/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Chlorite is reduced by bromide i... [FREE SOLUTION] | 91Ó°ÊÓ

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Chlorite is reduced by bromide in acidic solution according to the following balanced equation: \(\begin{aligned} \mathrm{ClO}_{2}^{-}(a q)+4 \mathrm{Br}^{-}(a q)+4 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+& 2 \mathrm{Br}_{2}(a q) \\ &+2 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned}\) (a) If \(\Delta\left[\mathrm{Br}_{2}\right] / \Delta t=4.8 \times 10^{-6} \mathrm{M} / \mathrm{s}\), what is the value of \(\Delta\left[\mathrm{ClO}_{2}^{-}\right] / \Delta t\) during the same time interval? (b) What is the average rate of consumption of \(\mathrm{Br}^{-}\) during the same time interval?

Short Answer

Expert verified
(a) \(2.4 \times 10^{-6} \ \mathrm{M/s}\); (b) \(19.2 \times 10^{-6} \ \mathrm{M/s}\).

Step by step solution

01

Understanding the Reaction

The given reaction is \( \mathrm{ClO}_{2}^{-}(aq)+4 \mathrm{Br}^{-}(aq)+4 \mathrm{H}^{+}(aq) \longrightarrow \mathrm{Cl}^{-}(aq)+2 \mathrm{Br}_{2}(aq)+2 \mathrm{H}_{2} \mathrm{O}(l) \). According to this balanced equation, 1 mole of \( \mathrm{ClO}_{2}^{-} \) reacts with 4 moles of \( \mathrm{Br}^{-} \) to produce 2 moles of \( \mathrm{Br}_{2} \).
02

Relate the Rate of Formation and Consumption Using Stoichiometry

The stoichiometry of the reaction provides a relationship between the rates of formation and consumption: - The rate of formation of \( \mathrm{Br}_{2} \) is related to the rate of consumption of \( \mathrm{ClO}_{2}^{-} \) by the stoichiometric coefficient in the balanced equation:\[ \frac{\Delta[\mathrm{Br}_2]}{\Delta t} = 2 \times \frac{\Delta [\mathrm{ClO}_{2}^{-}]}{\Delta t} \]
03

Calculate \( \Delta[\mathrm{ClO}_{2}^{-}]/\Delta t \)

Given \( \frac{\Delta[\mathrm{Br}_{2}]}{\Delta t} = 4.8 \times 10^{-6} \ \mathrm{M/s} \), and using the stoichiometry from Step 2:\[ 4.8 \times 10^{-6} = 2 \times \frac{\Delta [\mathrm{ClO}_{2}^{-}]}{\Delta t} \]Solving for \( \Delta [\mathrm{ClO}_{2}^{-}]/\Delta t \), we divide both sides by 2:\[ \frac{\Delta [\mathrm{ClO}_{2}^{-}]}{\Delta t} = 2.4 \times 10^{-6} \ \mathrm{M/s} \]
04

Understand Rate of Consumption from Stoichiometry

The rate of consumption of \( \mathrm{Br}^{-} \) is related to the rate of formation of \( \mathrm{Br}_{2} \) by the balanced equation:\[ \frac{\Delta [\mathrm{Br}^{-}]}{\Delta t} = 4 \times \frac{\Delta [\mathrm{Br}_2]}{\Delta t} \]
05

Calculate \( \Delta [\mathrm{Br}^{-}]/\Delta t \)

Using the given value \( \frac{\Delta[\mathrm{Br}_{2}]}{\Delta t} = 4.8 \times 10^{-6} \ \mathrm{M/s} \):\[ \frac{\Delta [\mathrm{Br}^{-}]}{\Delta t} = 4 \times 4.8 \times 10^{-6} \ \mathrm{M/s} \]Thus:\[ \frac{\Delta [\mathrm{Br}^{-}]}{\Delta t} = 19.2 \times 10^{-6} \ \mathrm{M/s} \]

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Key Concepts

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

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. In the context of reaction kinetics, stoichiometry allows us to understand how much of each reactant is required to form a certain amount of product. It relies on the balanced chemical equation to determine the molar ratios of these substances.

