Problem 16 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Compare the energy of interaction between two singly charged ions separated by a distance of \(0.1 \mathrm{nm}\) in (a) vacuum, (b) water (relative permittivity \(=78\) ), (c) a hydrocarbon oil (relative permittivity \(=2\) ). (Section 17.3 )

Step-by-Step Solution

Answer

Energy in vacuum: \(2.3 \times 10^{-18} \text{ J}\), in water: \(2.9 \times 10^{-20} \text{ J}\), in oil: \(1.1 \times 10^{-18} \text{ J}\).

Step 1: Understanding Coulomb's Law

Coulomb's Law describes the energy of interaction between two charged particles. It is given by the formula: \[ U = \frac{k_e \cdot q_1 \cdot q_2}{r \cdot \varepsilon} \] where \( U \) is the energy of interaction, \( k_e \) is Coulomb's constant \((8.9875 \times 10^9 \text{ N m}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the charges of the ions (in this case, each ion is singly charged, so \( q_1 = q_2 = 1.6 \times 10^{-19} \text{ C}\)), \( r \) is the distance separating the charges, and \( \varepsilon \) is the permittivity of the medium.

Step 2: Calculating Energy in Vacuum

For a vacuum, the permittivity \( \varepsilon = \varepsilon_0\), where \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2 \). Substitute these values into Coulomb’s Law:\[ U_{vacuum} = \frac{8.9875 \times 10^9 \text{ N m}^2/\text{C}^2 \cdot (1.6 \times 10^{-19} \text{ C})^2}{0.1 \times 10^{-9} \text{ m} \cdot 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2} \approx 2.3 \times 10^{-18} \text{ J} \]

Step 3: Calculating Energy in Water

For water, the relative permittivity \( \varepsilon_r = 78 \). Thus, the effective permittivity is \( \varepsilon = \varepsilon_0 \cdot \varepsilon_r = 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2 \cdot 78 \). The energy can be calculated as:\[ U_{water} = \frac{8.9875 \times 10^9 \text{ N m}^2/\text{C}^2 \cdot (1.6 \times 10^{-19} \text{ C})^2}{0.1 \times 10^{-9} \text{ m} \cdot 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2 \cdot 78} \approx 2.9 \times 10^{-20} \text{ J} \]

Step 4: Calculating Energy in Hydrocarbon Oil

For hydrocarbon oil, the relative permittivity \( \varepsilon_r = 2 \). Therefore, the effective permittivity becomes \( \varepsilon = \varepsilon_0 \cdot \varepsilon_r = 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2 \cdot 2 \). The energy is:\[ U_{oil} = \frac{8.9875 \times 10^9 \text{ N m}^2/\text{C}^2 \cdot (1.6 \times 10^{-19} \text{ C})^2}{0.1 \times 10^{-9} \text{ m} \cdot 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2 \cdot 2} \approx 1.1 \times 10^{-18} \text{ J} \]

Step 5: Compare the Energies

From the calculations, we find that the energy of interaction between the ions is highest in a vacuum \((2.3 \times 10^{-18} \text{ J})\), followed by in a hydrocarbon oil \((1.1 \times 10^{-18} \text{ J})\), and is lowest in water \((2.9 \times 10^{-20} \text{ J})\). This corresponds with the medium's permittivity values, where higher permittivity lowers the interaction energy.

Key Concepts

Energy of Interaction

The energy of interaction between charged particles, such as ions, is a key concept in physics. It denotes the potential energy resulting from the electrostatic force that these particles exert on each other. This interaction is largely governed by Coulomb's Law.

• According to Coulomb’s Law, the energy is inversely proportional to the distance between the particles and directly proportional to the magnitude of the charges.
• The interaction energy also depends on the medium in which the charges are present. This is because different media have different capabilities to reduce the electric field between the charges.

For instance, two ions in a vacuum interact with more energy compared to the same ions in a medium such as water or oil. This is because the permittivity of the medium affects the interaction strength.

Permittivity

Permittivity is a measure of how easily a medium allows the passage of an electric field. It is denoted by the symbol \( \varepsilon \) and determines how much the electric field generated by the charges is reduced.

• The basic permittivity in free space (vacuum) is known as \( \varepsilon_0 \), or the permittivity of free space.
• Permittivity in other materials is usually greater than \( \varepsilon_0 \), meaning these materials reduce the force between charges more than a vacuum does.

The presence of permittivity in the formula for interaction energy explains why the energy is highest in a vacuum. In mediums with higher permittivity, like water, the reduction in energy is significant due to the medium's ability to further diminish the electric field.

Relative Permittivity

Relative permittivity, also known as the dielectric constant, is a dimensionless number that describes how much a material can reduce the electric field compared to a vacuum.

• Relative permittivity is symbolized as \( \varepsilon_r \).
• It is calculated by dividing the permittivity of the medium by the permittivity of free space \( \varepsilon_0 \).

Materials with higher relative permittivity values significantly decrease the energy of interaction. For example, water, with a relative permittivity of 78, greatly lowers the interaction energy between ions, which is why the energy measured is much lower than in a vacuum or hydrocarbon oil.

Charged Ions

Charged ions are atoms or molecules that have lost or gained one or more electrons, resulting in a net charge. They play a crucial role in many physical and chemical processes.

• A singly charged ion typically has a charge magnitude equal to the charge of one electron, which is about \( 1.6 \times 10^{-19} \) C.
• The interaction energy between ions can depend on their charge. Higher charges result in stronger interactions.
• The distance separating ions also affects their interaction energy. Closer ions result in stronger electrostatic interaction.

Understanding the role of ions and how they interact in different environments helps to grasp concepts in fields as diverse as chemistry, physics, and even biology, where electrostatic interactions play a critical role.