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The decomposition of HOF occurs at \(25^{\circ} \mathrm{C}\) $$\mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ Using the data in the table below, determine the rate law, and then calculate the rate constant. $$\begin{array}{cc}{[\mathrm{HOF}](\mathrm{mol} / \mathrm{L})} & \mathrm{Time}(\mathrm{min}) \\ \hline 0.850 & 0 \\\0.810 & 2.00 \\\0.754 & 5.00 \\\0.526 & 20.0 \\\0.243 & 50.0 \\\\\hline\end{array}$$

Short Answer

Expert verified
The reaction is first-order with a rate constant of approximately 0.0246 min^{-1}.

Step by step solution

01

Determine Reaction Order

To determine the reaction order, we should plot the concentration of HOF against time. We have three common orders for reactions: zero-order, first-order, and second-order. Plot: - Zero-order: [HOF] vs. Time should be linear. - First-order: ln[HOF] vs. Time should be linear. - Second-order: 1/[HOF] vs. Time should be linear. By plotting, we find the first-order plot is linear, indicating a first-order reaction with respect to HOF.
02

Use the First-Order Rate Law

The rate law for a first-order reaction is given by \[ ext{rate} = k[ ext{HOF}] \] Where \(k\) is the rate constant. Since we've determined the reaction is first-order, we'll use this relationship to find the rate constant \(k\).
03

Apply the First-Order Integrated Rate Law

The integrated rate law for a first-order reaction is given by \[ ext{ln}([A]_t) = -kt + ext{ln}([A]_0) \]Substitute the initial concentration \([A]_0 = 0.850 \, \text{mol/L}\) and use the concentration at \(t = 5.00 \, \text{min}\) to solve for \(k\).
04

Calculate the Rate Constant

Using \[ ext{ln}(0.754) = ext{ln}(0.850) - k(5.00) \]Calculate:\[ k = \frac{ ext{ln}(0.850) - ext{ln}(0.754)}{5.00} \approx 0.0246 \, \text{min}^{-1} \]
05

Validate the Calculation and Result

Verify by checking if \(k\) remains consistent when using other data points. Substitute the concentration at \(t = 20.0 \, \text{min}\) and compare with initial concentration. The values should confirm that \(k \approx 0.0246 \, \text{min}^{-1}\).

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Key Concepts

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

Rate Law
In chemical reactions, the rate law expresses the relationship between the reaction rate and the concentrations of reactants. The rate at which a chemical reaction proceeds can often depend on the concentration of the reactants. The rate law provides the mathematical description of this dependency, serving as a crucial concept in reaction kinetics. When you determine a rate law, you identify the order of the reaction with respect to each reactant. This is done experimentally by observing how changes in concentration affect the reaction speed. There are several potential orders a reaction can have, including zero, first, and second order.
  • Zero-order: The rate does not change with concentration changes.
  • First-order: The rate changes linearly with concentration changes.
  • Second-order: The rate changes with the square of the concentration.
Understanding the rate law is essential as it helps chemists predict how a reaction behaves under different conditions, and it aids in determining all necessary parameters for controlling the reaction.
First-Order Reaction
A first-order reaction is a type of reaction where the rate depends linearly on the concentration of one reactant. This means that if you double the concentration of the reactant, the rate of the reaction also doubles. A first-order reaction is characterized by its straightforward mathematical relationship between concentration and time, which is expressed by the equation:\[\text{ln}([A]_t) = -kt + \text{ln}([A]_0)\]Here,
  • \([A]_t\) is the concentration at time \( t \).
  • \([A]_0\) is the initial concentration.
  • \(k\) is the rate constant.
This equation is the integrated rate law for first-order reactions. It tells us that the natural logarithm of the concentration of the reactant decreases linearly with time. When examining data to identify reagents with first-order behavior, plotting \(\text{ln}([A])\) versus time will yield a straight line. The slope of this line is equal to the negative of the rate constant \(k\). This linear relationship makes it easier to analyze and predict the behavior of chemical reactions over time.
Rate Constant Calculation
The rate constant \(k\) is a crucial parameter in the study of reaction kinetics. It indicates the speed of a reaction for a particular set of conditions and is specific to each reaction and temperature. In a first-order reaction, the rate constant can be determined using experimental data and the equation derived from the integrated rate law:\[k = \frac{\text{ln}([A]_0) - \text{ln}([A]_t)}{t}\]This formula allows you to calculate \(k\) by substituting the initial concentration \([A]_0\), the concentration at a particular time \([A]_t\), and the time \(t\). For accurate calculations, you'll typically use several data points to ensure consistency:
  • Check if the calculated \(k\) is consistent for all data points.
  • Any variation indicates potential errors in data or the assumption about the reaction order.
The value of \(k\) is essential because it informs us about the reaction's velocity and can predict how the reaction progresses under varying conditions. Additionally, performing calculations at different temperatures helps understand the effect of temperature on the reaction rate, using the Arrhenius equation to relate \(k\) to the temperature of the reaction.

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Most popular questions from this chapter

The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 minutes at \(500^{\circ} \mathrm{C}\) $$\mathrm{CH}_{3} \mathrm{OCH}_{3}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ (a) Starting with \(8.00 \mathrm{g}\) of dimethyl ether, what mass remains (in grams) after 125 minutes and after 145 minutes? (b) Calculate the time in minutes required to decrease \(7.60 \mathrm{ng}\) (nanograms) to 2.25 ng. (c) What fraction of the original dimethyl ether remains after 150 minutes?

Calculate the activation energy, \(E_{\alpha}\), for the reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=\) \(3.46 \times 10^{-5} s^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}.\)

Hundreds of different reactions can occur in the stratosphere, among them reactions that destroy the earth's ozone layer. The table below lists several (secondorder) reactions of Cl atoms with ozone and organic compounds; each is given with its rate constant. $$\begin{array}{ll}\text { Reaction } & \left(298 \mathrm{K}, \mathrm{cm}^{3} / \mathrm{molecule} \cdot \mathrm{s}\right) \\\\\hline \text { (a) } \mathrm{Cl}+\mathrm{O}_{3} \rightarrow \mathrm{ClO}+\mathrm{O}_{2} & 1.2 \times 10^{-11} \\\\\text {(b) } \mathrm{Cl}+\mathrm{CH}_{4} \rightarrow \mathrm{HCl}+\mathrm{CH}_{3} & 1.0 \times 10^{-13} \\\\\text {(c) } \mathrm{Cl}+\mathrm{C}_{3} \mathrm{H}_{8} \rightarrow \mathrm{HCl}+\mathrm{C}_{3} \mathrm{H}_{7} & 1.4 \times 10^{-10} \\\\\text {(d) } \mathrm{Cl}+\mathrm{CH}_{2} \mathrm{FCl} \rightarrow \mathrm{HCl}+\mathrm{CHFCl} &3.0 \times 10^{-18} \\\\\hline\end{array}$$ For equal concentrations of Cl and the other reactant, which is the slowest reaction? Which is the fastest reaction?

Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures, a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.

When heated, cyclopropane is converted to propene (see Example 15.5 ). Rate constants for this reaction at \(470^{\circ} \mathrm{C}\) and \(510^{\circ} \mathrm{C}\) are \(k=1.10 \times 10^{-4} \mathrm{s}^{-1}\) and \(k=\) \(1.02 \times 10^{-3} \mathrm{s}^{-1},\) respectively. Determine the activation energy, \(E_{\omega},\) from these data.

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