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. `49.43` kcal/mol
B. `41.51` kcal/mol
C. `71.82` kcal/mol
D. `11.76` kcal/mol

Answer

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

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

1. **Identify the given values:**
- First ionization enthalpy of Lithium (Li): \( \Delta H_{ionization} = 5.4 \, \text{eV} \)
- Electron gain enthalpy of Chlorine (Cl): \( \Delta H_{electron \, gain} = -3.6 \, \text{eV} \) (negative because energy is released when an electron is gained)

2. **Calculate ΔH for the reaction:**
The overall enthalpy change for the reaction can be calculated using the formula:
\[ \Delta H = \Delta H_{ionization} + \Delta H_{electron \, gain} \]
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 Joules:**
To convert electron volts to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Delta H = 1.8 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.88 \times 10^{-19} \, \text{J} \]

4. **Convert Joules to Calories:**
Using the conversion \( 1 \, \text{cal} \approx 4.18 \, \text{J} \):
\[ \Delta H = \frac{2.88 \times 10^{-19} \, \text{J}}{4.18 \, \text{J/cal}} \approx 0.689 \times 10^{-19} \, \text{cal} \]

5. **Convert to per mole:**
To find the enthalpy change per mole, multiply by Avogadro's number (\( N_A = 6.022 \times 10^{23} \)):
\[ \Delta H_{per \, mole} = 0.689 \times 10^{-19} \, \text{cal} \times 6.022 \times 10^{23} \approx 41.49 \, \text{kcal} \]

6. **Final Result:**
The change in enthalpy (ΔH) for the reaction in kcal/mole is approximately:
\[ \Delta H \approx 41.49 \, \text{kcal/mole} \]