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 B2O3(s) and H2O(l). The standard enthalpy of formation of B5H9(l) is 73.2 kJ/mol, the standard enthalpy of formation of B2O3(s) is -1272 kJ/mol and that of H2O(l) is -285.4 kJ/mol. Express your answer in kJ.

Answer 1

Answer: The amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ

Explanation:

The chemical equation for the reaction of $B_5H_9$ with oxygen gas follows:

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

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ$

To calculate the amount of heat released for the given amount of $B_5H_9(l)$, we use unitary method, we get:

When 2 moles of $B_5H_9(l)$ reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of $B_5H_9(l)$ will react, the amount of heat released will be = $\frac{5259}{2}\times 0.211=554.8kJ$

Hence, the amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