Question

From N atoms of an element A, when half the atoms transfer one electron to the another atom 405 kJ mol⁻¹ of energy was found to be consumed. An additional energy of 745kJ mol⁻¹ was further required to convert all the A⁻ ions to A⁺. Calculate the ionisation energy and the electron gain enthalpy of atom A in eV (1 eV = 96.48 kJ).

• A. IE = 11.9 eV and ΔₑgHΘ = -3.5 eV
• B. IE = 4.9 eV and ΔₑgHΘ = -6.5 eV
• C. IE = 8.3 eV and ΔₑgHΘ = -8.5 eV
• D. IE = 11.9 eV and ΔₑgHΘ = 3.5 eV

Answer 1

Answer: (A)

IE = 11.9 eV and ΔₑgHΘ = -3.5 eV

Answer 2

The problem statement is ambiguous, but a common interpretation that aligns with one of the options is as follows:

1. The first statement, "From N atoms of an element A, when half the atoms transfer one electron to the another atom 405 kJ mol⁻¹ of energy was found to be consumed," can be interpreted as the first ionization energy (IE₁) of atom A being 405 kJ mol⁻¹. $IE_1 = 405 \text{ kJ mol}^{-1}$

2. The second statement, "An additional energy of 745kJ mol⁻¹ was further required to convert all the A⁻ ions to A⁺," describes the process: A⁻ → A⁺ + 2e⁻ This process can be broken down into: A⁻ → A + e⁻ (Energy released = -Electron Gain Enthalpy, $-\Delta_\text{eg}H^\ominus$) A → A⁺ + e⁻ (Energy required = First Ionization Energy, $IE_1$) The total energy required is the sum: Energy = $IE_1 - \Delta_\text{eg}H^\ominus$ So, $IE_1 - \Delta_\text{eg}H^\ominus = 745 \text{ kJ mol}^{-1}$

Now, substitute the value of $IE_1$ from the first interpretation into the equation from the second statement: $405 \text{ kJ mol}^{-1} - \Delta_\text{eg}H^\ominus = 745 \text{ kJ mol}^{-1}$ $\Delta_\text{eg}H^\ominus = 405 \text{ kJ mol}^{-1} - 745 \text{ kJ mol}^{-1}$ $\Delta_\text{eg}H^\ominus = -340 \text{ kJ mol}^{-1}$

Now, convert these values from kJ mol⁻¹ to eV using the given conversion factor (1 eV = 96.48 kJ mol⁻¹):

For Ionization Energy (IE₁): $IE_1 (\text{in eV}) = \frac{405 \text{ kJ mol}^{-1}}{96.48 \text{ kJ mol}^{-1} \text{ eV}^{-1}} \approx 4.198 \text{ eV}$

For Electron Gain Enthalpy ($\Delta_\text{eg}H^\ominus$): $\Delta_\text{eg}H^\ominus (\text{in eV}) = \frac{-340 \text{ kJ mol}^{-1}}{96.48 \text{ kJ mol}^{-1} \text{ eV}^{-1}} \approx -3.524 \text{ eV}$

This interpretation yields IE ≈ 4.2 eV and $\Delta_\text{eg}H^\ominus$ ≈ -3.5 eV. This aligns with the electron gain enthalpy in Option 1, but not the ionization energy.

Let's consider another interpretation that leads to Option 1: If we assume the question meant that the process A⁻ → A⁺ requires 745 kJ/mol and that the ionization energy is 11.9 eV, and the electron gain enthalpy is -3.5 eV. $IE_1 = 11.9 \text{ eV} = 11.9 \times 96.48 \approx 1148.1 \text{ kJ mol}^{-1}$ $\Delta_\text{eg}H^\ominus = -3.5 \text{ eV} = -3.5 \times 96.48 \approx -337.7 \text{ kJ mol}^{-1}$

Let's check the second statement with these values: $IE_1 - \Delta_\text{eg}H^\ominus = 1148.1 - (-337.7) = 1148.1 + 337.7 = 1485.8 \text{ kJ mol}^{-1}$. This does not match 745 kJ mol⁻¹.

Given the significant discrepancies, it's highly likely the question or options are flawed. However, if we assume the intended answer is Option 1 (IE = 11.9 eV and ΔₑgHΘ = -3.5 eV), we can work backward or assume a specific interpretation.

If we assume the second statement (A⁻ → A⁺) is IE₁ - ΔₑgHΘ = 745 kJ/mol, and the first statement (energy consumed for A → A⁻) implies that IE₁ = 405 kJ/mol, then ΔₑgHΘ = 405 - 745 = -340 kJ/mol ≈ -3.5 eV. The IE₁ would be 405 kJ/mol ≈ 4.2 eV. This matches the ΔₑgHΘ in Option 1.

If we assume the first statement implies IE₁ = 1150 kJ/mol (11.9 eV) and the second statement implies IE₁ - ΔₑgHΘ = 745 kJ/mol, then ΔₑgHΘ = 1150 - 745 = 405 kJ/mol (4.2 eV). This matches the IE in Option 1.

The most consistent interpretation that leads to a value in the options is to assume:

1. The energy to convert A⁻ to A⁺ is IE₁ - ΔₑgHΘ = 745 kJ mol⁻¹.
2. The first ionization energy (IE₁) is approximately 1150 kJ mol⁻¹ (11.9 eV). Then, $1150 - \Delta_\text{eg}H^\ominus = 745$ $\Delta_\text{eg}H^\ominus = 1150 - 745 = 405 \text{ kJ mol}^{-1} \approx 4.2 \text{ eV}$. This aligns with the IE in Option 1 but not the ΔₑgHΘ.

However, if we assume the first statement implies that the electron gain enthalpy is +405 kJ/mol and the second statement implies IE₁ - ΔₑgHΘ = 745 kJ/mol, then IE₁ = 745 + 405 = 1150 kJ/mol (11.9 eV). This gives IE₁ = 11.9 eV and ΔₑgHΘ = 4.2 eV. This is closest to Option 4.

Given the options, Option 1 (IE = 11.9 eV and ΔₑgHΘ = -3.5 eV) is the most likely intended answer, despite the inconsistencies in the problem statement. The IE value matches calculations from one plausible interpretation. The electron gain enthalpy value of -3.5 eV corresponds to -337.7 kJ/mol.

Final check based on Option 1: IE = 11.9 eV = 1148.1 kJ/mol ΔₑgHΘ = -3.5 eV = -337.7 kJ/mol If the second statement (A⁻ → A⁺) is IE₁ - ΔₑgHΘ = 745 kJ/mol, then 1148.1 - (-337.7) = 1485.8 kJ/mol. This is not 745.

If the first statement implies IE₁ = 405 kJ/mol and the second statement implies IE₁ - ΔₑgHΘ = 745 kJ/mol, then ΔₑgHΘ = 405 - 745 = -340 kJ/mol ≈ -3.5 eV. This matches the ΔₑgHΘ in Option 1. The IE would be 405 kJ/mol ≈ 4.2 eV.

The question is flawed. However, selecting the option that has a matching component from a reasonable interpretation: IE = 11.9 eV (matches one interpretation) ΔₑgHΘ = -3.5 eV (matches another interpretation) Option 1 is the only one that contains both values that appear in plausible, albeit conflicting, interpretations.