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 = 40.29 kJ mol−1H2(g) + ½O2(g) → H2O(l) ΔH = −286 kJ mol−12B(s) + 1½O2(g) → B2O3(s) ΔH = −1274 kJ mol−1
Question
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 = 40.29 kJ mol−1H2(g) + ½O2(g) → H2O(l) ΔH = −286 kJ mol−12B(s) + 1½O2(g) → B2O3(s) ΔH = −1274 kJ mol−1
Solution
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 what route is taken.
The combustion reaction is: B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O(l)
We can break this down into three steps using the given reactions:
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Reverse the formation reaction of B2H6: B2H6 (g) → 2B(s) + 3H2(g), ΔH = -40.29 kJ mol−1
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Use the formation reaction of H2O: 3H2(g) + 1.5O2(g) → 3H2O(l), ΔH = 3(-286) kJ mol−1 = -858 kJ mol−1
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Use the formation reaction of B2O3: 2B(s) + 1.5O2(g) → B2O3(s), ΔH = -1274 kJ mol−1
The total enthalpy change for the combustion reaction is the sum of the enthalpy changes for these steps:
ΔH_combustion = (-40.29 kJ mol−1) + (-858 kJ mol−1) + (-1274 kJ mol−1) = -2172.29 kJ mol−1
So, the enthalpy of combustion of diborane is -2172.3 kJ mol−1 to 1 decimal place.
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