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Diborane is a potential rocket fuel with 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)(g), DeltaH = -1273 kJ "mol"^(-1)........(1) H_(2)(g) + 1/2O_(2) (g) to H_(2)O(l), DeltaH = -286 kJ"mol"^(-1).........(2) H_(2)O(l) to H_(2)O(g), DeltaH= 44 kJ "mol"^(-1) H_(2)O(l) to H_(2)O(g), DeltaH = 44 kJ "mol"^(-1).........(4) 2B(s) + 3H_(2)(g) to B_(2)H_(8)(g), DeltaH = 36 kJ"mol"^(-1)............(5)

Answer»

Solution :Finally, we have to get, `B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)O(g)`
MULTIPLY equation (2) and (3) by 3, add 1,2,3 and substract (4) from that
(i) `2B(s) + 3//2 O_(2)(g) to B_(2)O_(3)(s), DeltaH = -1273 KJ "mol"^(-1)`
(II) `3H_(2)(g) + 3//2O_(2)(g) to 3H_(2)O(l), DeltaH = -286 xx 3 kJ"mol"^(-1)`
(III) `3H_(2)O(l) to 3H_(2)O(g), DeltaH = 132 kJ"mol"^(-1)`
(iv) `B_(2)H_(6)(g) to 2B(s)+ 3H_(2)(g), DeltaH = -36 kJ"mol"^(-1)`
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`B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)+ 3H_(2)O(g), DeltaH= -2035 kJ "mol"^(-1)`


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