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) 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.

1 Answer

5 votes

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

Step-by-step 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* \Delta H_f_((B_2O_3(s))))+(9* \Delta H_f_((H_2O(l))))]-[(2* \Delta H_f_((B_5H_9(l))))+(12* \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* (-1272))+(9* (-285.4))]-[(2* (73.2))+(12* (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 =
(5259)/(2)* 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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