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

Options

A. `1273kJ mol ^(-1)`
B. `-1273 kJ mol ^(-1)`
C. `2035 kJ mol ^(-1)`
D. ` - 2035 kJ mol ^(-1)`

Explanation

From the given data, the heat of conbustion reaction of diborane can be calculated as
`B _(2) H _(6) (g) + 3 O _(2) (g) to B _(2) O _(3) (s) + 3 H _(2)O (g)`
`Delta _(r) H ^(@) = Delta H _(f) ^(@) (B _(2) O _(3)) + 3 xx [ Delta H _(f) ^(@)(H _(2) O ) + Delta H _(ap ) ^(@) (H_(2) O ) ]- Delta H _(f) ^(@) (B _(2) H _(6))`
Now, `Delta _( vap ) H ^(@) (H _(2) O ) = Delta H _(f) ^(@) ( H _(2) O ) + 44 = - 2 42 kJ mol ^(-1) Delta _(c) H ^(@) ( B _(2) H _(6)) =- 1273 + 3 ( - 242) - 36`
Therefore, `=- 2035 kJ mol ^(-1).`