Respuesta :
Answer: The amount of energy released per gram of [tex]B_5H_9[/tex] is -71.92 kJ
Explanation:
For the given chemical reaction:
[tex]2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}][/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})][/tex]
Taking the standard enthalpy of formation:
[tex]\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ[/tex]
We know that:
Molar mass of pentaborane -9 = 63.12 g/mol
By Stoichiometry of the reaction:
If 2 moles of [tex]B_5H_9[/tex] produces -9078.57 kJ of energy.
Or,
If [tex](2\times 63.12)g[/tex] of [tex]B_5H_9[/tex] produces -9078.57 kJ of energy
Then, 1 gram of [tex]B_5H_9[/tex] will produce = [tex]\frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ[/tex] of energy.
Hence, the amount of energy released per gram of [tex]B_5H_9[/tex] is -71.92 kJ
