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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 \(0.0012 \mathrm{mol} / \mathrm{L} ?\)

Short Answer

Expert verified
The rate equation is \( \text{Rate} = k [\mathrm{NO}]^2 [\mathrm{Br}_2] \). The rate triples with the new \([\mathrm{Br}_2]\) concentration and decreases to 1/4 with the new \([\mathrm{NO}]\) concentration.

Step by step solution

01

Understanding the Rate Equation

The rate equation for a reaction can be determined from the experimental orders of reaction with respect to each reactant. For the given reaction: \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})\), it is stated that the reaction is second order in \(\mathrm{NO}\) and first order in \(\mathrm{Br}_{2}\). Thus, the rate equation can be written as: \[\text{Rate} = k [\mathrm{NO}]^2 [\mathrm{Br}_2]^1\]where \(k\) is the rate constant.
02

Calculating the Change in Rate for Br2

To find how the rate changes when the concentration of \(\mathrm{Br}_2\) changes from \(0.0022 \mathrm{mol/L}\) to \(0.0066 \mathrm{mol/L}\), we analyze the rate equation with respect to \([\mathrm{Br}_2]\). Since the reaction is first order in \(\mathrm{Br}_2\), the rate is directly proportional to \([\mathrm{Br}_2]\). Thus, \[\frac{\text{New Rate}}{\text{Initial Rate}} = \frac{[\mathrm{Br}_2]_{\text{new}}}{[\mathrm{Br}_2]_{\text{initial}}}\]Substituting the given values: \[\frac{\text{New Rate}}{\text{Initial Rate}} = \frac{0.0066}{0.0022} = 3\]This means the rate triples when \([\mathrm{Br}_2]\) concentration increases from \(0.0022\) to \(0.0066 \mathrm{mol/L}\).
03

Calculating the Change in Rate for NO

To determine the change in rate if \([\mathrm{NO}]\) changes from \(0.0024 \mathrm{mol/L}\) to \(0.0012 \mathrm{mol/L}\), we apply the rate equation's dependency on \([\mathrm{NO}]^2\) since it is second order. So,\[\frac{\text{New Rate}}{\text{Initial Rate}} = \left(\frac{[\mathrm{NO}]_{\text{new}}}{[\mathrm{NO}]_{\text{initial}}}\right)^2\]Substituting the given values:\[\frac{\text{New Rate}}{\text{Initial Rate}} = \left(\frac{0.0012}{0.0024}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\]This means the rate is reduced to a quarter when \([\mathrm{NO}]\) concentration decreases from \(0.0024\) to \(0.0012 \mathrm{mol/L}\).

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Key Concepts

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

Order of Reaction
Understanding the order of reaction is key to determining how different reactants influence the rate of a chemical reaction. The order of reaction with respect to a specific reactant shows how the rate is affected by the concentration of that reactant. It is an experimental determination that can only be found through observation and experimentation.
  • For the reaction between \( \mathrm{NO} \) and \( \mathrm{Br}_2 \), it is specified that the reaction is second order in \( \mathrm{NO} \) and first order in \( \mathrm{Br}_2 \).
  • A second order reaction with respect to \( \mathrm{NO} \) means the rate is dependent on the square of \( \mathrm{NO} \)'s concentration.
  • A first order reaction in \( \mathrm{Br}_2 \) signifies the rate is directly proportional to the concentration of \( \mathrm{Br}_2 \).
This insight allows us to write rate equations that predict how changes in concentration can affect the overall reaction rate.
Initial Reaction Rate
The initial reaction rate reveals how fast a reaction begins and is crucial for understanding reaction dynamics. It's the rate of reaction at the very start, which means concentrations are at their original levels.
  • The initial reaction rate is determined by substituting the initial concentrations of reactants into the rate equation.
  • In practice, it's often measured by examining how quickly a product appears or a reactant disappears.
The concept is vital as it gives chemists a reference point to assess the reaction's behavior under controlled conditions.
Rate Constant
The rate constant, represented as \( k \), provides insight into the speed of a reaction. While it appears in the rate equation, it remains constant as long as temperature does not change.
  • The value of \( k \) is unique to each reaction and must be determined experimentally.
  • It relates to both the nature of the reactants and the conditions under which the reaction occurs.
Understanding how the rate constant works within the rate equation helps predict the reaction pace under various concentrations and conditions.
Concentration Change Effect
How the concentration of reactants changes is critical to predicting shifts in the reaction rate. In our example, changing the concentrations of \( \mathrm{NO} \) and \( \mathrm{Br}_2 \) directly influences the reaction rate due to their respective reaction orders.
  • Increasing the concentration of \( \mathrm{Br}_2 \) threefold will triple the reaction rate, as it's a first order reaction.
  • Reducing \( \mathrm{NO} \) by half decreases the rate to one-fourth, owing to its second order relationship.
Evaluating these changes illustrates how various factors impact reaction speed and efficiency, allowing for strategic adjustments in industrial and laboratory settings.

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Most popular questions from this chapter

Experimental data are listed here for the reaction \(A \rightarrow 2 B.\) $$\begin{array}{cc}\text { Time (s) } & {[\mathrm{B}](\mathrm{mol} / \mathrm{L})} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data; connect the points with a smooth line; and calculate the rate of change of [B] for each 10 -second interval from 0.0 to 40.0 seconds. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of [A] related to the rate of change of \([\mathrm{B}]\) in each time interval? Calculate the rate of change of \([\mathrm{A}]\) for the time interval from 10.0 to 20.0 seconds. (c) What is the instantaneous rate, \(\Delta[\mathrm{B}] / \Delta \mathrm{t},\) when \([\mathrm{B}]=0.750 \mathrm{mol} / \mathrm{L} ?\)

A reaction has the following experimental rate equation: Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}] .\) If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

Hydrogen iodide decomposes when heated, forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g}) .\) The rate law for this reaction is \(-\Delta[\mathrm{HI}] / \Delta t=k[\mathrm{HI}]^{2} .\) At \(443^{\circ} \mathrm{C}, k=30 . \mathrm{L} / \mathrm{mol} \cdot\) min. If the initial \(\mathrm{HI}(\mathrm{g})\) concentration is \(3.5 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) what concentration of HI (g) will remain after \(10 .\) minutes?

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.

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.

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