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Which oxide of nitrogen is the most stable? (1) \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{O}(\mathrm{g}) \mathrm{k}=6.7 \times 10^{6} \mathrm{~mol}\) litre (2) \(2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \mathrm{k}=2.2 \times 10^{30} \mathrm{~mol}\) litre (3) \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \mathrm{k}=1.2 \times 10^{34}\) mol litre (4) \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \mathrm{k}=3.5 \times 10^{23}\) mol litre

Short Answer

Expert verified
NO2 is the most stable oxide of nitrogen.

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01

Understanding the Concept of Stability

The stability of an oxide of nitrogen can be determined by the equilibrium constant (k). A lower equilibrium constant (k) suggests greater stability because it indicates that the equilibrium lies more towards the reactants, meaning that less product is being formed.
02

Identifying the Equilibrium Constants

For the given reactions, identify the equilibrium constants:1. For the reaction: 2 NO2(g) ⇌ N2(g) + 2 O(g), k = 6.7 × 10^6 mol/litre2. For the reaction: 2 NO(g) ⇌ N2(g) + O2(g), k = 2.2 × 10^30 mol/litre3. For the reaction: 2 N2O5(g) ⇌ N2(g) + 5 O2(g), k = 1.2 × 10^34 mol/litre4. For the reaction: 2 N2O(g) ⇌ 2 N2(g) + O2(g), k = 3.5 × 10^23 mol/litre
03

Comparing the Equilibrium Constants

Compare the values of the equilibrium constants. The reaction with the lowest value of k will correspond to the most stable oxide of nitrogen: k(NO2) = 6.7 × 10^6 mol/litrek(NO) = 2.2 × 10^30 mol/litrek(N2O5) = 1.2 × 10^34 mol/litrek(N2O) = 3.5 × 10^23 mol/litre
04

Determining the Most Stable Oxide

From the comparison, the equilibrium constant for the dissociation of NO2 (k = 6.7 × 10^6 mol/litre) is the lowest. Therefore, NO2 is the most stable oxide among the options provided.

Key Concepts

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

Equilibrium Constant
The equilibrium constant, represented as (k), plays a crucial role in understanding the stability of chemical compounds. It reflects the ratio of the concentration of products to reactants at equilibrium for a given reaction. When we talk about oxides of nitrogen, the value of (k) helps determine which oxide is more stable.

A lower equilibrium constant implies that the reaction does not proceed as extensively towards the products. This means more reactants remain unreacted, indicating higher stability of the compound. Conversely, a higher (k) value indicates that more reactants convert into products, suggesting lower stability.

For instance, in the provided reactions, the equilibrium constants are used to compare the stability of different nitrogen oxides. The oxide with the smallest equilibrium constant is the most stable because it produces the least amount of decomposition products.
Chemical Stability
Chemical stability refers to a substance's ability to resist chemical change. In the context of nitrogen oxides, it means how likely the oxide is to decompose into nitrogen and oxygen gases under given conditions.

Stability can be influenced by various factors, including temperature, pressure, and the nature of the chemical bonds within the compound. A compound with stronger bonds between its atoms requires higher energy to break these bonds, thus making it more stable.

For nitrogen oxides, this stability is reflected in their equilibrium constants. A lower equilibrium constant indicates that a smaller proportion of the oxide decomposes, making it chemically more stable. In the provided exercise, NO2 has the lowest equilibrium constant, highlighting its greater stability compared to other nitrogen oxides like NO, N2O5, and N2O.
Nitrogen Oxides
Nitrogen oxides are a group of gases composed of nitrogen and oxygen. They are significant both in environmental contexts and in industrial applications. Some common nitrogen oxides include NO (Nitric Oxide), NO2 (Nitrogen Dioxide), N2O (Nitrous Oxide), and N2O5 (Dinitrogen Pentoxide).

These compounds differ in their chemical composition, properties, and stability. For example, NO2 is a reddish-brown gas with a sharp, biting odor and is a major air pollutant. In contrast, N2O, also known as laughing gas, is a colorless, non-flammable gas with a slightly sweet odor and is used as an anesthetic.

Understanding the stability of these compounds is essential, especially when predicting their behavior in various chemical reactions and environmental impacts. By analyzing equilibrium constants, we can determine which of these oxides are more stable and thereby understand their potential reactivity and longevity in different contexts.

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Most popular questions from this chapter

For which reaction high pressure and high temperature is helpful in obtaining a high cquilibrium yicld? (1) \(2 \mathrm{NF}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \quad 54.40 \mathrm{kcal}\) (2) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{II}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NI}_{3}(\mathrm{~g})+22.08 \mathrm{kcal}\) (3) \(\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClO}_{3}(\mathrm{~g}) \quad 49.4 \mathrm{kcal}\) (4) \(2 \mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}_{2}(\mathrm{~g}) \times 7 \mathrm{O}_{2}(\mathrm{~g})+126.8 \mathrm{kcal}\)

When \(\mathrm{CaCO}_{3}\) is heated at a constant temperature in a closed container, the pressure due to \(\mathrm{CO}_{2}\) produced will (1) change with the amount of \(\mathrm{CaCO}_{3}\) taken (2) change with the size of the container (3) remain constant so long as the temperature is constant (4) remain constant even if the temperature is changea

The \(10^{4} \mathrm{Ka}\) values for the acids acetic, hydrofluoric, formic and nitrous are \(6.7,4.5,1.8\) and \(0.18\) but not in the correct order. The correct acid strengths arc (1) \(\mathrm{HF}=0.18, \mathrm{HNO}_{2}=1.8, \mathrm{HCOOH}=4.5\), \(\mathrm{CH}_{3} \mathrm{COOH}=6.7\) (2) \(\mathrm{HF}=6.7, \mathrm{HNO}_{2}=4.5, \mathrm{HCOOH}=1.8, \mathrm{CH}_{3} \mathrm{COOH}\) \(=0.18\) (3) \(\mathrm{HF}=1.8, \mathrm{HNO}_{2}=0.18, \mathrm{HCOOH}=4.5\), \(\mathrm{CH}_{3} \mathrm{COOH}=6.7\) (4) \(\mathrm{HF}=6.7, \mathrm{HNO}_{2}=0.18, \mathrm{HCOOH}=4.5\) \(\mathrm{CH}_{3} \mathrm{COOH}=1.8\)

\(\Lambda \mathrm{B}_{2}\) dissociates as \(\Lambda \mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons \Lambda \mathrm{B}(\mathrm{g})+\mathrm{B}(\mathrm{g})\). When the initial pressure of \(\Lambda \mathrm{B}_{2}\) is \(600 \mathrm{~mm} \mathrm{IIg}\), the total cquilibrium pressure is \(800 \mathrm{~mm} \mathrm{~kg}\). Calculate \(\mathrm{K}_{\mathrm{p}}\) for the reaction assuming that the volume of the system remains unchangcd. (1) 50 (2) 100 (3) \(166.8\) (4) 400

\(\Lambda\) chemist who is concerned with large-scale manufacture of useful compounds is primarily intcrested in (1) minimizing the cncrgy consumption (2) maximizing the backward reaction (3) minimizing the reverse reaction (4) decrcasing the acidity of the product

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