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Chapter 13: Problem 41

To obtain the rate of the reaction $$5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ you might follow the \(\mathrm{Br}^{-}\) concentration or the \(\mathrm{BrO}_{3}^{-}\) concentration. How are the rates in terms of these species related?

Short Answer

Expert verified
The rate of change of \(\mathrm{Br}^{-}\) is five times the rate of change of \(\mathrm{BrO}_{3}^{-}\).

Step by step solution

01

Write the Rate Expression for Each Reactant

For the reaction \(5 \mathrm{Br}^{-}(aq) + \mathrm{BrO}_{3}^{-}(aq) + 6 \mathrm{H}^{+}(aq) \rightarrow 3 \mathrm{Br}_{2}(aq) + 3 \mathrm{H}_{2}O(l)\), we start by writing the rate expression in terms of reactants. The rate of reaction can be determined based on the change in concentration of \(\mathrm{Br}^{-}\) and \(\mathrm{BrO}_{3}^{-}\) over time. Writing these expressions gives us:\[\text{Rate} = -\frac{1}{5} \frac{d [\mathrm{Br}^{-}]}{dt} \text{ and } \text{Rate} = - \frac{d [\mathrm{BrO}_{3}^{-}]}{dt}.\]
02

Equate the Rate Expressions

The rate of reaction is independent of the species used to measure it, meaning the rate calculated using \(\mathrm{Br}^{-}\) should equal the rate calculated using \(\mathrm{BrO}_{3}^{-}\). Thus, we equate both expressions:\[-\frac{1}{5} \frac{d [\mathrm{Br}^{-}]}{dt} = - \frac{d [\mathrm{BrO}_{3}^{-}]}{dt}.\]
03

Solve for Species Rate Relationship

From the equated expression, solve for the relationship between the rates of change of \(\mathrm{Br}^{-}\) and \(\mathrm{BrO}_{3}^{-}\). This becomes:\[\frac{d [\mathrm{Br}^{-}]}{dt} = 5 \cdot \frac{d [\mathrm{BrO}_{3}^{-}]}{dt}.\]This equation implies that for every change in the concentration of \(\mathrm{BrO}_{3}^{-}\), the concentration change for \(\mathrm{Br}^{-}\) is five times as much.

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Key Concepts

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

Rate Expression
In the context of chemical reactions, a rate expression is crucial as it defines how the rate of a reaction depends on the concentration of the reactants. By observing the change in concentration over time, we can determine how quickly a reaction proceeds. For a given reaction, the rate can be expressed in relation to any reactant or product. In the provided reaction, the rate expression focuses on the reactants:
  • For bromide ions, \( \mathrm{Br}^{-} \), the rate expression is given as \(-\frac{1}{5} \frac{d [\mathrm{Br}^{-}]}{dt} \). The negative sign indicates that the concentration of the reactant decreases over time.
  • For bromate ions, \( \mathrm{BrO}_{3}^{-} \), the rate expression is \(-\frac{d [\mathrm{BrO}_{3}^{-}]}{dt} \).
These expressions help quantify the reaction's progression by examining how fast each reactant is consumed.
Chemical Kinetics
Chemical kinetics is the area of chemistry that deals with understanding the speed or rate at which reactions occur. This includes examining factors such as temperature, concentration, and catalysts that influence the rate. By studying kinetics, we can develop a deeper understanding of reaction mechanisms and predict the behavior of chemical systems. In our example:
  • We see how the concentration of reactants \( \mathrm{Br}^{-} \) and \( \mathrm{BrO}_{3}^{-} \) changes over time to determine the reaction rate.
  • Equating the rates from the different reactants ensures that we understand that the actual speed of the reaction isn't dependent on which reactant's rate we are focusing on.
Kinetics forms a bridge connecting the theoretical world of chemical equations with practical, observable reaction speeds.
Reaction Stoichiometry
Stoichiometry in chemistry refers to the quantitative relationships between the amounts of reactants and products in a chemical reaction. This is derived from the balanced chemical equation that provides the mole ratio of reactants to products. In the given reaction:
  • 5 moles of \( \mathrm{Br}^{-} \) react with 1 mole of \( \mathrm{BrO}_{3}^{-} \).

  • The expression \( \frac{d [\mathrm{Br}^{-}]}{dt} = 5 \cdot \frac{d [\mathrm{BrO}_{3}^{-}]}{dt} \) shows that the rate at which \( \mathrm{Br}^{-} \) is consumed is 5 times that of \( \mathrm{BrO}_{3}^{-} \). This highlights the importance of stoichiometry in determining relative rates of consumption in a reaction.
Utilizing stoichiometry allows for accurate calculations of reactant consumption and product formation.
Reactant Concentration
Concentration refers to the amount of a substance present in a given volume of solution. In reactions, the concentration of reactants usually decreases as the reaction proceeds. For our specific reaction, focusing on reactant concentration is vital:
  • Changes in the concentrations of \( \mathrm{Br}^{-} \) and \( \mathrm{BrO}_{3}^{-} \) directly affect the observed reaction rate.

  • By monitoring these changes over time, we gain insights into the progression and efficiency of the reaction.
Reactant concentration is a key parameter in experimental kinetic studies, influencing how quickly or slowly a reaction reaches completion.

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Most popular questions from this chapter

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 firstand then second-order kinetics. [Data from H. Fukutomi and G. Gordon, J. Am. Chem. Soc., 89,1362 (1967).] \(\begin{array}{cl}\text { Time (s) } & {\left[\text { ClO }_{2}\right] \text { (moll } \text { L) }} \\ 0.00 & 4.77 \times 10^{-4} \\ 1.00 & 4.31 \times 10^{-4} \\ 2.00 & 3.91 \times 10^{-4} \\ 3.00 & 3.53 \times 10^{-4} \\ 5.00 & 2.89 \times 10^{-4} \\ 10.00 & 1.76 \times 10^{-4} \\ 30.00 & 2.4 \times 10^{-5} \\ 50.00 & 3.2 \times 10^{-6}\end{array}\)

Azomethane, \(\mathrm{CH}_{3} \mathrm{NNCH}_{3}\), decomposes according to the following equation: $$\mathrm{CH}_{3} \mathrm{NNCH}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{N}_{2}(g)$$ The initial concentration of azomethane was \(1.50 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) After \(7.00 \mathrm{~min}\), this concentration decreased to \(1.01 \times 10^{-2}\) \(\mathrm{mol} / \mathrm{L}\). Obtain the average rate of reaction during this time interval. Express the answer in units of \(\mathrm{mol} /(\mathrm{L} \cdot \mathrm{s})\).

At high temperature, the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ is thought to occur in a single step. What should be the rate law in that case?

The dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) into \(\mathrm{NO}_{2}\), $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ is believed to occur in one step. Obtain the concentration of \(\mathrm{N}_{2} \mathrm{O}_{4}\) in terms of the concentration of \(\mathrm{NO}_{2}\) and the rate constants for the forward and reverse reactions, when the reactions have come to equilibrium.

Ethyl chloride, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), used to produce tetraethyllead gasoline additive, decomposes, when heated, to give ethylene and hydrogen chloride. The reaction is first order. In an experiment, the initial concentration of ethyl chloride was \(0.00100 \mathrm{M}\). After heating at \(500^{\circ} \mathrm{C}\) for \(155 \mathrm{~s}\), this was reduced to \(0.00067 \mathrm{M}\). What was the concentration of ethyl chloride after a total of \(256 \mathrm{~s}\) ?
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