Asked by LaurenB
The standard enthalpy of formation of n-octane is -249.95 kJ/mol. Compute the amount of heat liberated when 5.24 g of n-octane is burned completely with excess oxygen to form carbon dioxide and liquid water.
Answers
Answered by
Katie
Hint:
Write the balanced combustion reaction. You can find the standard enthalpies of formation for O2, CO2, and H2O from many sources, including the CRC Handbook of Chemistry and Physics.
Solution:
Balanced combustion reaction: C8H18 + 12.5 O2 --> 8 CO2 + 9 H2O
From the CRC Handbook of Chemistry & PhysicsStandard enthalpies of formation (25C):
n-octane -249.95, oxygen 0.00, CO2 -393.5, H2O(l) -285.8 kJ/mol
ÄHrxn = -5470.25 kJ/mol
moles of octane burned = (5.24 g)/(114.231 g/mol) = X mol
Total heat released = 5470.25 kJ/mol * X mol = Answer kJ
Write the balanced combustion reaction. You can find the standard enthalpies of formation for O2, CO2, and H2O from many sources, including the CRC Handbook of Chemistry and Physics.
Solution:
Balanced combustion reaction: C8H18 + 12.5 O2 --> 8 CO2 + 9 H2O
From the CRC Handbook of Chemistry & PhysicsStandard enthalpies of formation (25C):
n-octane -249.95, oxygen 0.00, CO2 -393.5, H2O(l) -285.8 kJ/mol
ÄHrxn = -5470.25 kJ/mol
moles of octane burned = (5.24 g)/(114.231 g/mol) = X mol
Total heat released = 5470.25 kJ/mol * X mol = Answer kJ
Answered by
Anonymous
alchol releases 29.7kJ/g when it burns. Convert this value to the number of calories per gram
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