/*! 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 35 The decomposition of dinitrogen ... [FREE SOLUTION] | 91Ó°ÊÓ

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The decomposition of dinitrogen pentoxide is described by the chemical equation \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) If the rate of appearance of \(\mathrm{NO}_{2}\) is equal to \(0.560\) \(\mathrm{mol} / \mathrm{min}\) at a particular moment, what is the rate of appearance of \(\mathrm{O}_{2}\) at that moment? a. \(2.70 \mathrm{~mol} / \mathrm{min}\) b. \(3.40 \mathrm{~mol} / \mathrm{min}\) c. \(0.140 \mathrm{~mol} / \mathrm{min}\) d. \(1.14 \mathrm{~mol} / \mathrm{min}\)

Short Answer

Expert verified
The rate of appearance of \( \mathrm{O}_{2} \) is \( 0.140 \; \mathrm{mol/min} \), answer c.

Step by step solution

01

Analyze the Chemical Equation

The balanced chemical equation given is: \[ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2} + \mathrm{O}_{2} \] This equation shows that for every 2 moles of \( \mathrm{N}_{2} \mathrm{O}_{5} \) decomposed, 4 moles of \( \mathrm{NO}_{2} \) and 1 mole of \( \mathrm{O}_{2} \) are produced. We need to find the relationship between the rates of formation of \( \mathrm{NO}_2 \) and \( \mathrm{O}_2 \).
02

Determine Rate Relationships

From the balanced equation, the stoichiometry shows that 4 moles of \( \mathrm{NO}_{2} \) are produced for every mole of \( \mathrm{O}_{2} \). The rate of appearance of a substance is proportional to its stoichiometric coefficient in the balanced reaction.
03

Calculate the Rate of Appearance of O2

Given that the rate of appearance of \( \mathrm{NO}_{2} \) is \( 0.560 \; \mathrm{mol/min} \), we use stoichiometry to find the rate of \( \mathrm{O}_{2} \). Since the molar ratio of \( \mathrm{NO}_{2} \) to \( \mathrm{O}_{2} \) is 4:1, the rate of \( \mathrm{O}_{2} \) will be a quarter of \( 0.560 \; \mathrm{mol/min} \):\[ \frac{0.560 \; \mathrm{mol/min}}{4} = 0.140 \; \mathrm{mol/min} \]
04

Compare Answer to Choices

The calculated rate of appearance of \( \mathrm{O}_{2} \) is \( 0.140 \; \mathrm{mol/min} \). Among the provided options, option c corresponds to \( 0.140 \; \mathrm{mol/min} \). Therefore, the correct answer is c.

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Key Concepts

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

Rate of Reaction
Understanding the rate of reaction is crucial in chemical kinetics. It essentially indicates how quickly or slowly a reaction proceeds. The rate of reaction can be determined by measuring the change in concentration of a reactant or product over time. In the case of the decomposition of dinitrogen pentoxide into nitrogen dioxide and oxygen, the rate of reaction can be measured in terms of the appearance of the products. For example:
  • Rate of appearance of \( \mathrm{NO}_{2} \)
  • Rate of appearance of \( \mathrm{O}_{2} \)
If you know the rate at which one substance appears, you can calculate the rates of other substances in the reaction using stoichiometric relationships. This provides insight into how different species in the reaction are consumed or produced at different rates.
Stoichiometry
Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation, which provides the ratio of moles of each substance involved in the reaction. This is crucial for understanding the quantitative relationships in a chemical process.
For example, in the balanced decomposition reaction:\[ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2} + \mathrm{O}_{2} \]From this, you can see the stoichiometric relationships:
  • 2 moles of \( \mathrm{N}_{2} \mathrm{O}_{5} \) produce 4 moles of \( \mathrm{NO}_{2} \)
  • 2 moles of \( \mathrm{N}_{2} \mathrm{O}_{5} \) produce 1 mole of \( \mathrm{O}_{2} \)
This ratio allows us to determine the proportional rates at which products and reactants change during the reaction, making stoichiometry invaluable for chemical calculations.
Decomposition Reaction
Decomposition reactions are those in which a compound breaks down into two or more simpler substances. These reactions are common in chemistry and are often used for analysis and synthesis. The reaction's general form is:\[ AB \rightarrow A + B \]In the decomposition of dinitrogen pentoxide \( \mathrm{N}_{2} \mathrm{O}_{5} \), it breaks down to form nitrogen dioxide \( \mathrm{NO}_{2} \) and oxygen \( \mathrm{O}_{2} \). This process involves breaking the bonds in \( \mathrm{N}_{2} \mathrm{O}_{5} \), leading to smaller molecules. Decomposition reactions are crucial in processes like:
  • Manufacturing chemicals
  • Recycling waste
  • Analyzing substances
Understanding such reactions is essential for manipulating chemical processes.
Balanced Chemical Equation
A balanced chemical equation is crucial for accurately describing a chemical reaction, ensuring that the law of conservation of mass is respected. This means that the number of each type of atom on the reactant side matches the number on the product side. For instance, in the reaction:\[ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2} + \mathrm{O}_{2} \]Each element is balanced, maintaining the principle that matter cannot be created or destroyed in a chemical reaction. Balancing chemical equations is vital because:
  • It provides the stoichiometric coefficients needed to calculate reactant/product amounts
  • Helps in identifying limiting reactants
  • Ensures the reaction can be understood and utilized efficiently
Mastering this skill is important for anyone studying chemistry, as it applies to virtually all chemical analyses and reactions.

