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worty [1.4K]
3 years ago
11

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.
Chemistry
1 answer:
mote1985 [20]3 years ago
7 0

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

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

The equilibrium reaction follows:

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

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

\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]

We are given:

\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol

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

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat = \frac{-9078.57}{2}=-4539.28kJ

For per gram of compound:

Molar mass of B_5H_9 = 63.12 g/mole

\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g

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

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