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//`mole

Explanation

The concerned chemical reaction is
`B_(2) H_(6 (g)) + 3O_(2 (g)) rarr B_(2)O_((s)) + 3H_(2) O_((g)), Delta H = ?`
The enthalpy change can be calculated in the followingway.
`Delta H = [Delta H_(B_(2) O_(3) (s)) + 3 Delta H_(H_(2) O (g))] - Delta H_(B_(2)H_(6)(s)), ( :. Delta H_(f)^(0) " of " O_(2) = 0) Delta H_(B_(2)O (g))` can be obtained by adding `Delta H_(H_(2)O(l))` and `Delta H_(H_(2) O(g))`,
i.e. `-286 + 44 = - 242 kJ mol^(-1)`
`Delta H = [-1273 + 3 -242] - 36 kJ mol^(-1) = - 1273 - 726 - 36 = - 2035 kJ mol^(-1)`