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Relate the rate of decomposition of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) to the rate of formation of \(\mathrm{N}_{2}\) for the following reaction: $$ \mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$

Short Answer

Expert verified
The rate of decomposition of \( \mathrm{NH}_{4} \mathrm{NO}_{2} \) is equal to the rate of formation of \( \mathrm{N}_2 \).

Step by step solution

01

Understand the Reaction

In the given chemical reaction, ammonium nitrite \( \mathrm{NH}_{4} \mathrm{NO}_{2}(aq) \)decomposes to form \( \mathrm{N}_2 (g) \)and water \( \mathrm{H}_2 \mathrm{O} (l) \).The stoichiometry of the reaction is 1:1:2.
02

Define Rate of Decomposition

The rate of decomposition of \( \mathrm{NH}_{4} \mathrm{NO}_{2} \)is given by \(-\frac{d[\mathrm{NH}_{4} \mathrm{NO}_{2}]}{dt}\). This expression represents the decrease in concentration of \( \mathrm{NH}_{4} \mathrm{NO}_{2} \)per unit time.
03

Define Rate of Formation of Products

The rate of formation of \( \mathrm{N}_2 \)is given by \(\frac{d[\mathrm{N}_2]}{dt}\), representing the increase in concentration of \( \mathrm{N}_2 \)per unit time. The rate of formation of \( \mathrm{H}_2 \mathrm{O} \)is \( 2 \times \frac{d[\mathrm{N}_2]}{dt} \) since two moles of water are produced for each mole of \( \mathrm{N}_2 \).
04

Relate the Rates Using Stoichiometry

By analyzing the stoichiometry of the reaction, we observe that for every mole of \( \mathrm{NH}_{4} \mathrm{NO}_{2} \)decomposed, 1 mole of \( \mathrm{N}_2 \)forms. Therefore, we can relate the rates as \(-\frac{d[\mathrm{NH}_{4} \mathrm{NO}_{2}]}{dt} = \frac{d[\mathrm{N}_2]}{dt} \). This relationship indicates that the rate of decomposition of \( \mathrm{NH}_{4} \mathrm{NO}_{2} \)is equal to the rate of formation of \( \mathrm{N}_2 \).

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Key Concepts

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

Stoichiometry in Chemical Reactions
Stoichiometry is an essential concept in chemistry that involves the calculation of reactants and products in chemical reactions. It shows the relationship between the quantities of substances as expressed by a balanced chemical equation. For the reaction: \[ \mathrm{NH}_{4} \mathrm{NO}_{2}(aq) \rightarrow \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l) \], we see that 1 mole of ammonium nitrite produces 1 mole of nitrogen gas and 2 moles of water. This 1:1:2 ratio is crucial for determining the relationship between the rate of decomposition and the formation rate of products. **Key Points**
  • Stoichiometric coefficients (1:1:2) tell us the proportions in which reactants and products participate in the reaction.
  • They help in calculating how much of a reactant is needed or how much of a product will be formed.
  • A balanced equation is crucial, showing that no atoms are lost or gained, just rearranged.
Understanding this, you can relate how quickly a reactant disappears to how quickly products appear, thanks to stoichiometry.
Understanding Concentration Changes
Concentration refers to the amount of a substance in a given volume and is key in understanding how reactions proceed over time. In chemical reactions, the concentration of reactants often decreases while that of products increases. For example, in our given reaction, the concentration of \( \mathrm{NH}_{4} \mathrm{NO}_{2} \) decreases as it decomposes, while \( \mathrm{N}_2 \) increases as it forms. **Elements of Concentration in Reactions:**
  • Expressed in moles per liter (mol/L) or molarity \([\mathrm{M}]\).
  • Affects the reaction rate; higher concentrations typically lead to faster reactions as more molecules are available to collide and react.
  • Represented using notations: [-\frac{d[\text{Reactant}]}{dt}] and [\frac{d[\text{Product}]}{dt}] to show how concentrations change over time.
Keeping track of concentration changes allows chemists to measure how fast a reaction occurs and control it efficiently.
Chemical Reaction Rates
Reaction rates tell us how fast reactants turn into products. In any chemical equation, like \( \mathrm{NH}_{4} \mathrm{NO}_{2}(aq) \longrightarrow \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l) \), the rate can be described using changes in concentration over time, known as reaction rates. These rates are impacted by several factors including temperature, concentration, and the presence of catalysts. **Factors Influencing Reaction Rates:**
  • Concentration: Higher concentrations mean more collisions and a faster rate.
  • Temperature: Higher temperatures increase energy among molecules, leading to more collisions.
  • Catalysts: Substances that speed up reactions without being consumed themselves.
To calculate reaction rates, we use differential expressions such as \(-\frac{d[\mathrm{Reactant}]}{dt}\) and \(\frac{d[\mathrm{Product}]}{dt}\). For our reaction, the rates of decomposition of ammonium nitrite and formation of nitrogen are equal, demonstrating a 1:1 stoichiometric relationship. This equality helps quantify how fast one substance forms or depletes relative to another.

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Most popular questions from this chapter

A reaction of the form \(a \mathrm{~A} \longrightarrow\) Products is secondorder with a rate constant of \(0.725 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})\). If the initial concentration of \(\mathrm{A}\) is \(0.760 \mathrm{~mol} / \mathrm{L},\) what is the molar concentration of A after \(88.8 \mathrm{~s}\) ?

Cyclopropane, \(\mathrm{C}_{3} \mathrm{H}_{6},\) is converted to its isomer propylene, \(\mathrm{CH}_{2}=\mathrm{CHCH}_{3}\), when heated. The rate law is first order in cyclopropane, and the rate constant is \(6.0 \times 10^{-4} / \mathrm{s}\) at \(500^{\circ} \mathrm{C}\). If the initial concentration of \(\mathrm{cy}-\) clopropane is \(0.0226 \mathrm{~mol} / \mathrm{L},\) what is the concentration after \(525 \mathrm{~s}\) ?

The dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) into \(\mathrm{NO}_{2}\), $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ is believed to occur in one step. Obtain the concentration of \(\mathrm{N}_{2} \mathrm{O}_{4}\) in terms of the concentration of \(\mathrm{NO}_{2}\) and the rate constants for the forward and reverse reactions, when the reactions have come to equilibrium.

The hypothetical reaction \(\mathrm{A}+2 \mathrm{~B} \longrightarrow\) Products has the rate law Rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{3}\). If the reaction is run two separate times, holding the concentration of A constant while doubling the concentration of \(\mathrm{B}\) from one run to the next, how would the rate of the second run compare to the rate of the first run?

Azomethane, \(\mathrm{CH}_{3} \mathrm{NNCH}_{3},\) decomposes according to the following equation: $$ \mathrm{CH}_{3} \mathrm{NNCH}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{N}_{2}(g) $$ The initial concentration of azomethane was \(1.50 \times\) \(10^{-2} \mathrm{~mol} / \mathrm{L}\). After \(7.00 \mathrm{~min},\) this concentration decreased to \(1.01 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\). Obtain the average rate of reaction during this time interval. Express the answer in units of \(\mathrm{mol} /(\mathrm{L} \cdot \mathrm{s})\)

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