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Chapter 15: Problem 53

Formic acid decomposes at \(550^{\circ} \mathrm{C}\) according to the equation $$\mathrm{HCO}_{2} \mathrm{H}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ 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 \(t_{1 / 4}\) for this reaction.

Short Answer

Expert verified
\( t_{1/4} \approx 18.7 \, \text{s} \).

Step by step solution

01

Understanding First-Order Kinetics

A first-order reaction is described by the formula \( k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \), where \( k \) is the rate constant, \([A]_0\) is the initial concentration, and \([A]\) is the concentration at time \( t \).
02

Determine Rate Constant (k)

Given that 75% of the sample decomposed, 25% remains. Using the formula: \[ \ln \left( \frac{[A]_0}{[A]} \right) = \ln \left( \frac{1}{0.25} \right) = \ln(4) \]. With \( t = 72 \) seconds, calculate \( k \):\[ k = \frac{\ln(4)}{72} \approx 0.0193 \, \text{s}^{-1} \].
03

Determine Time for 25% Decomposition (\( t_{1/4} \))

The time for a quarter decomposition \( t_{1/4} \) implies 75% of the sample remains. Use the rearranged formula:\[ t = \frac{1}{k} \ln \left( \frac{1}{0.75} \right) \]. Substitute \( k = 0.0193 \, \text{s}^{-1} \) and solve:\[ t_{1/4} = \frac{1}{0.0193} \ln(\frac{4}{3}) \approx 18.7 \, \text{s} \].

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Key Concepts

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

Chemical Decomposition
Chemical decomposition is a process where a single compound breaks down into two or more simpler substances. In this exercise, formic acid (\(\text{HCO}_2\text{H}\)) decomposes into carbon dioxide (\(\text{CO}_2\)) and hydrogen gas (\(\text{H}_2\)). The reaction is represented by the equation:

\[ \text{HCO}_2\text{H} \rightarrow \text{CO}_2 + \text{H}_2 \]

This type of reaction is common in chemistry, where heat or another form of energy is often required to break the chemical bonds in the compound.
In our case, the reaction occurs at a high temperature of \(550^{\circ} \text{C}\).

Key points about chemical decomposition:
  • It requires energy input to break bonds.
  • It results in simpler products.
  • The process follows specific reaction kinetics, such as first-order.
Rate Constant
The rate constant, represented by \(k\), is a fundamental value in reaction kinetics. It indicates how fast a reaction proceeds. For first-order reactions, it is determined using the equation:

\[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \]

Here, \([A]_0\) is the initial concentration, and \([A]\) is the concentration at a given time \(t\).
In our problem, we calculated \(k\) for a reaction where 75% of \(\text{HCO}_2\text{H}\) decomposes in 72 seconds.

Substituting values:
\[ k = \frac{\ln(4)}{72} \approx 0.0193 \, \text{s}^{-1} \]

This value tells us about the speed and nature of the reaction. The larger the rate constant, the faster the reaction.

Keep in mind:
  • The units of \(k\) vary with the order of the reaction.
  • First-order reactions have \(k\) in units of \(\text{s}^{-1}\).
Reaction Time
Reaction time is a crucial aspect in studying chemical kinetics, reflecting how long it takes for a certain percentage of the reactants to transform into products. In the context of this exercise, we focus on the time needed for a specific fraction of decomposition.

For 25% decomposition (or 75% of the substance remaining), the reaction time \(t_{1/4}\) can be calculated using:
\[ t = \frac{1}{k} \ln \left( \frac{1}{0.75} \right) \]

Substituting in the given \(k = 0.0193 \, \text{s}^{-1}\):
\[ t_{1/4} \approx \frac{1}{0.0193} \ln\left(\frac{4}{3}\right) \approx 18.7 \, \text{s} \]

This shows that it takes approximately 18.7 seconds for 75% of the formic acid to remain.

Important points about reaction time:
  • Provides insight into reaction speed.
  • Varies with conditions like temperature and pressure.
  • Is essential for determining suitable reaction durations in industrial applications.

