Problem 154 - A Complete Resource Book in Chemistry for JEE Main

Question

'a' and 'b' are van der Wall's constants for gases. Chlorine is more easily liquefied than ethane because [2011] (a) a for \(\mathrm{Cl}_{2}<\mathrm{a}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) but \(\mathrm{b}\) for \(\mathrm{Cl}_{2}>\mathrm{b}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) (b) a and \(\mathrm{b}\) for \(\mathrm{Cl}_{2}<\mathrm{a}\) and \(\mathrm{b}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) a and \(\mathrm{b}\) for \(\mathrm{Cl}_{2}>\mathrm{a}\) and \(\mathrm{b}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) a for \(\mathrm{Cl}_{2}>\mathrm{a}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) but \(\mathrm{b}\) for \(\mathrm{Cl}_{2}<\mathrm{b}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\)

Step-by-Step Solution

Answer

Chlorine is more easily liquefied due to stronger intermolecular forces, so option (c) is correct.

Step 1: Understand van der Waals Constants

The van der Waals equation is given by \( \left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT \), where \( P \) is pressure, \( V_m \) is molar volume, \( R \) is the universal gas constant, \( T \) is temperature, \( a \) is the measure of interaction forces between particles, and \( b \) is the volume occupied by particles. A larger value of \( a \) indicates stronger intermolecular forces and a bigger \( b \) indicates larger molecular size.

Step 2: Evaluate Options' Implications

Options describe varying relationships between \( a \) and \( b \) for chlorine (\( \mathrm{Cl}_2 \)) and ethane (\( \mathrm{C}_2\mathrm{H}_6 \)). Chlorine being more easily liquefied suggests it has stronger intermolecular attractions and possibly a larger effective size due to its molecular structure. Thus, a higher \( a \) for \( \mathrm{Cl}_2 \) and a comparative \( b \) requirement should be determined.

Step 3: Check Chlorine's Properties with Each Option

**Option (a):** Implies weaker interaction strength (\( a \)) for \( \mathrm{Cl}_2 \) than \( \mathrm{C}_2\mathrm{H}_6 \), contradictory to liquefaction ease. \( b \) being more for \( \mathrm{Cl}_2 \) supports larger size, but weaker \( a \) is against the premise. **Option (b):** Suggests weaker interactions and smaller size for \( \mathrm{Cl}_2 \), inconsistent with being more easily liquefied. **Option (c):** Suggests stronger interactions and larger size for \( \mathrm{Cl}_2 \); consistent with being more easily liquefied. **Option (d):** Suggests stronger interactions for \( \mathrm{Cl}_2 \) but smaller size, partially fits but less aligned than (c) with liquefaction rationale.

Step 4: Determine the Correct Answer

Given chlorine (\( \mathrm{Cl}_2 \))'s ease of liquefaction due to stronger intermolecular forces and possibly larger interactions arising from lone electron pairs, option (c) is the most aligned with scientific rationale assuming stronger attractions aided by larger molecular potentials. Thus, option (c) is accurate.

Key Concepts

Intermolecular Forces

Intermolecular forces are the attractions between molecules. They play a crucial role in determining the physical properties of substances, such as boiling and melting points. These forces come in various types, such as dipole-dipole interactions, London dispersion forces, and hydrogen bonds.

When comparing gases like chlorine (\( \text{Cl}_2 \)) and ethane (\( \text{C}_2\text{H}_6 \)), intermolecular forces dictate how easily each gas can be liquefied. Chlorine, being a larger molecule with more electrons, often exhibits stronger London dispersion forces. These stronger forces make chlorine easier to liquefy compared to a smaller molecule like ethane.

The van der Waals constant \( a \) represents the strength of these intermolecular forces. A higher \( a \) value indicates stronger attractions between molecules, which is why a higher \( a \) for chlorine compared to ethane suggests that chlorine is more easily liquefied.

Liquefaction

Liquefaction refers to the process of converting a gas into a liquid. This process depends heavily on the strength of the intermolecular forces present within the gas. Stronger intermolecular forces facilitate the transition from gaseous to liquid state because they effectively pull molecules closer together.

In the context of chlorine and ethane, the ease with which chlorine is liquefied indicates not just stronger intermolecular forces, but also the presence of other factors that assist in this phase change. Factors like molecular structure and the resulting intermolecular attractions are key determinants in how readily a gas will liquefy.

The van der Waals constant, \( a \), provides insight into why chlorine is more easily liquefied. A higher \( a \) value for chlorine signifies stronger attractive forces. Thus, it becomes easier for chlorine molecules to overcome kinetic energy and condense into a liquid.

Molecular Size

Molecular size is an important factor in understanding how substances interact and behave. It influences the strength of intermolecular forces, as larger molecules generally exert more significant dispersion forces.

Molecules like chlorine have a large molecular size, which means they occupy more space and have a greater chance to interact with one another compared to smaller molecules like ethane. This is represented by the van der Waals constant \( b \), which indicates the volume occupied by the gas molecules themselves.

For chlorine (\( \text{Cl}_2 \)), this larger size often corresponds to a greater \( b \) value, suggesting that they occupy more volume compared to ethane. Thus, molecular size and the \( b \) constant provide critical insight into how gases might behave and why larger molecules may liquefy more readily.

van der Waals Equation

The van der Waals equation is an adjusted version of the ideal gas law that accounts for real gas behaviors. It introduces two constants, \( a \) and \( b \), which help account for the non-ideal interactions between gas molecules: \[(P + \frac{a}{V_m^2})(V_m - b) = RT\]
Here, \( P \) is pressure, \( V_m \) is molar volume, \( R \) is the universal gas constant, and \( T \) is temperature.

The constant \( a \) measures intermolecular attractions—the larger this value, the stronger the forces. Meanwhile, \( b \) accounts for the molecular size, representing the finite volume that gas molecules occupy. This is particularly useful in explaining why gases deviate from ideality—larger molecules with stronger intermolecular forces deviate more.

The equation offers a way to predict the behavior of real gases under various conditions by modifying the ideal gas law to include the effects of these molecular attractions and repulsions. For \( \text{Cl}_2 \) and \( \text{C}_2\text{H}_6 \), applying this equation justifies why chlorine, with higher \( a \) and \( b \) values, is more apt to condense into a liquid compared to ethane.