Answer: The enthalpy of burning of 1 mole of [tex]B_5H_9(s)[/tex] is -4543.05 kJ/mol
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/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]
For the given chemical reaction:
[tex]2B_5H_9(s)+12O_2\rightarrow 5B_2O_3(s)+9H_2O(l)[/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)})+(9\times \Delta H^o_f_{(H_2O)})]-[(2\times \Delta H^o_f_{(B_5H_9)})+(12\times \Delta H^o_f_{(O_2)})][/tex]
We are given:
[tex]\Delta H^o_f_{(B_2O_3)}=-1273.5kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_f_{(H_2O)}=-285.8kJ/mol\\\Delta H^o_f_{(B_5H_9)}=73.2kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(5\times (-1273.5))+(9\times (-285.8))]-[(2\times (73.2))+(12\times 0)]\\\\\Delta H^o_{rxn}=-9086.1kJ/mol[/tex]
The enthalpy of burning of 2 moles of [tex]B_5H_9(s)[/tex] is coming out to be -9086.1 kJ/mol
So, enthalpy of burning of 1 mole of [tex]B_5H_9[/tex] will be = [tex]\frac{-9086.1kJ/mol}{2}\times 1=-4543.05kJ/mol[/tex]
Hence, the enthalpy of burning of 1 mole of [tex]B_5H_9(s)[/tex] is -4543.05 kJ/mol
