/*! 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 26 (a) Consider the combustion of e... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\). If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.036 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? (b) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is 74 torr per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

Short Answer

Expert verified
In part (a), the rates of change for \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are both increasing at \(0.072 \;\text{M}/\text{s}\). In part (b), the rate of change for \(\mathrm{NH}_{3}\) partial pressure is increasing at \(148\;\text{torr}/\text{hour}\), while the rate of change of total pressure in the vessel is decreasing at \(222\;\text{torr}/\text{hour}\).

Step by step solution

01

The given reaction is \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g}) + 3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\), and the rate of decrease in the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is \(0.036 \mathrm{M} / \mathrm{s}\). #Step 1: Relate Rates of Change for Each Species to that of \(\mathrm{C}_{2} \mathrm{H}_{4}\)

Using stoichiometry, we can calculate the rates of change for the other species using the coefficients in the balanced chemical equation. Specifically, \[\frac{\Delta[\mathrm{CO}_2]}{\Delta[\mathrm{C}_2\mathrm{H}_4]}=\frac{2}{1}\] and \[\frac{\Delta[\mathrm{H}_2\mathrm{O}]}{\Delta[\mathrm{C}_2\mathrm{H}_4]}=\frac{2}{1}\] #Step 2: Calculate the Rates of Change for \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\)
02

Using the relationships we found in step 1, we can now calculate the rates of change for \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\): \[\Delta[\mathrm{CO}_2]=2 \times \Delta[\mathrm{C}_2\mathrm{H}_4] = 2 \times (-0.036 \;\text{M} / \text{s}) = -0.072 \;\text{M}/\text{s}\] \[\Delta[\mathrm{H}_2\mathrm{O}]=2 \times \Delta[\mathrm{C}_2\mathrm{H}_4] = 2 \times (-0.036 \;\text{M} / \text{s}) = -0.072 \;\text{M}/\text{s}\] Since the rates of change are negative, it indicates that the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are increasing at the rate of \(0.072 \;\text{M}/\text{s}\). #Part (b): Finding Rates of Change for \(\mathrm{NH}_{3}\) Partial Pressure and Total Pressure in the Vessel# #Step 1: Relate Rate of Change for \(\mathrm{NH}_{3}\) to that of \(\mathrm{N}_{2} \mathrm{H}_{4}\)

The given reaction is \(\mathrm{N}_{2} \mathrm{H}_{4}(g) + \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\). The rate of decrease in the partial pressure of \(\mathrm{N}_{2} \mathrm{H}_{4}\) is \(74 \;\text{torr}/\text{hour}\). To calculate the rate of change for \(\mathrm{NH}_{3}\) partial pressure, we'll use stoichiometry and relate it to the rate of decrease of \(\mathrm{N}_{2} \mathrm{H}_{4}\): \[\frac{\Delta P_{\mathrm{NH}_3}}{\Delta P_{\mathrm{N}_2\mathrm{H}_4}}=\frac{2}{1}\] #Step 2: Calculate the Rate of Change for \(\mathrm{NH}_{3}\) Partial Pressure
03

We can now calculate the rate of change for \(\mathrm{NH}_{3}\) partial pressure: \[\Delta P_{\mathrm{NH}_3}=2 \times \Delta P_{\mathrm{N}_2\mathrm{H}_4} = 2 \times (-74 \;\text{torr} / \text{hour}) = -148 \;\text{torr}/\text{hour}\] As \(\Delta P_{\mathrm{NH}_3}\) is negative, it indicates that the partial pressure of \(\mathrm{NH}_{3}\) is increasing at the rate of \(148 \;\text{torr}/\text{hour}\). #Step 3: Calculate the Rate of Change of Total Pressure in the Vessel

To determine the rate of change of total pressure, we can use the changes in partial pressures for each species: \[\Delta P_\text{total} = \Delta P_{\mathrm{N}_2\mathrm{H}_4} + \Delta P_{\mathrm{H}_2} + \Delta P_{\mathrm{NH}_3}\] Since there is no change in \(\mathrm{H}_{2}\) partial pressure, the rate of change of total pressure in the vessel is: \[\Delta P_\text{total} = (-74 \;\text{torr}/\text{hour}) + 0 + (-148\;\text{torr}/\text{hour}) = -222\;\text{torr}/\text{hour}\] Hence, the rates of change for \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) in part (a) are both increasing at \(0.072 \;\text{M}/\text{s}\), while the rates of change in part (b) for the \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel are increasing at \(148\;\text{torr}/\text{hour}\) and decreasing at \(222\;\text{torr}/\text{hour}\), respectively.

