Respuesta :
Answer : The heat released per gram of the compound reacted with oxygen is, 71.915 kJ
Solution :
The balanced chemical reaction is,
[tex]2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\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})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta H=[(9\times -285.83)+(5\times -1271.94)]-[(2\times 73.2)+(12\times 0)]\\\\\Delta H=-9078.57kJ[/tex]
Now we have to calculate the heat released per gram of the compound reacted with oxygen.
As we know that,
1 mole of [tex]B_5H_9[/tex] has 63.12 grams of mass
So, 2 mole of [tex]B_5H_9[/tex] has [tex]2\times 63.12=126.24[/tex] grams of mass
As, 126.24 g of [tex]B_5H_9[/tex] release heat = 9078.57 kJ
So, 1 g of [tex]B_5H_9[/tex] release heat = [tex]\frac{9078.57}{126.24}=71.915kJ[/tex]
Therefore, the heat released per gram of the compound reacted with oxygen is, 71.915 kJ
The heat released per gram of the compound reacted with oxygen is -71.92 KJ/mol per gram of B5H9 reacted.
The equation goes as follows;
2B5H9(l) + 12O2(g) → 5B2O3(s) + 9H2O(l)
We have the following information;
ΔH°f B5H9(l) = 73.2 kJ/mol
ΔH°f B2O3(s) = −1271.94 kJ/mol
ΔH°f H2O(l) = −285.83 kJ/mol
Note that;
ΔHrxn = ∑ΔH°f (products) - ΔH°f (reactants)
ΔHrxn = ∑(5 × ( −1271.94 kJ/mol)) + (9 × ( −285.83 kJ/mol)) - ∑(2 × (73.2 kJ/mol) + (12 × 0)
ΔHrxn = -9078.57 kJ/mol
Since 1 mole of B5H9 = 63.12 g/mol
Two moles of B5H9 reacted so 2 moles × 63.12 g/mol = 126.24 g
Heat released per gram of B5H9 reacted = -9078.57 kJ/mol/126.24 g
= -71.92 KJ/mol per gram of B5H9 reacted.
Learn more: https://brainly.com/question/4147359
