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Korvikt [17]
3 years ago
5

) B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 react

s 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.
Chemistry
1 answer:
Tom [10]3 years ago
7 0

<u>Answer:</u> The amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

<u>Explanation:</u>

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

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