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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
None of these. Percentage dissociation is approximately 67.4\%.

Step by step solution

01

Understanding Vapour Density

Vapour density is defined as the mass of a certain volume of a substance compared to the mass of an equal volume of hydrogen. The vapour density (VD) of a compound is equal to half of its molecular weight (MW). We'll use this relationship: \[ \text{VD} = \frac{\text{MW}}{2} \] to find the molecular weight if it wasn't dissociated.
02

Calculate Molecular Weight of \( \mathrm{N}_{2} \mathrm{O}_{4} \)

The molecular weight of \( \mathrm{N}_{2} \mathrm{O}_{4} \) can be calculated by summing the atomic masses: \[ \mathrm{N}_{2} \mathrm{O}_{4} = 2 \times 14 + 4 \times 16 = 92 \] So, if it wasn't dissociated, the vapour density would be:\[ \text{VD} = \frac{MW}{2} = \frac{92}{2} = 46 \]
03

Compare Given Vapour Density with Calculated

The given vapour density is 30. Comparing this with the undissociated value of 46:\[ 46 - 30 = 16 \] The difference suggests dissociation since the vapour density is lower than expected for a non-dissociated compound.
04

Relate Dissociation to Vapour Density

Assume that some of the \( \mathrm{N}_{2} \mathrm{O}_{4} \) dissociates into \( 2\mathrm{NO}_{2} \). Thus:\[ \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2\mathrm{NO}_{2} \] The mole ratio changes from 1:0 to 1-x:x, or to \( 1-x \) for \( \mathrm{N}_{2} \mathrm{O}_{4} \) and \( 2x \) for \( \mathrm{NO}_{2} \), making the total moles \( 1-x+2x=1+x \).
05

Derive Formula for Percentage Dissociation

Since vapour density changes with moles and dissociation:\[ \text{Effective vapour density} = \frac{M}{1+x} \]Using given vapour density and assuming equilibrium we get:\[ 30 = \frac{92}{1+x} \]Rearrange:\[ 1+x = \frac{92}{30} \approx 3.067 \]Then, solving for \( x \):\[ x = 3.067 - 1 \approx 2.067 \]
06

Calculate Percentage Dissociation

Percentage dissociation is calculated using:\[ \text{Percentage dissociation} = \frac{x}{1+x} \times 100 \% \approx \frac{2.067}{3.067} \times 100 = 67.4 \% \]

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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 that provides a measure of a gas's density compared to hydrogen. It is calculated by taking the mass of a given volume of the substance and comparing it to the mass of an equivalent volume of hydrogen. This is expressed through:
  • Vapour density is defined as half the molecular weight of a compound.
  • The formula is: \[ \text{VD} = \frac{\text{MW}}{2} \]
For example, if we consider a substance that is not undergoing any changes, such as dissociation, the vapour density can help determine its molecular weight. In the case where the vapour density value is lower than expected compared to the non-dissociated compound, it suggests some changes like dissociation might have occurred in the gas.
Molecular Weight
Molecular weight, also known as molecular mass, is the sum of the atomic masses of all the atoms in a molecule. This is an important factor in predicting how different molecules will behave under given conditions. For example:
  • To find the molecular weight of \( \mathrm{N}_{2} \mathrm{O}_{4} \), sum up the atomic masses: 2 atoms of nitrogen (N) and 4 atoms of oxygen (O).
  • The formula is: \[ \mathrm{N}_{2} \mathrm{O}_{4} = 2 \times 14 + 4 \times 16 = 92 \]
Knowing the molecular weight of a compound helps predict various properties like its vapour density when the gas is not dissociating. If a compound undergoes partial dissociation, as often occurs with gases, molecular weight appears different, thus altering its vapour density.
Dissociation Percentage
Dissociation percentage helps us understand how much of a compound has broken down into its constituent parts at a certain condition, commonly under heat or pressure. When a compound dissociates, it shifts the balance in the molecular setup, directly affecting the vapour density.
  • For \( \mathrm{N}_{2} \mathrm{O}_{4} \), dissociation into \( 2\mathrm{NO}_{2} \) occurs, altering the compound's properties.
  • Use the formula: \[ \text{Percentage dissociation} = \frac{x}{1+x} \times 100 \% \] where \( x \) is the fraction of dissociation.
  • In practice, after determining the effective vapour density and rearranging equations, solving gives \( x \), allowing us to calculate the percentage dissociation.
Through the steps of calculation, understanding the dissociation percentage is critical for knowing how much of the compound remains intact and helps predict further chemical behavior under given conditions.

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Most popular questions from this chapter

The equilibrium constant for the reaction \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})\) is \(K_{c}=4.9 \times 10^{-2}\). the value of \(K_{c}\) for the reaction \(2 \mathrm{SO}_{2}\) \((\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(300^{\circ} \mathrm{C}\) in a closed one litre vessel till the following equilibrium is reached. \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) The equilibrium constant of this reaction at \(300^{\circ} \mathrm{C}\) is 4 . What is the concentration of \(\mathrm{B}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium? (a) \(0.2\) (b) \(0.6\) (c) \(0.75\) (d) \(0.1\)

Consider the reaction equilibrium, \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}^{\circ}=-198 \mathrm{~kJ}\) on the basis of Le Chatelier's principle, the condition favourable for the forward reaction is (a) lowering of temperature as well as pressure (b) increasing temperature as well as pressure (c) lowering the temperature and increasing the pressure (d) any value of temperature and pressure

If the concentrations of two monobasic acids are same, their relative strengths can be compared by (a) \(\left(\frac{\mathrm{K}_{\mathrm{l}}}{\mathrm{K}_{2}}\right)\) (b) \(\left(\frac{\alpha_{1}}{\alpha_{2}}\right)\) (c) \(\left(\sqrt{\frac{K_{1}}{K_{2}}}\right)\) (d) \(\frac{\left[\mathrm{H}^{+}\right]_{1}}{\left[\mathrm{H}^{+}\right]_{2}}\)

A flask containing \(0.5\) atm pressure of \(\mathrm{A}_{2}(\mathrm{~g})\), some solid AB added into flask which undergoes dissociation according to \(2 \mathrm{AB}(\mathrm{s}) \rightleftharpoons \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \mathrm{K}_{\mathrm{P}}=\) \(0.06 \mathrm{~atm}^{2} .\) The total pressure (in \(\mathrm{atm}\) ) at equilibrium is: (a) \(0.70\) (b) \(0.6\) (c) \(0.10\) (d) None of these
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