The amount of energy when million atoms of iodine are completely converted into `I^(-)` ions in the vapour state according to the equation,`I_((g))+e^(-)`to`I^(-)_((g)) ` is `4.9X10^(-13)` J.What would be the electron gain enthalpy of iodine in terms of KJ `mol^(-1)` and eV per atom? (A)295 , 3.06 (B)− 295, − 3.06 (C)439 , 5.09 (D)− 356 , − 7.08

Options

A. `295,3.06`
B. `-295-3.06`
C. `439,5.09`
D. `-356,-7.08`

Explanation

To find the electron gain enthalpy of iodine based on the given information, we can follow these steps:

### Step 1: Understand the given data
We know that the energy required to convert 1 million atoms of iodine (I) into I⁻ ions is given as \(4.9 \times 10^{-13}\) J.

### Step 2: Calculate the energy per atom
To find the energy per atom, we divide the total energy by the number of atoms:
\[ \text{Energy per atom} = \frac{4.9 \times 10^{-13} \text{ J}}{1,000,000 \text{ atoms}} = 4.9 \times 10^{-19} \text{ J/atom} \]

### Step 3: Convert energy from Joules to kilojoules
To convert the energy from joules to kilojoules, we divide by 1000:
\[ \text{Energy per atom in kJ} = \frac{4.9 \times 10^{-19} \text{ J}}{1000} = 4.9 \times 10^{-22} \text{ kJ/atom} \]

### Step 4: Convert energy per atom to kJ/mol
Since 1 mole of atoms contains \(6.022 \times 10^{23}\) atoms (Avogadro's number), we can calculate the energy per mole:
\[ \text{Energy per mole} = 4.9 \times 10^{-19} \text{ J/atom} \times 6.022 \times 10^{23} \text{ atoms/mol} = 295 \text{ kJ/mol} \]

### Step 5: Determine the sign of the electron gain enthalpy
Since energy is released when an electron is gained (the process is exothermic), the electron gain enthalpy will be negative:
\[ \text{Electron gain enthalpy} = -295 \text{ kJ/mol} \]

### Step 6: Convert energy per atom to eV
To convert from joules to electron volts (eV), we use the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\):
\[ \text{Energy per atom in eV} = \frac{4.9 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 3.06 \text{ eV/atom} \]

### Final Answer
Thus, the electron gain enthalpy of iodine is:
- In kJ/mol: \(-295 \text{ kJ/mol}\)
- In eV per atom: \(-3.06 \text{ eV/atom}\)

The correct answer is (B) \(-295, -3.06\).

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