Dilithium is crucial to the propulsion system of some starships. Dilithium is fomred by the adhesion of two lithium atoms in the gas phase:
`Li(g)+Li(g)toLi_(2)(g)`
The enthalpy of formation of dilithium is not easily measurable by direct means. However, th following thermochemicalparameters are known.
`DeltaH_(f)^(@)` of `Li_(g)=150kJ//mol`
IE of `Li_(g)=5eV//"atom"[1eV//"atom"=100kJ//mol]`
`BE` of `Li_((g))^(+)=130kJ//mol`
`IE` of `Li_(2(g))=5eV//"molecule"`
`[IE=` ionisation energy] `[BE=` bond energy]
Q. What is bond energy of `Li_(2(g))`?

Options

A. `-230kJ//mol`
B. `230kJ//mol`
C. `170kJ//mol`
D. `130kJ//mol`

Explanation

To find the bond energy of dilithium (Li₂(g)), we can use Hess's law and the given thermochemical parameters. Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Processes:**
We have several processes that we need to consider:
- **Process A:** Formation of lithium gas from solid lithium.
- **Process B:** Ionization of one lithium atom to form Li⁺.
- **Process C:** Formation of Li₂ from two lithium atoms.
- **Process D:** Ionization of Li₂ to form Li₂⁺.
- **Process E:** Dissociation of Li₂ into two lithium atoms.
- **Process F:** Formation of Li₂ from Li and Li⁺.

2. **Write the Known Enthalpy Changes:**
- **ΔH_f (Li(g)) = 150 kJ/mol**
- **IE (Li(g)) = 5 eV/atom = 500 kJ/mol** (since 1 eV = 100 kJ/mol)
- **BE (Li⁺(g)) = 130 kJ/mol**
- **IE (Li₂(g)) = 5 eV/molecule = 500 kJ/mol**

3. **Apply Hess's Law:**
According to Hess's law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can set up the equation:
\[ ΔH_A + ΔH_C + ΔH_D + ΔH_F = 0 \]

4. **Substituting Known Values:**
- **ΔH_A = 2 * ΔH_f (Li(g)) = 2 * 150 kJ/mol = 300 kJ/mol**
- **ΔH_B = +500 kJ/mol** (ionization of Li)
- **ΔH_C = -BE (Li₂(g)) = -BE**
- **ΔH_D = -500 kJ/mol** (ionization of Li₂)
- **ΔH_F = 0** (since it is the formation of Li₂ from its elements)

Plugging these into the equation gives:
\[ 300 + (-BE) + (-500) + 0 = 0 \]

5. **Solving for BE:**
Rearranging the equation:
\[ -BE = -300 + 500 \]
\[ -BE = 200 \]
Therefore:
\[ BE = 200 \text{ kJ/mol} \]

### Final Answer:
The bond energy of Li₂(g) is **200 kJ/mol**.