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Chapter 14: Problem 24

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(\mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g)\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{NH}_{3}(g)\)

Short Answer

Expert verified
Write rate expressions using stoichiometric coefficients: (a) Rate = \(-\frac{d[\mathrm{O}_3]}{dt} = -\frac{d[\mathrm{H}_2\mathrm{O}]}{dt} = \frac{1}{2}\frac{d[\mathrm{O}_2]}{dt} = \frac{d[\mathrm{H}_2]}{dt}\); (b) Rate = \(-\frac{1}{4}\frac{d[\mathrm{NH}_3]}{dt} = -\frac{1}{5}\frac{d[\mathrm{O}_2]}{dt} = \frac{1}{4}\frac{d[\mathrm{NO}]}{dt} = \frac{1}{6}\frac{d[\mathrm{H}_2\mathrm{O}]}{dt}\); (c) Rate = \(-\frac{1}{2}\frac{d[\mathrm{C}_2\mathrm{H}_2]}{dt} = -\frac{1}{5}\frac{d[\mathrm{O}_2]}{dt} = \frac{1}{4}\frac{d[\mathrm{CO}_2]}{dt} = \frac{1}{2}\frac{d[\mathrm{H}_2\mathrm{O}]}{dt}\); (d) Rate = \(-\frac{d[\mathrm{C}_3\mathrm{H}_7\mathrm{NH}_2]}{dt} = \frac{d[\mathrm{C}_3\mathrm{H}_6]}{dt} = \frac{d[\mathrm{NH}_3]}{dt}\).

Step by step solution

01

Understanding Rate Expressions for Reactions

A rate expression relates the rate of reaction to the concentration of reactants and products. It is expressed in terms of the rate of appearance or disappearance of the reactants and products. For a general reaction \( aA + bB \rightarrow cC + dD \), the rate can be expressed as:\[\text{-Rate} = \frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}\]
02

Writing Rate Expressions for Reaction (a)

In the reaction \( \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) \), the rate expression is given by:\[\text{-Rate} = -\frac{d[\mathrm{O}_{3}]}{dt} = -\frac{d[\mathrm{H}_{2} \mathrm{O}]}{dt} = \frac{1}{2}\frac{d[\mathrm{O}_{2}]}{dt} = \frac{d[\mathrm{H}_{2}]}{dt}\]This indicates that the rate of disappearance of \( \mathrm{O}_{3} \) and \( \mathrm{H}_{2}\mathrm{O} \) is equal, and the rate of appearance of \( \mathrm{O}_{2} \) is half the rate of disappearance of \( \mathrm{O}_{3} \).
03

Writing Rate Expressions for Reaction (b)

For the reaction \( 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \), the rate expression is given by:\[\text{-Rate} = -\frac{1}{4} \frac{d[ \mathrm{NH}_{3} ]}{dt} = -\frac{1}{5} \frac{d[ \mathrm{O}_{2} ]}{dt} = \frac{1}{4} \frac{d[ \mathrm{NO} ]}{dt} = \frac{1}{6} \frac{d[ \mathrm{H}_{2} \mathrm{O} ]}{dt}\]This indicates how each species contributes to the overall reaction rate.
04

Writing Rate Expressions for Reaction (c)

In the reaction \( 2 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \), the rate is expressed as:\[\text{-Rate} = -\frac{1}{2} \frac{d[ \mathrm{C}_{2} \mathrm{H}_{2} ]}{dt} = -\frac{1}{5} \frac{d[ \mathrm{O}_{2} ]}{dt} = \frac{1}{4} \frac{d[ \mathrm{CO}_{2} ]}{dt} = \frac{1}{2} \frac{d[ \mathrm{H}_{2} \mathrm{O} ]}{dt}\]This uses stoichiometric coefficients to balance the rate for each molecule in the equation.
05

Writing Rate Expressions for Reaction (d)

For the reaction \( \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{NH}_{3}(g) \), the rate can be written as:\[\text{-Rate} = -\frac{d[ \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} ]}{dt} = \frac{d[ \mathrm{C}_{3} \mathrm{H}_{6} ]}{dt} = \frac{d[ \mathrm{NH}_{3} ]}{dt}\]Since each reactant and product appears in a 1:1 ratio, their rate expressions are directly related.

