Answer : The heat released per gram of the compound reacted with oxygen is, 71.915 kJ
Solution :
The balanced chemical reaction is,
![2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)](https://tex.z-dn.net/?f=2B_5H_9%28l%29%2B12O_2%28g%29%5Crightarrow%205B_2O_3%28s%29%2B9H_2O%28l%29)
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{H_2O}\times \Delta H_{H_2O})+(n_{B_2O_3}\times \Delta H_{B_2O_3})]-[(n_{B_5H_9}\times \Delta H_{B_5H_9})+(n_{O_2}\times \Delta H_{O_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%2B%28n_%7BB_2O_3%7D%5Ctimes%20%5CDelta%20H_%7BB_2O_3%7D%29%5D-%5B%28n_%7BB_5H_9%7D%5Ctimes%20%5CDelta%20H_%7BB_5H_9%7D%29%2B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(9\times -285.83)+(5\times -1271.94)]-[(2\times 73.2)+(12\times 0)]\\\\\Delta H=-9078.57kJ](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%289%5Ctimes%20-285.83%29%2B%285%5Ctimes%20-1271.94%29%5D-%5B%282%5Ctimes%2073.2%29%2B%2812%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%3D-9078.57kJ)
Now we have to calculate the heat released per gram of the compound reacted with oxygen.
As we know that,
1 mole of
has 63.12 grams of mass
So, 2 mole of
has
grams of mass
As, 126.24 g of
release heat = 9078.57 kJ
So, 1 g of
release heat = ![\frac{9078.57}{126.24}=71.915kJ](https://tex.z-dn.net/?f=%5Cfrac%7B9078.57%7D%7B126.24%7D%3D71.915kJ)
Therefore, the heat released per gram of the compound reacted with oxygen is, 71.915 kJ