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Key Concepts

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

Understanding Stoichiometry in Chemical Reactions
Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction. It involves the use of balanced chemical equations to calculate the amounts of substances consumed and produced. In educational content, the importance of stoichiometry can never be overstated, as it is fundamental to most calculations in chemistry.

Let's consider a reaction where ethylene reacts with oxygen to produce carbon dioxide and water: \[\mathrm{C}_{2}\mathrm{H}_{4}(\mathrm{g}) + 3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)\].Here, the stoichiometric coefficients indicate the proportion at which substances react and are formed. For every mole of ethylene, three moles of oxygen are required, and two moles each of carbon dioxide and water are produced. To grasp the concept fully, students can imagine stoichiometry as a recipe that dictates exactly how much of each ingredient is needed and how much of a product will be made. It's the balancing act that ensures atoms are conserved and reactions go to completion.
Partial Pressure and Its Relation to Reaction Rates
Partial pressure is a measure of the pressure that a single gas in a mixture of gases would exert if it occupied the entire volume on its own. In a chemical reaction, the rates at which partial pressures of reactants decrease and products increase can be crucial for understanding how fast a reaction occurs.

Applying this to an example, the rate at which the partial pressure of \(\mathrm{N}_{2}\mathrm{H}_{4}\) decreases, during its reaction with \(\mathrm{H}_{2}\), demonstrates the dynamic nature of gases within a closed system. The rate of change of the partial pressure for ammonia \(\mathrm{NH}_{3}\) can be determined using stoichiometry, reflecting the balanced equation and providing insights into gas behavior over time. It's essential for students to understand that changes in partial pressure are directly related to the amounts of substances in the gaseous state, and hence, integral to reaction kinetics and stoichiometric calculations.
Reaction Kinetics: The Study of Reaction Rates
Reaction kinetics delves into the speed of chemical reactions and the factors that influence this rate. It encompasses the study of how quickly reactants are converted into products. Reaction rates can be expressed in terms of changes in concentration over time, and these rates are influenced by factors such as temperature, concentration, surface area, and catalysts.

In the context of our example, the rate at which the concentration of ethylene decreases is directly connected to the rates at which carbon dioxide and water are produced. To put it simply for students, reaction kinetics is akin to timing a race; it tells us how fast each 'runner' (reactant or product) is moving toward the finish line (completion of the reaction). Understanding kinetics allows chemists to control reactions more effectively — speeding them up or slowing them down as necessary.
Chemical Equilibrium: A Dynamic Balance
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. However, this does not mean that the reaction has stopped — it is a dynamic equilibrium where reactions continue to occur in both directions at equal rates. Equilibrium can be affected by changes in concentration, temperature, and pressure, as described by Le Chatelier's Principle.

It is essential for students to understand that equilibrium does not signify that reactants and products are present in equal amounts but that their rates of formation are balanced. Chemical equilibrium is the culmination of reaction kinetics and stoichiometry, where the dance of atoms and molecules settles into a steady rhythm. Understanding this concept is vital for predicting the outcomes of reactions and for the manufacturing of chemicals in the industry.

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Most popular questions from this chapter

Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

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The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). Catalysts that convert these species to \(\mathrm{N}_{2}\). are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of \(\mathrm{NO}, \mathrm{NO}_{2}\), and \(\mathrm{N}_{2}\) - (b) Using a resource such as Table \(8.4\), look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2}\), describing what wavelengths of light need to be monitored as a function of time.

Indicate whether each statement is true or false. (a) If you compare two reactions with similar collision factors, the one with the larger activation energy will be faster. (b) A reaction that has a small rate constant must have a small frequency factor. (c) Increasing the reaction temperature increases the fraction of successful collisions between reactants.

Platinum nanoparticles of diameter \(\sim 2 \mathrm{~nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(3.924 \AA\). (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \dot{A}=1 \times 10^{-10} \mathrm{~m}\) and \(1 \mathrm{~nm}=1 \times 10^{-9} \mathrm{~m}\). (b) Estimate how many platinum atoms are on the surface of a \(2.0\)-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(2.8 \mathrm{~A}\). (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a \(5.0\)-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

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