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Most popular questions from this chapter

The equation of tris( 1,10 -phenanthroline) iron(II) in acid solution takes place according to the equation:\(\mathrm{Fe}(\text { phen })_{3}^{2+}+3 \mathrm{H}_{3} \mathrm{O}^{+}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\) \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}+3\) (phen) \(\mathrm{H}^{+}\) If the activation energy (Ea) is \(126 \mathrm{~kJ} / \mathrm{mol}\) and the rate constant at \(30^{\circ} \mathrm{C}\) is \(9.8 \times 10^{-3} \mathrm{~min}^{-1}\), what is the rate constant at \(50^{\circ} \mathrm{C}\) ? a. \(2.2 \times 10^{-1} \mathrm{~min}^{-1}\) b. \(3.4 \times 10^{-2} \mathrm{~min}^{-1}\) c. \(0.23 \times 10^{-1} \mathrm{~min}^{-1}\) d. \(1.2 \times 10^{-1} \mathrm{~min}^{-1}\)

If ' \(\mathrm{I}\) ' is the intensity of absorbed light and ' \(\mathrm{C}\) ' is the concentration of \(A B\) for the photochemical process \(\mathrm{AB}+\mathrm{hv} \rightarrow \mathrm{AB}^{*}\), the rate of formation of \(\mathrm{AB}^{*}\) is directly proportional to a. \(\mathrm{C}\) b. I c. \(\mathrm{I}^{2}\) d. C.I.

\text { For a first order reaction, }a. The degree of dissociation is equal to \(\left(1-\mathrm{e}^{-\mathrm{k} 1}\right)\) b. The pre-exponential factor in the Arrhenius equation has the dimensions of time \(\mathrm{T}^{-1}\). c. The time taken for the completion of \(75 \%\) reaction is thrice the \(t \frac{1}{2}\) of the reaction. d. both (a) and (b)

At \(380^{\circ} \mathrm{C}\), half life period for the first order decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(360 \mathrm{~min}\). The energy of activation of the reaction is \(200 \mathrm{~kJ} \mathrm{~mol}^{-1} .\) Calculate the time required for \(75 \%\) decomposition at \(450^{\circ} \mathrm{C}\) if half life for decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(10.17 \mathrm{~min}\) at \(450^{\circ} \mathrm{C}\). a. \(20.4 \mathrm{~min}\) b. \(408 \mathrm{~min}\) c. \(10.2 \mathrm{~min}\) d. none

The decomposition of hydrogen peroxide is given by the following reaction: \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})\) In the presence of KI the reaction is thought to occur by the following mechanism: Step \(\mathrm{I}: \mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{I}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{IO}^{-}\) Step II: \(\mathrm{IO}^{-}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+\mathrm{I}^{-}\) What is the role of \(\mathrm{I}^{-}\)in this mechanism? a. Intermediate b. Transition state c. Catalyst d. Frequency factor

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