Diborane is a potential rocket fuel which undergoes combustion according to the reaction,
`B_(2)H_(6)(g)+3O_(2)(g)toB_(2)O_(3)(s)+3H_(2)O(g)`
from the following data, the enthalpy change for the combustion of diborane will be :
`2B(s)+(3)/(2)O_(2)(g)toB_(2)O_(3)(s)," "DeltaH=-1273 " kJ"`
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)," "DeltaH=-286 " kJ"`
`H_(2)O(l)toH_(2)O(g)," "DeltaH=44" kJ"`
`2B(s)+3H_(2)(g)toB_(2)H_(6)(g)," "DeltaH=46" kJ"`

Options

A. `-2079 " kJ" " mol"^(-1)`
B. `-1091 " kJ" " mol"^(-1)`
C. `-2045 " kJ" " mol"^(-1)`
D. `-762 " kJ" " mol"^(-1)`

Answer

To find the enthalpy change for the combustion of diborane (B₂H₆), we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction. We will manipulate the given reactions to derive the desired reaction.

### Step-by-Step Solution:

1. **Write the target reaction**:
\[ \text{B}_2\text{H}_6(g) + 3\text{O}_2(g) \rightarrow \text{B}_2\text{O}_3(s) + 3\text{H}_2\text{O}(g) \]

2. **List the given reactions and their enthalpy changes**:
- Reaction 1:
\[ 2\text{B}(s) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{B}_2\text{O}_3(s), \quad \Delta H = -1273 \text{ kJ} \]
- Reaction 2:
\[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l), \quad \Delta H = -286 \text{ kJ} \]
- Reaction 3:
\[ \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g), \quad \Delta H = 44 \text{ kJ} \]
- Reaction 4:
\[ 2\text{B}(s) + 3\text{H}_2(g) \rightarrow \text{B}_2\text{H}_6(g), \quad \Delta H = 46 \text{ kJ} \]

3. **Manipulate the reactions**:
- We need to reverse Reaction 4 to get B₂H₆ on the reactant side:
\[ \text{B}_2\text{H}_6(g) \rightarrow 2\text{B}(s) + 3\text{H}_2(g), \quad \Delta H = -46 \text{ kJ} \]
- We will use Reaction 1 as is:
\[ 2\text{B}(s) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{B}_2\text{O}_3(s), \quad \Delta H = -1273 \text{ kJ} \]
- We will use Reaction 2 three times to get 3 moles of H₂O:
\[ 3\left(\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)\right), \quad \Delta H = 3 \times (-286) \text{ kJ} = -858 \text{ kJ} \]
- We will use Reaction 3 three times to convert H₂O(l) to H₂O(g):
\[ 3\left(\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)\right), \quad \Delta H = 3 \times 44 \text{ kJ} = 132 \text{ kJ} \]

4. **Combine the enthalpy changes**:
Now we can sum the enthalpy changes:
\[ \Delta H_{\text{total}} = (-46) + (-1273) + (-858) + 132 \]
\[ \Delta H_{\text{total}} = -46 - 1273 - 858 + 132 = -2045 \text{ kJ} \]

5. **Final result**:
The enthalpy change for the combustion of diborane is:
\[ \Delta H = -2045 \text{ kJ} \]