Calculate the kJ of heat per gram released by Pentaborane-9?
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 42.84, -1271.94, and -285.83 kJ/mol, respectively.
Homework Answers
ΔH = (5*-1271.94+9*-285.83-2*42.84)
ΔH = 9.017*103kJ/mole
so 12 moles of Oxygen is needed to get this energy,
for 1gm it will b = 9017/(12*32) = 23.48kJ/mole gm
Step2 =[ -6359.7 -2572.47+85.68]/2 =-8846.49/2=-4423.245 kJ/mole
Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to...
Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is: 2 B5H9 (l)+ 12 O2 (g)-----> 5 B2O3 (s)+ 9 H2O (l) ∆Hfº = -8890.2 kJ/mol Calculate the kilojoules of heat released per gram of the compound reacted with oxygen
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