Diborane is a potential rocket fuel which undergoes combustion according to the reaction,
`B_(2)H_(6) (g) + 3O_(2) (g) rarr B_(2) O_(3) (s) + 3H_(2) O (g)`
From the following data, calculate the enthalpy change for the combustion of diborane:
(i) `2 B (s) + ((3)/(2)) O_(2) (g) rarr B_(2) O_(3) (s) , Delta H = 0 1273 k J//mol`
(ii) `H_(2) (g) + ((1)/(2)) O_(2) (g) rarr H_(2) O (l), Delta H = -286 kJ//mol`
(iii) `H_(2) O (l) rarr H_(2)O (g) Delta H = 44 kJ//mol`
(iv) `2B (s) + 3H_(2) (G) rarr B_(2) H_(6) (g), Delta H = 36 kJ//mol`

Answer

`-2035 kJ mol^(-1)`

Explanation

Given reaction can be obtained as follows:
Add
`{:(B_(2)H_(6)(g)rarr2B(s)+3H_(2)(d)" " DeltaH= -36kJ mol^(-1)),(2B(s)+(3//2)O_(2)rarrB_(2)O_(3)(s)" " DeltaH= -1273kJmol^(-1)),(3[H_(2)O(1)rarrH_(2)O(g)]" " DeltaH= 3xx44 kJ mol^(-1)),(3"["H_(9)(s)+(1//2)O_(2)(g)rarrH_(2)O(l)" " DeltaH= -3xx286kJ mol^(-1)),(bar(B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(3)(s)+3H_(2)O(g) DeltaH= -2035 kJ mol^(-1))):}`