Problem 148 - A Complete Resource Book in Chemistry for JEE Main

Question

Viscosity coefficients of some liquids are given below: Liquid \(\eta\) in millipoise at \(30^{\circ} \mathrm{C}\) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{3}\) \(2.11\) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}\) \(2.89\) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CH}_{3}\) \(3.68\)The order of viscosity coefficient of the liquids (1) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (2) CCC(C)CC (3) CC(C)C(C)C (a) \(1>2>3\) (b) \(1<2<3\) (c) \(1>2=3\) (d) remains same

Step-by-Step Solution

Answer

The answer is (a) 1>2>3.

Step 1: Understand Exercise Context

The problem provides the viscosity coefficients for three alkanes, given in a list format, and asks to determine the order of their viscosity from highest to lowest. The coefficients are given in millipoise at a temperature of 30°C.

Step 2: Identify Alkane Structures

1 refers to the linear alkane hexane with the structure CH₃(CH₂)₄CH₃, which is a straight chain, n-hexane. 2 and 3 are given in SMILES notation: 2 corresponds to CH₃CH(CH₃)CH₂CH₂CH₃ (isopentane); 3 corresponds to (CH₃)₂CHCH₂CH(CH₃)₂ (2,3-dimethylbutane).

Step 3: Analyze Viscosity Data Provided

The provided viscosity data at 30°C shows linear alkanes: n-pentane (CH₃(CH₂)₃CH₃) has a viscosity of 2.11 millipoise, n-hexane (CH₃(CH₂)₄CH₃) has a viscosity of 2.89 millipoise, and n-heptane (CH₃(CH₂)₅CH₃) has a viscosity of 3.68 millipoise. Generally, increased chain length in alkanes increases viscosity.

Step 4: Compare and Arrange Based on Viscosity

Using the identified structures: 1 (n-hexane) with 2.89 millipoise. 2 (isopentane) is a branched form of pentane, likely to have lower viscosity than straight chains so estimated less than 2.11 millipoise. 3, a more branched 2,3-dimethylbutane possibly has even lower viscosity than isopentane as branching typically reduces viscosity. Therefore, organized by decreasing viscosity: 1 > 2 > 3.

Step 5: Choose Correct Option

From the analysis in Step 4, the correct order is 1 > 2 > 3, which matches option (a). So (a) 1>2>3 is correct choice.

Key Concepts

Alkanes

Alkanes are saturated hydrocarbons, composed exclusively of carbon (C) and hydrogen (H) atoms arranged in various configurations.
These can be visualized as a backbone structure made up only of carbon atoms linked by single covalent bonds, with hydrogen atoms filling the remaining valences around each carbon atom.

Here are some key characteristics of alkanes:

• Simplicity in structure: They contain only single bonds (C-C and C-H) making them fully saturated.
• General formula: Their formula follows the pattern \( C_nH_{2n+2} \) where \( n \) is the number of carbon atoms.
• Boiling and melting points: These increase with an increase in the number of carbon atoms due to greater surface area.
• Non-polar: Alkanes are non-polar molecules, which means they do not dissolve well in water, a polar solvent.

Understanding the structure of alkanes is instrumental in comprehending their behavior, including the effect on properties like viscosity, as seen later.
The physical properties of alkanes such as viscosity are influenced heavily by how the carbon atoms are connected.

Viscosity Order

Viscosity, simply put, is a measure of a liquid's resistance to flow.
Imagine honey and water; honey flows more slowly because it is more viscous than water.
Similarly, for alkanes, viscosity is influenced by the length of the carbon chain.

• Linear Alkanes: The longer the carbon chain, the greater the dispersion forces (Van der Waals forces) between the molecules, leading to higher viscosity.
• Measurement: In the given exercise, viscosity is measured in millipoise at \( 30^{\circ}C \). As the molecular size increases (i.e., more carbon atoms), the viscosity also increases because larger molecules experience more intermolecular forces.

This results in a predictable order of viscosity among different alkanes. For example, n-hexane, which has more carbon atoms than n-pentane, is more viscous.
This principle can be applied to determine the viscosity order: longer, linear chains exhibit higher viscosity.

Branching Effect

Branching refers to the arrangement of carbon atoms in a non-linear format, where some carbon atoms are bonded not just in a straight line.
This structure has significant implications for the physical properties of alkanes, including viscosity.

• Less Surface Area: Branched alkanes have a more compact shape which decreases the surface area exposed to other molecules.
• Weaker Intermolecular Forces: Due to their compact shape, they experience weaker Van der Waals forces than linear alkanes. This happens because there is less surface-to-surface contact between molecules.
• Impact on Viscosity: As a result, branched alkanes typically exhibit lower viscosity compared to their straight-chain counterparts with a similar molecular weight.

For example, consider isopentane and n-pentane.
Despite having the same molecular formula, isopentane with branched structure has lower viscosity.
This concept explains the observed order in the given exercise, where more branching leads to lower viscosity: hence, the order 1 > 2 > 3 is driven by these branching effects.