pentaborane-9, bshg; is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. the reaction is 2bshg(l) 1202(g) sbzos(s) 9hzo(l) calculate the kilojoules of heat released per gram of the compound reacted with oxygen. the standard enthalpy of formation of shg is 73.2 kj mol.

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21-10-2022
Chemistry

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pentaborane-9, bshg; is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. the reaction is 2bshg(l) 1202(g) sbzos(s) 9hzo(l) calculate the kilojoules of heat released per gram of the compound reacted with oxygen. the standard enthalpy of formation of shg is 73.2 kj mol.

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kjuli1766 kjuli1766 · Answer 1 · 2022-10-26T08:01:43+02:00

The kilojoules of heat released per gram of the compound reacted with oxygen is -72.052 kj/g

The standard enthalpy of formation is a measure of the strength launched or fed on while one mole of a substance is created under trendy situations from its natural elements. The symbol of the standard enthalpy of formation is ΔH f. = A degree means that it's a fashionable enthalpy change.

Enthalpy of a formation refers to the trade in enthalpy for the formation of one mole of a substance from the maximum solid form of its constituent factors. change in enthalpy is symbolized by means of delta H and the f stands for formation.

Given,

compound ΔH°f

B₅H₉ 73.2

B₂O₃ -1271.94

H₂O -285.83

Balanced chemical equation:-

2B₅H₉ (l) + 12O₂ +(g) => 5B₂O₃ + 9H₂O(l)

From Hess's law

ΔH°min = Σ enthalpies of products - Σ enthalpies of reactants

= 5× (-1271.94) + 9×( -285.83) - 2(73.2) kj/mol

ΔH°min = -9078.59 kj/mol

It is seen that 2 moles of B₅H₉ react with O₂ is -9078.59 kj/mol

For 1 mole = -9078.59 kj/mol/2

= 4539.295 KJ/mol

The energy released per gram = energy released per mol/molar mass of B₅H₉

= 4539.295 kj/mol/63 g/mol

energy released per gram = -72.052 kj/g

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