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Key Concepts

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

Understanding Reaction Kinetics
Reaction kinetics is the branch of chemistry that deals with the rates at which chemical reactions occur. It provides valuable insights into how a reaction progresses and helps in understanding the dynamics of reactants transforming into products. This area of study is crucial because it informs how quickly products appear and reactants disappear, which can be important in various sectors such as pharmaceuticals, manufacturing, and environmental science.

To study reaction kinetics, scientists often focus on several key aspects:
  • Rate Expressions: These are mathematical formulas that describe the speed of a reaction. Rate expressions are derived from the stoichiometry of the reaction and reflect how the concentration of reactants and products changes over time.
  • Order of Reaction: The order defines the relationship between the concentration of reactants and the rate of reaction. It is determined experimentally and usually indicates how the rate changes with varying concentrations.
  • Activation Energy: This is the minimum energy needed for a reaction to proceed. Understanding activation energy is vital for controlling reactions in industrial processes.
Reaction kinetics is foundational in designing chemical processes and optimizing conditions for desired reaction outcomes.
Defining the Rate of Reaction
The rate of reaction, simply put, is a measure of how fast or slow a chemical reaction takes place. It is typically expressed in terms of molarity per second (M/s), indicating how the concentration of a reactant or product changes over time.

In formulating rate expressions, we use the stoichiometric coefficients from the balanced chemical equation. These coefficients help relate the rate at which each species in the reaction grows or diminishes.
  • Rate of Reaction Formula: For a generic reaction formula like \( aA + bB \rightarrow cC + dD \), the rate expression would be: \[ -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt} \]This equation states that the rate of decrease of reactants \([A]\) and \([B]\) is proportional to the rate of increase of products \([C]\) and \([D]\).
  • Interpretation of Rates: A high rate implies a rapid reaction where reactants quickly transform into products. Conversely, a low rate suggests a slow reaction.
Understanding the rate of reaction allows chemists to control and predict the speed of chemical processes.
The Basics of Chemical Reactions
Chemical reactions involve the transformation of substances, called reactants, into different substances known as products. These transformations occur through the breaking and forming of chemical bonds, leading to the rearrangement of atoms.

Understanding the nature of chemical reactions is fundamental in chemistry as it provides insights into how substances interact to form new compounds. This knowledge is essential for developing new materials, understanding biological processes, and improving industrial chemical processes.
  • Types of Chemical Reactions: Reactions are classified based on how reactants are converted to products, such as synthesis, decomposition, single-replacement, and double-replacement reactions.
  • Reaction Conditions: Various factors can influence a reaction's progress, including temperature, pressure, and the presence of catalysts. These conditions can alter the speed of the reaction and sometimes the products formed.
  • Conservation of Mass: In all chemical reactions, the mass of the reactants is equal to the mass of the products, adhering to the law of conservation of mass.
Chemical reactions are the building blocks of chemical knowledge, paving the way for innovative solutions in multiple scientific fields.

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Most popular questions from this chapter

Draw a graph showing the reaction pathway for an overall exothermic reaction with two intermediates that are produced at different rates. On your graph indicate the reactants, products, intermediates, transition states, and activation energies. [Sections 14.6 and 14.7\(]\)

(a) Consider the combustion of hydrogen, \(2 \mathrm{H}_{2}(g)+\) \(\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) .\) If hydrogen is burning at a rate of \(0.5 \mathrm{~mol} / \mathrm{s},\) what is the rate of consumption of oxygen? What is the rate of formation of water vapor? (b) The reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of \(60 \mathrm{kPa} / \mathrm{min},\) what is the rate of change of the total pressure of the vessel?

Urea (NH_2CONH \(_{2}\) ) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{MHCl}\) occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M} ?\) (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C} ?\)

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\) The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l} \text { Step } 1: A+B \longrightarrow C+X \\ \text { Step } 2: A+X \longrightarrow C+D \end{array} $$ \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(30^{\circ} \mathrm{C}\) is \(4.0 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.5 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2}\) is \(2.0 \times 10^{-2} \mathrm{M},\) what is the rate of formation of \(\mathrm{H}^{+} ?\)
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