Question

B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 reacts with excess oxygen where the products are B2H3(s) and H2O(l). The standard enthalpy of formation of B5H9(l) is 73.2 kJ/mol, the standard enthalpy of formation of B2H3(s) is -1272 kJ/mol and that of H2O(l) is -285.4 kJ/mol. Express your answer in kJ.

Answer 1

Answer: 958 kJ  heat is released when 0.211 mol of $B_5H_9$ reacts with excess oxygen.

Explanation:

The balanced chemical reaction is,

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2H_3(s)+9H_2O(l)$

The expression for enthalpy change is,

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

$\Delta H=[(n_{B_2H_3}\times \Delta H_{B_2H_3})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{B_5H_9}\times \Delta H_{B_5H_9})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

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

$\Delta H=[(5\times -1272)+(9\times -285.5]-[(12\times 0)+(2\times 73.2)]$

$\Delta H=-9075.9kJ$

Thus 2 moles of $B_5H_9$ release heat = 9075.9 kJ

0.211 moles of $B_5H_9$ release heat =$\frac{9075.9}{2}\times 0.211=958 kJ$