Determine the enthalpy of formation of `B_(2)H_(6)`(g) in kJ/mol of the following reaction :
`B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(3)(s)+3H_(2)O(g)`,
Given : `Delta_(r )H^(@)=-1941 " kJ"//"mol", " "DeltaH_(f)^(@)(B_(2)O_(3),s)=-1273" kJ"//"mol,"`
`DeltaH_(f)^(@)(H_(2)O,g)=-241.8 " kJ"//"mol"`

Options

A. `-75.6`
B. `+75.6`
C. `-57.4`
D. `-28.4`

Explanation

To determine the enthalpy of formation of \( B_2H_6(g) \) from the given reaction:

\[ B_2H_6(g) + 3O_2(g) \rightarrow B_2O_3(s) + 3H_2O(g) \]

we can use the following information provided:

- The enthalpy change of the reaction, \( \Delta_r H^\circ = -1941 \, \text{kJ/mol} \)
- The enthalpy of formation of \( B_2O_3(s) \), \( \Delta H_f^\circ(B_2O_3, s) = -1273 \, \text{kJ/mol} \)
- The enthalpy of formation of \( H_2O(g) \), \( \Delta H_f^\circ(H_2O, g) = -241.8 \, \text{kJ/mol} \)

### Step-by-step solution:

1. **Identify the enthalpy of formation of the products:**
- For \( B_2O_3(s) \): The enthalpy of formation is \( -1273 \, \text{kJ/mol} \).
- For \( H_2O(g) \): Since there are 3 moles of water produced, we multiply the enthalpy of formation by 3:
\[ \Delta H_f^\circ(H_2O, g) = 3 \times (-241.8) = -725.4 \, \text{kJ/mol} \]

2. **Calculate the total enthalpy of formation for the products:**
- The total enthalpy of formation for the products is:
\[ \Delta H_{products} = \Delta H_f^\circ(B_2O_3, s) + 3 \times \Delta H_f^\circ(H_2O, g) \]
\[ = -1273 + (-725.4) = -1998.4 \, \text{kJ/mol} \]

3. **Set up the equation using the enthalpy change of the reaction:**
- The enthalpy change of the reaction can be expressed as:
\[ \Delta_r H^\circ = \Delta H_{products} - \Delta H_{reactants} \]
- Rearranging gives:
\[ \Delta H_{reactants} = \Delta H_{products} - \Delta_r H^\circ \]

4. **Substituting the known values:**
- We know \( \Delta_r H^\circ = -1941 \, \text{kJ/mol} \):
\[ \Delta H_{reactants} = -1998.4 - (-1941) \]
\[ = -1998.4 + 1941 = -57.4 \, \text{kJ/mol} \]

5. **Conclusion:**
- The enthalpy of formation of \( B_2H_6(g) \) is:
\[ \Delta H_f^\circ(B_2H_6, g) = -57.4 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of formation of \( B_2H_6(g) \) is \( -57.4 \, \text{kJ/mol} \).

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