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Chapter 7: Problem 69

The vapour density of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at a certain temperature is 30\. What is the percentage dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at this temperature? (a) \(53.3\) (b) \(106.6\) (c) \(26.7\) (d) none of these

Short Answer

Expert verified
(d) none of these

Step by step solution

01

Identify Molecular Weights

Determine the molecular weights of the compounds involved. The molecular weight of \( \mathrm{N}_{2} \mathrm{O}_{4} \) is \( 2 \times (14 \text{ for N}) + 4 \times (16 \text{ for O}) = 92 \). For \( \mathrm{NO}_{2} \), it is \( 14 + 2 \times 16 = 46 \).
02

Use Vapour Density Formula

Vapour density is half the molecular mass of a substance. Given vapour density of \( \mathrm{N}_{2} \mathrm{O}_{4} \) is 30, calculate the effective molecular mass: \( 2 \times 30 = 60 \).
03

Calculate Degree of Dissociation

Use the equation for dissociation: \( \alpha = \frac{(n-1)}{n} \), where \( n \) is the ratio of the molecular weight of \( \mathrm{N}_{2} \mathrm{O}_{4} \) to the effective molecular weight: \( n = \frac{92}{60} \approx 1.53 \). Thus, \( \alpha = \frac{1.53 - 1}{1.53} \approx 0.346 \).
04

Convert to Percentage Dissociation

The percentage dissociation \( \alpha \) is converted to percentage by multiplying by 100: \( 0.346 \times 100 \approx 34.6 \).
05

Compare and Choose the Closest Option

Compare the calculated percentage dissociation (\( 34.6 \)) with the given options. None of the options exactly match this value.

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Key Concepts

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

Vapour Density
Vapour density is a concept used in chemistry to compare the density of a vapor for a given substance relative to hydrogen. Calculated by taking the ratio of the mass of a certain volume of gas to the mass of the same volume of hydrogen under identical conditions, it provides insight into the molecular weight of substances.

In our problem, vapour density is given for \( \mathrm{N}_{2} \mathrm{O}_{4} \)\ as 30. This value helps us determine the effective molecular weight because vapour density is defined as half of the molecular weight: \( \text{Effective Molecular Weight} = 2 \times \text{Vapour Density} = 2 \times 30 = 60 \)\.
Percentage Dissociation
Percentage dissociation measures the extent to which a compound separates into two or more simpler substances. It is usually expressed as a percentage to make it more comprehensible.

In our exercise, we calculated the degree of dissociation to be around 0.346, or 34.6% when converted to a percentage. This value indicates that 34.6% of the original compound \( \mathrm{N}_{2} \mathrm{O}_{4} \)\ has dissociated into \( \mathrm{NO}_{2} \). \ However, none of the multiple choice options correspond precisely to this percentage, highlighting the importance of recalculating during assessments.
Degree of Dissociation
The degree of dissociation is a fraction that indicates the amount of the original substance that has dissociated compared to the total amount initially present. This value is crucial in chemical equilibrium because it explains the stability or reactivity of a compound under certain conditions.

In our example, we calculated the degree of dissociation as \( \alpha = \frac{(n-1)}{n} \), where \( n \) is the ratio of the original molecular weight over the effective molecular weight calculated from vapour density. The actual calculation was \( n = \frac{92}{60} \approx 1.53 \), \ resulting in a degree of dissociation: \( \alpha = \frac{1.53 - 1}{1.53} \approx 0.346 \).\ Knowing the degree of dissociation allows chemists to predict how much reactant is converted into products at a given temperature.
Molecular Weight Calculation
Determining the molecular weight of a compound is fundamental in chemistry because it affects how substances react and interact with each other. For a compound like \( \mathrm{N}_{2} \mathrm{O}_{4} \), \ the molecular weight is calculated by summing the weights of all atoms: \[ 2 \times \text{atomic weight of N (14)} + 4 \times \text{atomic weight of O (16)} = 92. \] This precise value is leveraged in further calculations, like finding effective molecular weights and establishing ratios with vapour density.
  • Molecular weight directly influences vapour density findings.
  • It also acts as a reference for calculating the degree of dissociation.
  • Accurate molecular weights ensure correct stoichiometric calculations in broader chemical reactions.
Mastering the art of molecular weight calculation is crucial for students to excel in chemistry.

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Most popular questions from this chapter

The equilibrium constant \(K_{P}\) and \(K_{p}\), for the reactions, \(\mathrm{X}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Y}(\mathrm{g})\) and \(\mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})+\mathrm{Q}(\mathrm{g})\) respectively are in the ratio of \(1: 9\). If the degree of dissociation of \(\mathrm{X}\) and \(\mathrm{Z}\) be equal then the ratio of total pressures at these equilibria is: (a) \(1: 36\) (b) \(1: 9\) (c) \(1: 3\) (d) \(1: 1\)

At constant temperature, the equilibrium constant \(\left(\mathrm{K}_{p}\right)\) for the decomposition reaction, \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) is expressed by \(\mathrm{K}_{\mathrm{p}}=\left(4 \mathrm{x}^{2} \mathrm{P}\right) /\left(1-\mathrm{x}^{2}\right)\), where \(\mathrm{P}=\) pressure, \(\mathrm{x}=\) extent of decomposition. Which one of the following statements is true? (a) \(\mathrm{K}_{\text {p increases with increase of } \mathrm{P}}\) (b) \(K_{p}\) increases with increase of \(x\) (c) \(\mathrm{K}_{\mathrm{p}}\) increases with decrease of \(\mathrm{x}\) (d) \(\mathrm{K}_{\text {p }}\) remains constant with change in \(\mathrm{P}\) and \(\mathrm{x}\)

For the reversible reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) At \(500{ }^{\circ} \mathrm{C}\), the value of \(\mathrm{K}_{\mathrm{p}}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmospheres. The corresponding value of \(\mathrm{K}_{c}\), with concentration in mole \(\mathrm{L}^{-1}\), is (a) \(1.44 \times 10^{-5} /(0.082 \times 500)^{-2}\) (b) \(1.44 \times 10^{-5} /(8.314 \times 773)^{-2}\) (c) \(1.44 \times 10^{-5}(0.082 \times 773)^{2}\) (d) \(1.44 \times 10^{-5} /(0.082 \times 773)^{-2}\)

In what manner will increase of pressure affect the following equation? \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})\) (a) shift in the reverse direction (b) shift in the forward direction (c) increase in the yield of hydrogen (d) no effect

A saturated solution of non-radioactive sugar was taken and a little radioactive sugar was added to it. A small amount of it gets dissolved in solution and an equal amount of sugar was precipitated. This proves (a) the equilibrium has been established in the solution (b) radioactive sugar can displace non-radioactive sugar from its solution. (c) Equilibrium is dynamic in nature (d) none of the above
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