The compound diborane (B2H6) is used as a rocket fuel. The equation for the combustion of diborane is shown below.B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O(l)Calculate the enthalpy of combustion of diborane (B2H6) in kJ mol−1 to 1 d.p, using the following data.2B(s) + 3H2(g) → B2H6 (g) ΔH = 31.33 kJ mol−1H2(g) + ½O2(g) → H2O(l) ΔH = −286 kJ mol−12B(s) + 1½O2(g) → B2O3(s) ΔH = −1274 kJ mol−1

To calculate the enthalpy of combustion of diborane, we need to use Hess's Law, which states that the total enthalpy change for a reaction is the same no matter how many steps it takes.

The combustion reaction is: B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O(l)

We can break this down into three steps using the given reactions:

1. Formation of B2H6: 2B(s) + 3H2(g) → B2H6 (g) ΔH = 31.33 kJ mol−1
2. Formation of H2O: H2(g) + ½O2(g) → H2O(l) ΔH = −286 kJ mol−1
3. Formation of B2O3: 2B(s) + 1½O2(g) → B2O3(s) ΔH = −1274 kJ mol−1

We need to adjust these reactions to match the combustion reaction:

1. Formation of B2H6: 2B(s) + 3H2(g) → B2H6 (g) ΔH = 31.33 kJ mol−1 (no change)
2. Formation of H2O: 3[H2(g) + ½O2(g) → H2O(l)] ΔH = 3 * -286 kJ mol−1 = -858 kJ mol−1 (multiplied by 3 to get 3 moles of H2O)
3. Formation of B2O3: 2B(s) + 1½O2(g) → B2O3(s) ΔH = −1274 kJ mol−1 (no change)

Now, add up the adjusted ΔH values to get the total enthalpy change for the combustion reaction:

ΔH_combustion = ΔH1 + ΔH2 + ΔH3 = 31.33 kJ mol−1 - 858 kJ mol−1 - 1274 kJ mol−1 = -2100.67 kJ mol−1

So, the enthalpy of combustion of diborane is -2100.7 kJ mol−1 to 1 decimal place.