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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: \(\begin{array}{l}{\text { (a) } 2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)} \\ {\text { (b) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)} \\\ {\text { (c) } 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)}\end{array}\)

Short Answer

Expert verified
\(a)\) Rate = -\(\frac{1}{2}\) × (rate of disappearance of \(H_2O\)) = \(\frac{1}{2}\) × (rate of appearance of \(H_2\)) = (rate of appearance of \(O_2\)) \(b)\) Rate = -\(\frac{1}{2}\) × (rate of disappearance of \(SO_2\)) = - (rate of disappearance of \(O_2\)) = \(\frac{1}{2}\) × (rate of appearance of \(SO_3\)) \(c)\) Rate = -\(\frac{1}{2}\) × (rate of disappearance of \(NO\)) = -\(\frac{1}{2}\) × (rate of disappearance of \(H_2\)) = (rate of appearance of \(N_2\)) = \(\frac{1}{2}\) × (rate of appearance of \(H_2O\)) \(d)\) Rate = - (rate of disappearance of \(N_2\)) = -\(\frac{1}{2}\) × (rate of disappearance of \(H_2\)) = (rate of appearance of \(N_2H_4\))

Step by step solution

01

(a) Reaction and rate expression for H2O -> H2 + O2

Given reaction: \(2 H_2O(g) \longrightarrow 2 H_2(g) + O_2(g)\) Rate expression: Rate = -\(\frac{1}{2}\) × (rate of disappearance of \(H_2O\)) = \(\frac{1}{2}\) × (rate of appearance of \(H_2\)) = (rate of appearance of \(O_2\))
02

(b) Reaction and rate expression for SO2 + O2 -> SO3

Given reaction: \(2 SO_2(g) + O_2(g) \longrightarrow 2 SO_3(g)\) Rate expression: Rate = -\(\frac{1}{2}\) × (rate of disappearance of \(SO_2\)) = - (rate of disappearance of \(O_2\)) = \(\frac{1}{2}\) × (rate of appearance of \(SO_3\))
03

(c) Reaction and rate expression for NO + H2 -> N2 + H2O

Given reaction: \(2 NO(g) + 2 H_2(g) \longrightarrow N_2(g) + 2 H_2O(g)\) Rate expression: Rate = -\(\frac{1}{2}\) × (rate of disappearance of \(NO\)) = -\(\frac{1}{2}\) × (rate of disappearance of \(H_2\)) = (rate of appearance of \(N_2\)) = \(\frac{1}{2}\) × (rate of appearance of \(H_2O\))
04

(d) Reaction and rate expression for N2 + H2 -> N2H4

Given reaction: \(N_2(g) + 2 H_2(g) \longrightarrow N_2H_4(g)\) Rate expression: Rate = - (rate of disappearance of \(N_2\)) = -\(\frac{1}{2}\) × (rate of disappearance of \(H_2\)) = (rate of appearance of \(N_2H_4\))

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Key Concepts

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

Understanding Reaction Rates
Reaction rates describe how quickly a reaction occurs in a chemical process. This involves both the disappearance of reactants and the appearance of products over time. For example, if you have a gas-phase reaction like \[2 \mathrm{H}_2\mathrm{O}(g) \longrightarrow 2 \mathrm{H}_2(g) + \mathrm{O}_2(g)\]you can express the rate as the change in concentration of hydrogen peroxide ( \mathrm{H}_2\mathrm{O} ) over time. The rate is often represented by negative values when describing reactants because their concentrations decrease. For products, the rate is positive.
  • Rate of disappearance is negative and shows how quickly reactants are consumed.
  • Rate of appearance is positive and reflects how fast products are formed.
This foundational understanding helps to establish rate expressions in reactions, ensuring precise scientific analysis of chemical reactions.
The Role of Gas-Phase Reactions
Gas-phase reactions occur entirely in the gaseous state and involve reactants and products that are gases. These reactions are significant because of their applications in industrial processes, pollution control, and natural phenomena like weather changes. For instance, in the reaction \[2 \mathrm{SO}_2(g) + \mathrm{O}_2(g) \longrightarrow 2 \mathrm{SO}_3(g)\]both reactants \mathrm{SO}_2 and \mathrm{O}_2 are gases, forming gaseous \mathrm{SO}_3 . Some characteristics of gas-phase reactions include:
  • Fast Mixing: Gases mix rapidly due to high kinetic energy.
  • Pressure and Temperature Influence: These factors can greatly affect the reaction rate and equilibrium.
By considering these factors, scientists can control reaction environments for desired outcomes, like maximizing product yield or minimizing harmful emissions.
Exploring Chemical Kinetics
Chemical kinetics is the study of reaction rates and the steps occurring during chemical reactions. It examines the factors affecting rates, such as concentration, temperature, and presence of catalysts. Consider the reaction\[2 \mathrm{NO}(g) + 2 \mathrm{H}_2(g) \longrightarrow \mathrm{N}_2(g) + 2 \mathrm{H}_2\mathrm{O}(g)\] Chemical kinetics helps to understand how fast nitrogen \mathrm{N}_2 forms and how the concentration of \mathrm{NO} and \mathrm{H}_2 changes over time. Key aspects of chemical kinetics include:
  • Rate Laws: Mathematical relationships that describe how rate depends on concentration.
  • Reaction Mechanisms: Detailed pathways showing individual steps of reactions.
Mastering these concepts helps chemists to predict reaction behavior and optimize processes for practical applications.
The Importance of Stoichiometry
Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It ensures substances are combined in precise ratios for reactions to proceed properly. In the reaction \[\mathrm{N}_2(g) + 2 \mathrm{H}_2(g) \longrightarrow \mathrm{N}_2\mathrm{H}_4(g)\] stoichiometry ensures that two molecules of hydrogen \mathrm{H}_2 react with one molecule of nitrogen \mathrm{N}_2 to form hydrazine \mathrm{N}_2\mathrm{H}_4 .Some benefits of understanding stoichiometry include:
  • Accurate Calculations: Determines exact amounts of reactants needed and predicts product yields.
  • Balancing Equations: Vital for understanding chemical reactions and ensuring mass conservation.
With a grasp of stoichiometry, students and professionals can efficiently plan and execute chemical experiments while predicting their outcomes effectively.

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Most popular questions from this chapter

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

A reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) obeys the following rate law: Rate \(=k[\mathrm{B}]^{2}\) . (a) If [A] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B} ?\) What is the overall reaction order? (c) What are the units of the rate constant?

You have studied the gas-phase oxidation of HBr by \(\mathrm{O}_{2}\) : $$4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g)$$ You find the reaction to be first order with respect to HBr and first order with respect to \(\mathrm{O}_{2}\) . You propose the following mechanism: $$ \begin{array}{c}{\operatorname{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{HOOBr}(g)} \\ {\operatorname{HOOBr}(g)+\operatorname{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)} \\\ {\operatorname{HOBr}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)}\end{array}$$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

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 \(28^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1}\) . If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2}\) is \(0.025 \mathrm{M},\) what is the rate of formation of \(\mathrm{Cl}^{-} ?\)

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: \(\begin{array}{l}{\text { (a) } \mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)} \\ {\text { (b) } 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)} \\ {\text { (c) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {\text { (d) } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{NH}_{3}(g)}\end{array}\)

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