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Most popular questions from this chapter

Nitramide, \(\mathrm{NO}_{2} \mathrm{NH}_{2},\) decomposes slowly in aqueous solution according to the following reaction: $$\mathrm{NO}_{2} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ The reaction follows the experimental rate law $$\text { Rate }=\frac{k\left[\mathrm{NO}_{2} \mathrm{NH}_{2}\right]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}$$ (a) What is the apparent order of the reaction in a buffered solution? (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. Mechanism 1 \(\mathrm{NO}_{2} \mathrm{NH}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Mechanism 2 \(\begin{array}{l}\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \frac{k_{2}}{k_{2}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} \\\\\mathrm{NO}_{2} \mathrm{NH}_{3}+\stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}\mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+}(\text {rapid equilibrium })\end{array}\) Mechanism 3 \(\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \frac{k_{4}}{k_{4}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) \(\mathrm{NO}_{2} \mathrm{NH}^{-} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{OH}^{-} \quad\) (rapid equilibrium) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \stackrel{k_{6}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O} \quad\) (very fast reaction) (c) Show the relationship between the experimentally observed rate constant, \(k\), and the rate constants in the selected mechanism. (Note that when writing the expression for \(K\), the equilibrium constant, \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) is not involved. \(\rightarrow\) Chapter \(16 .\) ) (d) Show that hydroxide ions catalyze the decomposition of nitramide.

We want to study the hydrolysis of the beautiful green, cobalt-based complex called trans-dichlorobis(ethylenediamine) cobalt(III) ion, (Check your book to see figure) In this hydrolysis reaction, the green complex ion trans\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en \(=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) ). The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray. Changes in color with time as \(\mathrm{Cl}^{-}\) ion is replaced by \(\mathrm{H}_{2} \mathrm{O}\) in a cobalt(III) complex. The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker. Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co-Cl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water. Slow: $$\begin{aligned}\operatorname{trans}-\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) & \rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\end{aligned}$$ Fast: $$\begin{aligned}\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) & \rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})\end{aligned}$$ (a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting In \(k\) versus \(1 / T .\) However, here we do not need to measure \(k\) directly. Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\) the time needed to achieve the gray color is a measure of \(k .\) Use the data below to find the activation energy. $$\begin{array}{cc}\text { Temperature }^{\circ} \mathrm{C} & \text { (for the Same Initial Concentration) } \\\\\hline 56 & 156 \mathrm{s} \\\60 & 114 \mathrm{s} \\\65 & 88 \mathrm{s} \\\75 & 47 \mathrm{s} \\\\\hline\end{array}$$

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}\) is a first-order reaction. If \(2.56 \mathrm{mg}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is present initially and \(2.50 \mathrm{mg}\) is present after 4.26 minutes at \(55^{\circ} \mathrm{C},\) what is the value of the rate constant, \(k ?\)

Chlorine atoms contribute to the destruction of the earth's ozone layer by the following sequence of reactions: $$\begin{array}{l}\mathrm{Cl}+\mathrm{O}_{3} \rightarrow \mathrm{ClO}+\mathrm{O}_{2} \\ \mathrm{ClO}+\mathrm{O} \rightarrow \mathrm{Cl}+\mathrm{O}_{2}\end{array}$$ where the O atoms in the second step come from the decomposition of ozone by sunlight: $$\mathrm{O}_{3}(\mathrm{g}) \rightarrow \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ What is the net equation on summing these three equations? Why does this lead to ozone loss in the stratosphere? What is the role played by Cl in this sequence of reactions? What name is given to species such as ClO?

The reaction between ozone and nitrogen dioxide at \(231 \mathrm{K}\) is first order in both \(\left[\mathrm{NO}_{2}\right]\) and \(\left[\mathrm{O}_{3}\right].\) $$2 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ (a) Write the rate equation for the reaction. (b) If the concentration of \(\mathrm{NO}_{2}\) is tripled (and \(\left[\mathrm{O}_{3}\right]\) is not changed), what is the change in the reaction rate? (c) What is the effect on reaction rate if the concentration of \(\left.\mathbf{O}_{3} \text { is halved (with no change in }\left[\mathrm{NO}_{2}\right]\right) ?\)
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