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True or false: (a) \(\mathrm{CBr}_{4}\) is more volatile than \(\mathrm{CCl}_{4}\). (b) \(\mathrm{CBr}_{4}\) has a higher boiling point than \(\mathrm{CCl}_{4}\) - (c) \(\mathrm{CBr}_{4}\) has weaker intermolecular forces than \(\mathrm{CCl}_{4}\) - (d) \(\mathrm{CBr}_{4}\) has a higher vapor pressure at the same temperature than \(\mathrm{CCl}_{4}\).

Short Answer

Expert verified
In summary, the correct answers are: (a) False (CBr4 is less volatile than CCl4 because it has stronger intermolecular forces due to dispersion forces)\ (b) True (CBr4 has a higher boiling point than CCl4 because of its stronger intermolecular forces)\ (c) False (CBr4 has stronger intermolecular forces than CCl4)\ (d) False (CBr4 has a lower vapor pressure at the same temperature than CCl4 due to its stronger intermolecular forces)

Step by step solution

01

(a) CBr4 is more volatile than CCl4.

Volatility refers to the tendency of a substance to vaporize. A compound with weaker intermolecular forces is more likely to evaporate, making it more volatile. Comparing CBr4 and CCl4, we can see that Br is heavier than Cl, and this will result in stronger dispersion forces in CBr4, meaning that it has stronger intermolecular forces than CCl4. Therefore, CBr4 should be less volatile than CCl4. So, this statement is \(\textbf{false}\). \( \)
02

(b) CBr4 has a higher boiling point than CCl4.

Boiling point is directly related to the strength of intermolecular forces: the stronger the intermolecular forces, the higher the boiling point. We have already established that CBr4 has stronger intermolecular forces than CCl4 due to stronger dispersion forces. Therefore, CBr4 should have a higher boiling point than CCl4. So, this statement is \(\textbf{true}\). \( \)
03

(c) CBr4 has weaker intermolecular forces than CCl4.

As mentioned earlier, CBr4 has stronger dispersion forces compared to CCl4 due to the heavier mass of Br compared to Cl. Thus, CBr4 has stronger, not weaker, intermolecular forces than CCl4. This statement is \(\textbf{false}\). \( \)
04

(d) CBr4 has a higher vapor pressure at the same temperature than CCl4.

Vapor pressure is inversely related to the strength of intermolecular forces. Substances with weaker intermolecular forces have a higher vapor pressure at a given temperature. Since CBr4 has stronger intermolecular forces than CCl4, it should have a lower vapor pressure at the same temperature. Therefore, this statement is \(\textbf{false}\). \( \) In summary, the correct answers are: (a) False \ (b) True \ (c) False \ (d) False

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Key Concepts

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

Volatility in Chemistry
Volatility in chemistry refers to the tendency of a substance to vaporize or become a gas. It's an important concept to understand as it influences how we handle and use chemicals on a daily basis. Imagine for a moment that you have two similar substances and you want to compare their volatilities. The one that turns into a gas more readily at a given temperature is considered to be more volatile.

Why does this matter? Because volatility affects everything from the smell of perfume to the evaporation of fuel in an engine. In an educational context, comparing the volatility of CBr4 and CCl4 helps illustrate the principle that molecular mass impacts volatility. Generally, heavier molecules have stronger intermolecular forces, particularly London dispersion forces, so they are less volatile. Understanding this, we can see why the statement that CBr4 is more volatile than CCl4 is incorrect; CBr4, with its heavier bromine atoms, actually has stronger dispersion forces and thus lower volatility.

As a takeaway, when considering volatility, think about intermolecular forces at play, and remember, lighter molecules generally escape into the gaseous phase more easily.
Boiling Point Comparison
Comparing boiling points is akin to assessing the thermal endurance of a molecule. The boiling point is the temperature at which a substance transitions from a liquid to a gas, and it's intimately tied to the strength of forces holding the molecules together in the liquid state.

When analyzing CBr4 and CCl4, their boiling point comparison can reveal much about their molecular interactions. CBr4 has a higher boiling point, which indicates stronger intermolecular forces within its structure. This is because the mass of the bromine atoms contributes to stronger London dispersion forces than in the chlorine-containing CCl4.

Therefore, in a learning environment, it's crucial to link the boiling point directly to intermolecular forces. Recognize that a higher molecular weight compounds like CBr4 tends to have a higher boiling point due to its stronger intermolecular forces.
Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its liquid phase at a given temperature. It may sound complex, but it's really just a measure of a liquid's propensity to evaporate. Imagine a sealed container half-filled with liquid; the space above the liquid is where you'd find vapor pressure at work.

In a classroom setting, understanding vapor pressure is essential when dealing with evaporation and boiling. The substances with weaker intermolecular forces will have higher vapor pressures since their molecules can escape to the gas phase more readily. Considering CBr4 versus CCl4, the stronger intermolecular forces in CBr4 result in a lower vapor pressure.

This inverse relationship between vapor pressure and intermolecular forces is a key point to grasp. It's a common misconception to link high vapor pressure to more substantial intermolecular forces. In fact, the opposite is true, as seen with CBr4, which has stronger intermolecular forces and therefore a lower vapor pressure compared to CCl4.

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Most popular questions from this chapter

At room temperature, \(\mathrm{Si}\) is a solid, \(\mathrm{CCl}_{4}\) is a liquid, and \(\mathrm{Ar}\) is a gas. List these substances in order of (a) increasing intermolecular energy of attraction, (b) increasing boiling point.

Which of the following affects the vapor pressure of a liquid? (a) Volume of the liquid, (b) surface area, (c) intermolecular attractive forces, (d) temperature, (e) density of the liquid.

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) melts at \(-114^{\circ} \mathrm{C}\) and boils at \(78^{\circ} \mathrm{C}\). The enthalpy of fusion of ethanol is \(5.02 \mathrm{~kJ} / \mathrm{mol}\), and its enthalpy of vaporization is \(38.56 \mathrm{~kJ} / \mathrm{mol}\). The specific heats of solid and liquid ethanol are \(0.97\) and \(2.3 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. (a) How much heat is required to convert \(42.0 \mathrm{~g}\) of ethanol at \(35^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C}\) ? (b) How much heat is required to convert the same amount of ethanol at \(-155^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C}\) ?

True or false: (a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable. (b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, dipole- dipole interactions, when present, are always greater than dispersion forces. (d) All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical.

Acetone \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\right]\) is widely used as an industrial solvent. (a) Draw the Lewis structure for the acetone molecule and predict the geometry around each carbon atom. (b) Is the acetone molecule polar or nonpolar? (c) What kinds of intermolecular attractive forces exist between acetone molecules? (d) 1-Propanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) has a molecular weight that is very similar to that of acetone, yet acetone boils at \(56.5^{\circ} \mathrm{C}\) and 1 -propanol boils at \(97.2^{\circ} \mathrm{C}\). Explain the difference.

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