Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g
Explanation :
Enthalpy change : It 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 ![\Delta H^o](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo)
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
The equilibrium reaction follows:
![2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)](https://tex.z-dn.net/?f=2B_5H_9%28l%29%2B12O_2%28g%29%5Crightleftharpoons%205B_2O_3%28s%29%2B9H_2O%28l%29)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%28n_%7B%28B_2O_3%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_2O_3%29%7D%29%2B%28n_%7B%28H_2O%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%29%7D%29%5D-%5B%28n_%7B%28B_5H_9%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_5H_9%29%7D%29%2B%28n_%7B%28O_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%29%7D%29%5D)
We are given:
![\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_f_%7B%28B_5H_9%28l%29%29%7D%3D73.2kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%3D0kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28B_2O_3%28s%29%29%7D%3D-1271.94kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28H_2O%28l%29%29%7D%3D-285.83kJ%2Fmol)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20-1271.94%29%2B%289%5Ctimes%20-285.83%29%5D-%5B%282%5Ctimes%2073.2%29%2B%2812%5Ctimes%200%29%5D%3D-9078.57kJ%2Fmol)
Now we have to calculate the heat released per gram of the compound reacted with oxygen.
From the reaction we conclude that,
As, 2 moles of compound released heat = -9078.57 kJ
So, 1 moles of compound released heat = ![\frac{-9078.57}{2}=-4539.28kJ](https://tex.z-dn.net/?f=%5Cfrac%7B-9078.57%7D%7B2%7D%3D-4539.28kJ)
For per gram of compound:
Molar mass of
= 63.12 g/mole
![\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Cfrac%7B-4539.28%7D%7B63.12%7D%3D-71.915kJ%2Fg)
Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g