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Write the reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products: $$ \begin{array}{l} \text { (a) } \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g) \\ \text { (b) } 5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \\ \quad 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$

Short Answer

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(a) Rate = \(-\frac{d[\mathrm{H}_2]}{dt} = -\frac{d[\mathrm{I}_2]}{dt} = \frac{1}{2}\frac{d[\mathrm{HI}]}{dt}\) (b) Rate = \(-\frac{1}{5}\frac{d[\mathrm{Br}^-]}{dt} = -\frac{1}{1}\frac{d[\mathrm{BrO}_3^-]}{dt} = -\frac{1}{6}\frac{d[\mathrm{H}^+]}{dt} = \frac{1}{3}\frac{d[\mathrm{Br}_2]}{dt}\)

Step by step solution

01

Understand Reaction Rates

The rate of a chemical reaction can be described in terms of the disappearance of reactants or the appearance of products. For the reaction \( aA + bB \rightarrow cC + dD \), the rate can be expressed for each component with respect to time. It's important to note that the rate at which a reactant disappears is negative, whereas the rate of product formation is positive.
02

Reaction Rate Expression for Reaction (a)

The reaction is \( \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g) \). For this reaction: \[ \text{Rate} = -\frac{1}{1} \frac{d[\mathrm{H}_2]}{dt} = -\frac{1}{1} \frac{d[\mathrm{I}_2]}{dt} = \frac{1}{2} \frac{d[\mathrm{HI}]}{dt} \] Each term equates because the stoichiometry of the reaction governs the proportionate rate of change for reactants and products.
03

Reaction Rate Expression for Reaction (b)

The reaction is \( 5 \mathrm{Br}^{-}(aq) + \mathrm{BrO}_{3}^{-}(aq) + 6 \mathrm{H}^{+}(aq) \rightarrow 3 \mathrm{Br}_{2}(aq) + 3 \mathrm{H}_{2} \mathrm{O}(l) \). The expression for the reaction rates is: \[ \text{Rate} = -\frac{1}{5} \frac{d[\mathrm{Br}^-]}{dt} = -\frac{1}{1} \frac{d[\mathrm{BrO}_3^-]}{dt} = -\frac{1}{6} \frac{d[\mathrm{H}^+]}{dt} = \frac{1}{3} \frac{d[\mathrm{Br}_2]}{dt} \]The water is not included in the rate expression since it is a liquid in this reaction.

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Key Concepts

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

Disappearance of Reactants
In a chemical reaction, reactants are substances that start a reaction and are consumed as the reaction progresses. By monitoring the disappearance of these reactants, we gain insight into how fast a reaction is proceeding. This is typically expressed as the rate of decrease in concentration of the reactants over time. Mathematically, this can be represented as a rate of change, specifically:
  • For a reactant A, the rate of disappearance is \(-\frac{d[A]}{dt}\), where the negative sign highlights that the quantity is decreasing.
This calculation helps to quantify the speed at which reactants are disappearing, allowing us to understand the kinetics of the reaction. For example, in the reaction of hydrogen and iodine to form hydrogen iodide, both hydrogen (\(H_2\) ) and iodine (\(I_2\) ) are disappearing over time, and this disappearance is a crucial part of determining the reaction rate.
Appearance of Products
As reactants disappear in a chemical reaction, products are formed. This formation or 'appearance' of products can also be described in terms of rate. It is common to monitor how quickly products form by observing the change in concentration over time. The rate of appearance is represented as:
  • For a product C, the rate of appearance is \(\frac{d[C]}{dt}\), and it is always positive since the concentration of a product is increasing as the reaction proceeds.
This relationship is key to understanding the efficiency and progress of chemical reactions. For instance, in a reaction forming hydrogen iodide (\(HI\) ), the concentration of \(HI\) increases as \(H_2\) and \(I_2\) react. Understanding and calculating the rate at which products form is essential for controlling reaction conditions and ensuring desired outcomes.
Chemical Reaction Rates
The rate of a chemical reaction is a measure of how quickly a reaction occurs. It involves both the disappearance of reactants and the appearance of products. Reaction rates can be affected by several factors, including temperature, concentration, and the presence of catalysts.
  • Reaction rate expressions capture these changes quantitatively and help compare different reactions.
For any reaction like \(aA + bB \rightarrow cC + dD\), the rate expression is based on stoichiometry, providing a relation between the change in concentration of reactants and products over time. The beauty of this expression lies in its simplicity and insight into how different reaction components relate to each other in the reaction equation.
Stoichiometry
Stoichiometry involves the relative quantities of reactants and products in a chemical reaction. When writing rate expressions, stoichiometric coefficients from the balanced equation play a crucial role. Each term in the rate expression involves dividing by these stoichiometric coefficients to account for the proportional contribution of each species to the overall reaction rate.
  • For instance, if 3 moles of \(Br_2\) are formed for every 5 moles of \(Br^-\) consumed in a reaction, these coefficients are integral to establishing a proper rate relation.
Stoichiometry ensures that the rate expressions correctly reflect the balanced chemical equation, making this concept indispensable for accurate reaction analysis and understanding.
Rate of Change with Time
The rate of change with time is a critical aspect of reaction kinetics, reflecting how concentrations of reactants and products change as the reaction progresses. This rate can be calculated as a time derivative, helping us capture the speed of reaction phases.
  • The expressions \(\frac{d[A]}{dt}\) or \(\frac{d[C]}{dt}\) are derivatives representing this temporal change.
In many reactions, keeping track of these rates is essential for predicting product yields, optimizing reactions, and controlling reactor conditions. By understanding how fast different components of a reaction system change over time, chemists can effectively manage reactions to achieve desired outcomes in both laboratory and industrial settings.

