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 and data, we can follow these steps:

### Step 1: Write the reaction and identify the given data
The reaction is:
\[ B_2H_6(g) + 3O_2(g) \rightarrow B_2O_3(s) + 3H_2O(g) \]
Given data:
- \( \Delta_r H^\circ = -1941 \, \text{kJ/mol} \) (enthalpy change for the reaction)
- \( \Delta H_f^\circ (B_2O_3, s) = -1273 \, \text{kJ/mol} \) (enthalpy of formation of \( B_2O_3 \))
- \( \Delta H_f^\circ (H_2O, g) = -241.8 \, \text{kJ/mol} \) (enthalpy of formation of \( H_2O \))

### Step 2: Write the enthalpy change equation
The enthalpy change for the reaction can be expressed using the enthalpies of formation:
\[ \Delta_r H^\circ = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]
Substituting the known values:
\[ -1941 \, \text{kJ/mol} = \left( \Delta H_f^\circ (B_2O_3) + 3 \times \Delta H_f^\circ (H_2O) \right) - \Delta H_f^\circ (B_2H_6) \]

### Step 3: Substitute the values into the equation
Substituting the values of \( \Delta H_f^\circ \):
\[ -1941 = \left( -1273 + 3 \times (-241.8) \right) - \Delta H_f^\circ (B_2H_6) \]

### Step 4: Calculate the total enthalpy of products
Calculating \( 3 \times (-241.8) \):
\[ 3 \times (-241.8) = -725.4 \]
Now substituting this back into the equation:
\[ -1941 = \left( -1273 - 725.4 \right) - \Delta H_f^\circ (B_2H_6) \]
Calculating the sum:
\[ -1273 - 725.4 = -1998.4 \]
So the equation now looks like:
\[ -1941 = -1998.4 - \Delta H_f^\circ (B_2H_6) \]

### Step 5: Solve for \( \Delta H_f^\circ (B_2H_6) \)
Rearranging the equation to isolate \( \Delta H_f^\circ (B_2H_6) \):
\[ \Delta H_f^\circ (B_2H_6) = -1998.4 + 1941 \]
Calculating the right side:
\[ \Delta H_f^\circ (B_2H_6) = -57.4 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of formation of \( B_2H_6(g) \) is:
\[ \Delta H_f^\circ (B_2H_6) = -57.4 \, \text{kJ/mol} \]