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Diborane is a potential rocket fuel whic...

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

Text Solution

Verified by Experts

The equation is :
`B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)O(g) DeltaH = ?`
(i)`3B(s) + 3/2 O_(2)(g) to B_(2)O_(3)(s) DeltaH = -1273 kJ "mol"^(-1)`
(ii)` H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) DeltaH = -286 kJ "mol"^(-1)`
(iii) `H_(2)O(l) to H_(2)O(g) DeltaH = 44 kJ "mol"^(-1)`
(iv)`2B(s) + 3H_(2)(g) to B_(2)H_(6)(g) DeltaH = 36 kJ "mol"^(-1)`
Subtract eq.(iv) from eq.(i)
(v) `B_(2)H_(6) (g) + 3/2O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)(g) DeltaH = -1309 kJ "mol"^(-1)`
Multiply eq.(ii) by 3
(vi) `3H_(2)(g) + 3/2O_(2)(g) to 3H_(2)O(l) DeltaH = -858 kJ "mol"^(-1)`
Add`B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)O(l) DeltaH = -2167 kJ "mol"^(-1)`
Multiply eq.(iii) by 3
(vii) `3H_(2)O(l) to 3H_(2)O(g) DeltaH = 132 kJ "mol"^(-1)`
Add `B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)O(g) Delta = -2035 kJ "mol"^(-1)`.
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