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 `DeltaH_(f)^(@)` of `Li_(2(g))`?

Options

A. `130kJ//mol`
B. `170kJ//mol`
C. `270kJ//mol`
D. `-270kJ//mol`

Answer

To find the enthalpy of formation of dilithium (Li₂(g)), we can use the given thermochemical parameters and apply Hess's law. Let's break down the steps systematically.

### Step-by-Step Solution:

1. **Identify the Reaction and Parameters**:
The reaction we are considering is:
\[ \text{Li(g)} + \text{Li(g)} \rightarrow \text{Li}_2(g) \]
We are given the following parameters:
- \(\Delta H_f^\circ\) of \(\text{Li(g)} = 150 \, \text{kJ/mol}\)
- Ionization energy (IE) of \(\text{Li(g)} = 5 \, \text{eV/atom} = 500 \, \text{kJ/mol}\)
- Bond energy (BE) of \(\text{Li}^+(g) = 130 \, \text{kJ/mol}\)
- Ionization energy of \(\text{Li}_2(g) = 5 \, \text{eV/molecule} = 500 \, \text{kJ/mol}\)

2. **Calculate the Enthalpy Change for Each Step**:
- **Step A**: Formation of 2 moles of \(\text{Li(g)}\) from solid lithium:
\[ A = 2 \times \Delta H_f^\circ(\text{Li(g)}) = 2 \times 150 \, \text{kJ/mol} = 300 \, \text{kJ} \]

- **Step B**: Ionization of 2 moles of \(\text{Li(g)}\) to form 2 moles of \(\text{Li}^+(g)\):
\[ B = 2 \times \text{IE}(\text{Li(g)}) = 2 \times 500 \, \text{kJ/mol} = 1000 \, \text{kJ} \]

- **Step C**: Formation of 1 mole of \(\text{Li}_2(g)\) from 2 moles of \(\text{Li}^+(g)\):
\[ C = -\text{BE}(\text{Li}_2) = -130 \, \text{kJ/mol} \]

- **Step D**: Ionization of \(\text{Li}_2(g)\) to form 2 moles of \(\text{Li}^+(g)\):
\[ D = \text{IE}(\text{Li}_2) = 500 \, \text{kJ/mol} \]

3. **Apply Hess's Law**:
According to Hess's law, the total enthalpy change for the reaction can be calculated as:
\[ \Delta H_f^\circ(\text{Li}_2(g)) = A + B + C - D \]
Substituting the values:
\[ \Delta H_f^\circ(\text{Li}_2(g)) = 300 \, \text{kJ} + 1000 \, \text{kJ} - 130 \, \text{kJ} - 500 \, \text{kJ} \]
\[ \Delta H_f^\circ(\text{Li}_2(g)) = 300 + 1000 - 130 - 500 = 670 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta H_f^\circ(\text{Li}_2(g)) = 670 \, \text{kJ/mol} \]