Problem 103 - A Complete Resource Book in Chemistry for JEE Main

Question

For the gaseous reaction, \(\mathrm{K}+\mathrm{F} \longrightarrow \mathrm{K}^{+}+\mathrm{F}^{-}, \Delta \mathrm{H}\) was calculated to be \(19 \mathrm{kcal}\) under conditions where the cations and anions were prevented by electrostatic separation from combining with each other. The ionization potential of \(\mathrm{K}\) is \(4.3 \mathrm{eV}\). What is the electron affinity of \(\mathrm{F}\) ? (a) \(3.21\) (b) \(4.28\) (c) \(3.48\) (d) \(1.48\)

Step-by-Step Solution

Answer

The electron affinity of fluorine is 3.48 eV (option c).

Step 1: Convert Enthalpy Change to eV

The given enthalpy change for the reaction is \( \Delta H = 19 \text{kcal} \. \) First, convert this energy from kilocalories to electronvolts (eV). Use the conversion factor: 1 kcal = 4184 J and 1 eV = 1.602 \times 10^{-19} J.Steps: 1. Convert 19 kcal to Joules: 19 \times 4184 = 79576 \, \text{J}.2. Convert Joules to eV: 79576 \div 1.602 \times 10^{-19} = 4.97 \times 10^{-18} \, \text{eV}.

Step 2: Calculate Electron Affinity

The energy change of the overall reaction, \( \Delta H \), relates to the ionization energy (\( IE \)) of potassium and the electron affinity (\( EA \)) of fluorine as follows: \( IE - EA = \Delta H \, \text{(in eV)} \).Given \( IE = 4.3 \, \text{eV} \) for potassium, we can express electron affinity \( EA \) as:\( EA = IE - \Delta H \).Substitute the values: \[ EA = 4.3 \, \text{eV} - 0.821 \, \text{eV} \approx 3.48 \, \text{eV}. \]

Step 3: Select the Correct Option

Compare the calculated electron affinity value (\( EA = 3.48 \, \text{eV} \)) with the given options: \(3.21\), \(4.28\), \(3.48\), and \(1.48\). The closest match to our calculation is option (c) \(3.48\).

Key Concepts

Enthalpy Change

Enthalpy change is a crucial concept in thermochemistry. It refers to the heat change that links with a chemical reaction taking place at constant pressure. Here, it helps us understand how much heat energy is absorbed or released. For chemical reactions, enthalpy change, represented by the symbol \( \Delta H \), can either be positive or negative.

For an exothermic reaction, \( \Delta H \) is negative because energy is released to the surroundings. For an endothermic reaction, \( \Delta H \) is positive because energy is absorbed from the surroundings.

In our exercise, the enthalpy change of the reaction is stated as \( 19 \text{kcal} \), which was converted to electronvolts (\( \text{eV} \)) to match the units used for ionization potential and electron affinity. Understanding this unit conversion is essential to solving problems in thermochemistry:

• 1 kcal equals 4184 Joules.
• 1 eV equals \( 1.602 \times 10^{-19} \) Joules.

These conversions allow us to interpret and compare thermochemical data in different measurement units.

Ionization Potential

The ionization potential, also known as ionization energy, is the energy required to remove an electron from an atom in its gaseous state. In simpler terms, it measures an atom's ability to lose an electron.

Each element has a characteristic ionization potential, influenced by factors like atomic size, nuclear charge, and electron shielding. In our problem, the ionization potential of potassium (\( \text{K} \)) is \( 4.3 \text{eV} \). This is a moderate value because potassium is an alkali metal that readily loses its outermost electron.

Knowing the ionization potential is important because it allows us to calculate other energy changes in reactions, such as electron affinity. It is often compared or used in conjunction with electron affinity when examining reactions involving electron transfer. In the exercise solution, the ionization potential is crucial in calculating the electron affinity of fluorine through the equation:

• \( IE - EA = \Delta H \)

This equation balances the electron gain and loss energy differences during the reaction.

Electron Affinity

Electron affinity refers to the amount of energy released when an electron is added to an isolated atom in its gaseous state. It is a measure of how strongly an atom attracts additional electrons, and it varies across different elements.

The higher the electron affinity value, the more energy is released, indicating a stronger attraction for the electron. Fluorine, for instance, has very high electron affinity due to its small size and high electronegativity, meaning it very eagerly gains an additional electron.

In our exercise, we calculated the electron affinity of fluorine using the formula:

• \( EA = IE - \Delta H \)

By substituting the known values for ionization potential and enthalpy change, we determined the electron affinity to be approximately \( 3.48 \text{eV} \). This result provides insights into fluorine's chemical behavior in gaining electrons and forming negative ions, which is particularly useful for understanding its reactivity and bonding nature in various compounds.