Problem 24 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

\(a)\) 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 1: (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 \(O_2\))

Step 2: (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 \(SO_3\))

Step 3: (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 \(H_2O\))

Step 4: (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 \(N_2H_4\))

Key Concepts

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.