Question

Enter your answer in the provided box. Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is 2B5H9(l) 12O2(g) → 5B2O3(s) 9H2O(l) Calculate the kilojoules of heat released per gram of the compound reacted with oxygen. The standard enthalpy of formations of B5H9(l), B2O3(s), and H2O(l) are 73.2, −1271.94, and −285.83 kJ/mol, respectively.

Answer 1

Answer : The heat released per gram of the compound reacted with oxygen is, 71.915 kJ

Solution :

The balanced chemical reaction is,

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{H_2O}\times \Delta H_{H_2O})+(n_{B_2O_3}\times \Delta H_{B_2O_3})]-[(n_{B_5H_9}\times \Delta H_{B_5H_9})+(n_{O_2}\times \Delta H_{O_2})]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(9\times -285.83)+(5\times -1271.94)]-[(2\times 73.2)+(12\times 0)]\\\\\Delta H=-9078.57kJ$

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

As we know that,

1 mole of $B_5H_9$ has 63.12 grams of mass

So, 2 mole of $B_5H_9$ has $2\times 63.12=126.24$ grams of mass

As, 126.24 g of $B_5H_9$ release heat = 9078.57 kJ

So, 1 g of $B_5H_9$ release heat = $\frac{9078.57}{126.24}=71.915kJ$

Therefore, the heat released per gram of the compound reacted with oxygen is, 71.915 kJ