Diborane is a potentail rocket fuel which undergoes combusion according to the reaction
`B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(3)(s)+3H_(2)O(g)`
From the following data, calculate the enthalpy change for the combustion of diborane
`2B(s)+(3//2)O_(2)(g)rarrB_(2)O_(3)(s) DeltaH= -1273 kJ mol^(-1)`
`H_(2)(g)+(1//2)O_(2)(g)rarrH_(2)O(1) " "Delta H= -286 kJ mol^(-1)`
`H_(2)O(1)rarrH_(2)O(g)" "DeltaH=44 kJ mol^(-1)`
`2B(s)+3H_(2)(g)rarrB_(2)H_(6)(g)" "Delta H= 36kJ mol^(-1)`

Explanation

To calculate 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.

### Step-by-Step Solution:

1. **Write the combustion reaction for diborane:**
\[ B_2H_6(g) + 3O_2(g) \rightarrow B_2O_3(s) + 3H_2O(g) \]

2. **Identify the given reactions and their enthalpy changes:**
- \(2B(s) + \frac{3}{2}O_2(g) \rightarrow B_2O_3(s) \quad \Delta H = -1273 \, \text{kJ/mol}\)
- \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H = -286 \, \text{kJ/mol}\)
- \(H_2O(l) \rightarrow H_2O(g) \quad \Delta H = 44 \, \text{kJ/mol}\)
- \(2B(s) + 3H_2(g) \rightarrow B_2H_6(g) \quad \Delta H = 36 \, \text{kJ/mol}\)

3. **Calculate the enthalpy change for the formation of water vapor (H₂O(g)):**
- Combine the second and third reactions:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H = -286 \, \text{kJ/mol} \]
\[ H_2O(l) \rightarrow H_2O(g) \quad \Delta H = 44 \, \text{kJ/mol} \]
- The total enthalpy change for the formation of water vapor is:
\[ \Delta H_{H_2O(g)} = -286 + 44 = -242 \, \text{kJ/mol} \]

4. **Use Hess's law to calculate the enthalpy change for the combustion of diborane:**
- The overall reaction can be expressed in terms of the enthalpy changes:
\[ \Delta H = \Delta H_{B_2O_3} + 3 \Delta H_{H_2O(g)} - \Delta H_{B_2H_6} \]
- Substitute the known values:
\[ \Delta H = (-1273) + 3(-242) - (36) \]
- Calculate:
\[ \Delta H = -1273 - 726 - 36 = -2035 \, \text{kJ/mol} \]

5. **Final Result:**
- The enthalpy change for the combustion of diborane is:
\[ \Delta H = -2035 \, \text{kJ/mol} \]