/*! 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 56 Questions 54-56 refer to the fol... [FREE SOLUTION] | 91Ó°ÊÓ

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Questions 54-56 refer to the following. GRAPH CAN'T COPY Between propane and ethene, which will likely have the higher boiling point and why? (A) Propane, because it has a greater molar mass (B) Propane, because it has a more polarizable electron cloud (C) Ethene, because of the double bond (D) Ethene, because it is smaller in size

Short Answer

Expert verified
The correct answer is (A). Propane, because it has a greater molar mass

Step by step solution

01

Identify the Type of Interactions in Both Compounds

Firstly, identify both propane and ethene as non-polar molecules due to their linear configuration and thus can provide only London Dispersion Forces (which is a type of Van der Waals force is the weakest intermolecular interaction).
02

Comparing Molecular Mass

The molecular mass of propane (C3H8) is greater than that of ethene (C2H4). More massive molecules tend to have higher boiling points since they possess more electrons and thereby can generate larger London dispersion forces.
03

Analyze Option(A) and(B)

In this case, option (A) is correct since it accurately indicates that propane will have a higher boiling point due to its greater molar mass. As far as option (B) is concerned, Propane being more polarizable is true, however, this is not the most crucial factor resulting in the higher boiling point. Therefore option (B) is incorrect.
04

Analyze Option(C) and (D)

Options (C) and (D) that suggest ethene has a higher boiling point is incorrect. While ethene does indeed possess a double bond, this does not influence its boiling point, as the double bond does not enhance intermolecular forces and ethene's smaller size and lighter mass would allow it to have lower boiling point versus Propane.

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Key Concepts

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

London Dispersion Forces
London Dispersion Forces are fascinating yet quite simple to understand. They are a type of weak intermolecular force that occurs between non-polar molecules. These forces arise because electrons in atoms or molecules are constantly moving, creating temporary dipoles.

When the electrons in a molecule momentarily align in such a way that creates a slight charge difference, they induce dipoles in neighboring molecules, causing a weak attraction. Think of it as a fleeting handshake between molecules.

These forces are the only type of intermolecular interaction present between non-polar molecules like propane and ethene. Because these forces are temporary and weak, they do not require much energy to overcome, typically resulting in lower boiling points for substances primarily held together by them.
Molecular Mass
The molecular mass of a substance plays a vital role in determining its boiling point, especially in non-polar molecules wherein London Dispersion Forces predominate. Let's consider the case of propane and ethene.

Propane has a higher molecular mass than ethene due to its extra carbon atom and hydrogens.
  • Propane: C\(_3\)H\(_8\) = 44 g/mol approximately
  • Ethene: C\(_2\)H\(_4\) = 28 g/mol approximately
A heavier molecule like propane has more electrons, enhancing its ability to form instantaneous dipoles. Consequently, these stronger dipoles lead to stronger London dispersion forces resulting in a higher boiling point. In simpler terms, heavier molecules are like anchors that don't easily lift off the ground.

Therefore, when comparing molecular masses, the higher the mass, the greater the dispersion forces, and hence, higher the boiling point.
Intermolecular Forces
Intermolecular forces are the forces that hold molecules together, and they play a crucial role in determining the physical properties of substances, such as boiling points, melting points, and viscosity.

The strength of these forces varies among different substances and depends on the type of molecules involved. For non-polar molecules like propane and ethene, the primary force at play is the London Dispersion Force.

Typically, there are three types of intermolecular forces:
  • London Dispersion Forces
  • Dipole-Dipole Interactions
  • Hydrogen Bonding
The London Dispersion Forces are the weakest, whereas hydrogen bonds are the strongest. In our scenario, since propane and ethene are non-polar, only dispersion forces are relevant, thus driving the discussion and comparison of their boiling points based on their respective molecular masses.
Non-polar Molecules
Non-polar molecules are molecules that do not have distinct positive and negative poles. In simple terms, the electrons are distributed evenly, or nearly evenly, in the molecule, causing no permanent dipole.

Propane and ethene are classic examples of non-polar molecules. Their symmetric structures mean that any slight charge differences are balanced out. Because they are non-polar, these molecules don't engage in stronger types of intermolecular interactions, such as dipole-dipole interactions or hydrogen bonding.

Instead, they rely on the considerably weaker London Dispersion Forces. Understanding the nature of non-polar molecules helps explain why their boiling points are influenced more by their molecular mass than by the presence of additional bonds like double bonds in ethene. Remember, without polar regions, these substances remain relatively indifferent in their interactions, unlike their polar counterparts.

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Most popular questions from this chapter

Choose the correct net ionic equation representing the reaction that occurs when solutions of potassium carbonate and copper (I) chloride are mixed. (A) \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+2 \mathrm{CuCl}(a q) \rightarrow 2 \mathrm{KCl}(a q)+\mathrm{Cu}_{2} \mathrm{CO}_{3}(s)\) (B) \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+2 \mathrm{CuCl}(a q) \rightarrow 2 \mathrm{KCl}(\mathrm{s})+\mathrm{Cu}_{2} \mathrm{CO}_{3}(a q)\) (C) \(\mathrm{CO}_{3}^{2-}+2 \mathrm{Cu}^{+} \rightarrow \mathrm{Cu}_{2} \mathrm{CO}_{3}\) (D) \(\mathrm{CO}_{3}^{2-}+\mathrm{Cu}^{2+} \rightarrow \mathrm{CuCO}_{3}(s)\)

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