If the first ionization enthalpy of Li is 5.4 eV and the elec- tron gain enthalpy of chlorine is 3.6 eV then the `Delta H` in kcal/mole for the reaction will be
Li(g)+Cl(g)`to Li^(+) (g) Cl^(-) (g)` (Presume that the pressure is so low that the ions do not combine with each other)

Options

A. 70
B. 100
C. 170
D. 270

Answer

To solve the problem, we need to calculate the change in enthalpy (ΔH) for the reaction:

\[ \text{Li(g)} + \text{Cl(g)} \rightarrow \text{Li}^+(g) + \text{Cl}^-(g) \]

### Step-by-Step Solution:

1. **Identify the Given Values**:
- First ionization enthalpy of Lithium (Li): \( I_E = 5.4 \, \text{eV} \)
- Electron gain enthalpy of Chlorine (Cl): \( E_A = -3.6 \, \text{eV} \) (Note: This value is negative because energy is released when an electron is gained.)

2. **Calculate the Change in Enthalpy (ΔH)**:
The change in enthalpy for the reaction can be calculated using the formula:
\[ \Delta H = I_E + E_A \]
Substituting the values:
\[ \Delta H = 5.4 \, \text{eV} + (-3.6 \, \text{eV}) = 5.4 \, \text{eV} - 3.6 \, \text{eV} = 1.8 \, \text{eV} \]

3. **Convert ΔH from eV to kJ**:
To convert electron volts to kilojoules, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Therefore,
\[ 1.8 \, \text{eV} = 1.8 \times 1.6 \times 10^{-19} \, \text{J} = 2.88 \times 10^{-19} \, \text{J} \]

4. **Convert Joules to kJ**:
\[ 2.88 \times 10^{-19} \, \text{J} = 2.88 \times 10^{-22} \, \text{kJ} \]

5. **Calculate ΔH per mole**:
Using Avogadro's number (\( N_A = 6.02 \times 10^{23} \)):
\[ \Delta H \text{ (per mole)} = 2.88 \times 10^{-22} \, \text{kJ} \times 6.02 \times 10^{23} = 172 \, \text{kJ/mol} \]

6. **Convert kJ to kcal**:
Since \( 1 \, \text{kcal} = 4.184 \, \text{kJ} \):
\[ \Delta H \text{ (in kcal)} = \frac{172 \, \text{kJ}}{4.184} \approx 41.1 \, \text{kcal/mol} \]

### Final Answer:
The change in enthalpy (ΔH) for the reaction is approximately **41.1 kcal/mol**.