Question

Pentaborane-9,

, is a colorless, highly reactive liquid thatwill burst into flame
when exposed to oxygen. the reactionis:

Calculate the kilojoules of heat released per gram of
thecompound reacted with oxygen. The standard enthalpy of formation
ofB5 H9 is 73.2 kJ/mol.

Answer 1

Given question is incomplete. The complete question is as follows.

Pentaborane (
`B_(5)H_(9)`) is a colorless highly reactive liquid that will burst into flames when exposed to oxygen.the reaction is:

`2B_(5)H_(9)(l) + 12O_(2) \rightarrow 5B_(2)O_(2)(s) + 9H_(2)O(l)`

Calculate the kilojoules of heat released per gram of the compound reacted with oxygen.the standard enthalpy of formation
`B_(5)H_(9)(l)` ,
`B_(2)O_(3)(s)`, and
`H_(2)O(l)` are 73.2, -1271.94, and -285.83 kJ/mol, respectively.

Step-by-step explanation:

As the given reaction is as follows.

`2B_(5)H_(9)(l) + 12O_(2) \rightarrow 5B_(2)O_(2)(s) + 9H_(2)O(l)`

Therefore, formula to calculate the heat energy released is as follows.

`\Delta H = \sum H_(products) - \Delta H_(reactants)`

Hence, putting the given values into the above formula is as follows.

`\Delta H = \sum H_(products) - \Delta H_(reactants)`

=
`5 * (-1271.94 kJ/mol) + 9 * (-285.83 kJ/mol) - 2 * (73.2 kJ/mol) - 12(0)`

= -9078.59 kJ/mol

Since, 2 moles of Pentaborane reacts with oxygen. Therefore, heat of reaction for 2 moles of Pentaborane is calculated as follows.

`\frac{\Delta H}{2 * \text{molar mass of pentaborane}}`

`(-9078.59 kJ/mol)/(2 * 63.12 g/mol)`

= -71.915 kJ/g

Thus, we can conclude that heat released per gram of the compound reacted with oxygen is 71.915 kJ/g.