For instance, in the given reaction, the stoichiometry is captured by the equation:
  • 1 mole of \( \mathrm{ClO}_{2}^{-} \) reacts with 4 moles of \( \mathrm{Br}^{-} \)
  • This produces 2 moles of \( \mathrm{Br}_{2} \) and other products.
Here, each stoichiometric coefficient (numbers before the chemical formulas) gives the ratio of moles of each substance. This ratio helps to relate the rate of consumption of \( \mathrm{Br}^{-} \) to the rate of production of \( \mathrm{Br}_{2} \), ensuring that we can calculate changes in concentrations over time efficiently.
Rate of Formation
The rate of formation is a measure of how quickly a product is formed in a chemical reaction. In kinetic studies, this is typically expressed as the change in concentration of a product per unit time, written as \( \Delta [\text{Product}]/\Delta t \).

In our example, we focus on the formation of \( \mathrm{Br}_{2} \), with the rate given as \( 4.8 \times 10^{-6} \ \mathrm{M/s} \). This means that during every second of the reaction, the concentration of \( \mathrm{Br}_{2} \) increases by \( 4.8 \times 10^{-6} \ \mathrm{mol/L} \).

This speed of formation is directly linked to the stoichiometry of the reaction, indicating a proportional change to other substances involved. Understanding the rate of formation is crucial for predicting how quickly a product will appear, which is vital in both laboratory environments and industrial applications.
Rate of Consumption
The rate of consumption is similar to the rate of formation but measures how quickly a reactant is being used up in a chemical reaction. This is expressed as \( \Delta [\text{Reactant}]/\Delta t \) and is viewed as a negative rate since reactant concentration decreases over time.

In the provided reaction, the consumption of \( \mathrm{Br}^{-} \) is given by \( \frac{\Delta [\mathrm{Br}^{-}]}{\Delta t} = 19.2 \times 10^{-6} \ \mathrm{M/s} \). This implies that every second, the concentration of \( \mathrm{Br}^{-} \) decreases by \( 19.2 \times 10^{-6} \ \mathrm{mol/L} \).

Maintaining an understanding of the rate of consumption helps chemists control reactions to ensure they proceed under optimal conditions, avoiding unnecessary waste of reactants and achieving desired product yields.
Balancing Chemical Equations
Balancing chemical equations is the process of ensuring that the number of atoms for each element is equal on both sides of the equation. This reflects the principle of the conservation of mass, which states that matter can neither be created nor destroyed in a chemical reaction.

For our reaction:\[\mathrm{ClO}_{2}^{-}(aq) + 4\mathrm{Br}^{-}(aq) + 4\mathrm{H}^{+}(aq) \rightarrow \mathrm{Cl}^{-}(aq) + 2\mathrm{Br}_{2}(aq) + 2\mathrm{H}_{2} \mathrm{O}(l)\]each molecule of chlorine, bromide, hydrogen, and oxygen must have the same amount as reactants and products.

This balanced state is crucial because it ensures that stoichiometric calculations, such as those for rates of formation and consumption, are accurate. An unbalanced equation could lead to errors, hampering the reliability of kinetic data and any derived conclusions.

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Most popular questions from this chapter

The initial rates listed in the following table were measured for the reaction between nitrogen dioxide and carbon monoxide that occurs in engine exhaust.\(\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$$ \begin{array}{cccc} \hline \text { Experiment } & \text { Initial }\left[\mathrm{NO}_{2}\right] & \text { Initial }[\mathrm{CO}] & \text { Initial Reaction Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.100 & 0.100 & 5.00 \times 10^{-3} \\ 2 & 0.150 & 0.100 & 1.13 \times 10^{-2} \\ 3 & 0.200 & 0.200 & 2.00 \times 10^{-2} \\ \hline \end{array} $$(a) What is the rate law? (b) What is the value of the rate constant and its units? (c) What is the initial rate of consumption of \)\mathrm{NO}_{2}\( when the concentrations of both reactants are \)0.150 \mathrm{M} ?$

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