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Chapter 11: Problem 34

Based on their composition and structure, list $\mathrm{CH}_{3} \mathrm{COOH},\( \)\mathrm{CH}_{3} \mathrm{COOCH}_{3}\(, and \)\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ in order of (a) increasing intermolecular forces, \((\mathbf{b})\) increasing viscosity, \((\mathbf{c})\) increasing surface tension.

Short Answer

Expert verified
Based on their composition and structure, the order of the molecules CH3COOH, CH3COOCH3, and CH3CH2OH is: a) increasing intermolecular forces: CH3COOCH3 (Methyl Acetate) < CH3CH2OH (Ethanol) < CH3COOH (Acetic Acid) b) increasing viscosity: CH3COOCH3 (Methyl Acetate) < CH3CH2OH (Ethanol) < CH3COOH (Acetic Acid) c) increasing surface tension: CH3COOCH3 (Methyl Acetate) < CH3CH2OH (Ethanol) < CH3COOH (Acetic Acid)

Step by step solution

01

Identify Intermolecular Forces

First, let's identify the types of intermolecular forces present in the given molecules 1. CH3COOH (Acetic Acid) has hydrogen bonding (due to the presence of OH group), dipole-dipole forces (due to polar C=O bond), and London dispersion forces. 2. CH3COOCH3 (Methyl Acetate) has dipole-dipole forces (due to polar C=O bond) and London dispersion forces 3. CH3CH2OH (Ethanol) has hydrogen bonding (due to the presence of OH group), dipole-dipole forces (due to polar C-O bond), and London dispersion forces.
02

Compare the Strength of Intermolecular Forces

Now, let's compare the strengths of the intermolecular forces present in these molecules. Generally, hydrogen bonding is the strongest intermolecular force followed by dipole-dipole forces and London dispersion forces. Between CH3COOH and CH3CH2OH, since both have hydrogen bonding, their strengths depend on the extent of hydrogen bonding. Acetic acid (CH3COOH) has a stronger hydrogen bonding because of its more electronegative oxygen atom in the C=O bond. Ethanol (CH3CH2OH) has weaker hydrogen bonding in comparison. Methyl acetate (CH3COOCH3) does not have hydrogen bonding and has the weakest intermolecular forces among these three molecules.
03

List Molecules in Increasing Order of Intermolecular forces

Using the information above, we can list the molecules in increasing order of intermolecular forces: CH3COOCH3 (Methyl Acetate) < CH3CH2OH (Ethanol) < CH3COOH (Acetic Acid)
04

List Molecules in Increasing Order of Viscosity

Viscosity depends on the strength of intermolecular forces present in the molecules. Stronger intermolecular forces lead to higher viscosity. Thus, the order of increasing viscosity is: CH3COOCH3 (Methyl Acetate) < CH3CH2OH (Ethanol) < CH3COOH (Acetic Acid)
05

List Molecules in Increasing Order of Surface Tension

Surface tension is also determined by the strength of intermolecular forces. Molecules with stronger intermolecular forces will have higher surface tension. Therefore, the order of increasing surface tension is: CH3COOCH3 (Methyl Acetate) < CH3CH2OH (Ethanol) < CH3COOH (Acetic Acid) To summarize, based on their composition and structure, the order of the molecules CH3COOH, CH3COOCH3, and CH3CH2OH is: a) increasing intermolecular forces： CH3COOCH3 < CH3CH2OH < CH3COOH b) increasing viscosity： CH3COOCH3 < CH3CH2OH < CH3COOH c) increasing surface tension： CH3COOCH3 < CH3CH2OH < CH3COOH

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Most popular questions from this chapter

Suppose you have two colorless molecular liquids A and B whose boiling points are \(78^{\circ} \mathrm{C}\) and \(112^{\circ} \mathrm{C}\) respectively and both are at atmospheric pressure. Which of the following statements is correct? For each statement that is not correct, modify the statement so that it is correct. (a) Both A and \(B\) are liquids with identical vapor pressure at room temperature of \(25^{\circ} \mathrm{C} .(\mathbf{b})\) Liquid A must consist of nonpo- (c) Both lar molecules with lower molecular weight than B. liquids A and \(B\) have higher total intermolecular forces than water. (d) Liquid \(\mathrm{A}\) is more volatile than liquid \(\mathrm{B}\) because it has a lower boiling point. (e) At \(112^{\circ} \mathrm{C}\) both liquids have a vapor pressure of $1 \mathrm{~atm}$.

At standard temperature and pressure, the molar volumes of \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) gases are 22.06 and \(22.40 \mathrm{~L}\), respectively. (a) Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same? (b) On cooling to $160 \mathrm{~K}$, both substances form crystalline solids. Do you expect the molar volumes to decrease or increase on cooling the gases to \(160 \mathrm{~K} ?\) (c) The densities of crystalline \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) at \(160 \mathrm{~K}\) are 2.02 and \(0.84 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. Calculate their molar volumes. (d) Are the molar volumes in the solid state as similar as they are in the gaseous state? Explain. (e) Would you expect the molar volumes in the liquid state to be closer to those in the solid or gaseous state?

List the three states of matter in order of (a) increasing molecular disorder and \((\mathbf{b})\) increasing intermolecular attraction. (c) Which state of matter is most easily compressed?

You are high up in the mountains and boil water to make some tea. However, when you drink your tea, it is not as hot as it should be. You try again and again, but the water is just not hot enough to make a hot cup of tea. Which is the best explanation for this result? (a) High in the mountains, it is probably very dry, and so the water is rapidly evaporating from your cup and cooling it. (b) High in the mountains, it is probably very windy, and so the water is rapidly evaporating from your cup and cooling it. (c) High in the mountains, the air pressure is significantly less than \(101.3 \mathrm{kPa}\), so the boiling point of water is much lower than at sea level. (d) High in the mountains, the air pressure is significantly less than \(101.3 \mathrm{kPa}\), so the boiling point of water is much higher than at sea level.

Due to the environmental concern of fluorocarbons as refrigerants, a refrigerant based on a mixture of hydrocarbons was used as a replacement. It is a patented blend of ethane, propane, butane, and isobutane. Isobutane has a normal boiling point of \(-12^{\circ} \mathrm{C}\). The molar specific heat of liquid phase and gas phase isobutane are $129.7 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\( and \)95.2 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ respectively. The heat of vaporization for this compound is $21.3 \mathrm{~kJ} / \mathrm{mol}\(. Calculate the heat required to convert \)25.0 \mathrm{~g}$ of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at $40^{\circ} \mathrm{C}$.
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