \(0.293 \mathrm{~mol} / \mathrm{L}\), what is the molar concentration of A after \(35.4 \mathrm{~s}\) ?

Butadiene can undergo the following reaction to form a dimer (two butadiene molecules hooked together). $$ 2 \mathrm{C}_{4} \mathrm{H}_{8}(g) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(g) $$ The half-life for the reaction at a given temperature is \(5.92 \times\) \(10^{-2} \mathrm{~s}\). The reaction is second-order. a. If the initial concentration of \(\mathrm{C}_{4} \mathrm{H}_{8}\) is \(0.50 \mathrm{M}\), what is the rate constant for the reaction? b. If the initial concentration of \(\mathrm{C}_{4} \mathrm{H}_{8}\) is \(0.010 \mathrm{M}\), what will be the concentration of \(\mathrm{C}_{4} \mathrm{H}_{8}\) after \(3.6 \times 10^{2} \mathrm{~s}\) ?
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