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Most popular questions from this chapter

Consider the reaction: $$ \mathrm{A}+\mathrm{B} \longrightarrow \text { products } $$ From the following data obtained a a certain temperature, determine the order of the reaction and calculate the rate constant. $$ \begin{array}{ccc} {[\mathbf{A}](\boldsymbol{M})} & {[\mathbf{B}](\boldsymbol{M})} & \text { Rate }(\boldsymbol{M} / \mathbf{s}) \\ \hline 1.50 & 1.50 & 3.20 \times 10^{-1} \\ 1.50 & 2.50 & 3.20 \times 10^{-1} \\ 3.00 & 1.50 & 6.40 \times 10^{-1} \end{array} $$

A flask contains a mixture of compounds \(\mathrm{A}\) and \(\mathrm{B}\). Both compounds decompose by first-order kinetics. The half-lives are 50.0 min for \(\mathrm{A}\) and 18.0 min for \(\mathrm{B}\). If the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are equal initially, how long will it take for the concentration of \(\mathrm{A}\) to be four times that of \(\mathrm{B}\) ?

Polyethylene is used in many items, including water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a polymer, a molecule with a very high molar mass made by joining many ethylene molecules together. (Ethylene is the basic unit, or monomer, for polyethylene.) The initiation step is: \(\mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot\) (initiation) The \(\mathrm{R}\). species (called a radical) reacts with an ethylene molecule (M) to generate another radical: $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} $$ The reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: \(\mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad\) (propagation) This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine: $$ \mathbf{M}^{\prime} \cdot+\mathbf{M}^{\prime \prime} \cdot \stackrel{k_{\mathrm{t}}}{\longrightarrow} \mathbf{M}^{\prime}-\mathbf{M}^{\prime \prime} \quad \text { (termination) } $$ The initiator frequently used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]\) : $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} $$ This is a first-order reaction. The half-life of benzoyl peroxide at \(100^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~min}\). (a) Calculate the rate constant (in \(\min ^{-1}\) ) of the reaction. (b) If the half-life of benzoyl peroxide is \(7.30 \mathrm{~h}\), or \(438 \mathrm{~min}\), at \(70^{\circ} \mathrm{C},\) what is the activation energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) for the decomposition of benzoyl peroxide? (c) Write the rate laws for the elementary steps in the preceding polymerization process, and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long, high-molar-mass polyethylenes?

At \(25^{\circ} \mathrm{C}\), the rate constant for the ozone-depleting reaction: $$ \mathrm{O}(g)+\mathrm{O}_{3}(g) \longrightarrow 2 \mathrm{O}_{2}(g) $$ is \(7.9 \times 10^{-15} \mathrm{~cm}^{3} / \mathrm{molecule} \cdot \mathrm{s}\). Express the rate constant in units of \(1 / M \cdot \mathrm{s}\).

The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent \(\left(\mathrm{CCl}_{4}\right)\) at a certain temperature: $$ \begin{array}{cc} 2 \mathrm{~N}_{2} \mathrm{O}_{5} & \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} \\ {\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 0.92 & 0.95 \times 10^{-5} \\ 1.23 & 1.20 \times 10^{-5} \\ 1.79 & 1.93 \times 10^{-5} \\ 2.00 & 2.10 \times 10^{-5} \\ 2.21 & 2.26 \times 10^{-5} \end{array} $$ Determine graphically the rate law for the reaction, and calculate the rate constant.

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