What is the #DeltaH_f^@# of liquid water ? Show calculation with bond energies
1 Answer
281.2kJ/mol
Explanation:
This is such example.. Under normal conditions hydrogen and oxygen combine to water vapor(gaseous form) but under nonstandard conditions hydrogen and water combine to form iquid water. Thus
So let us calculate the
The reaction to form water vapor is
Product is water
Bonds in water are 2O-H bonds
O-H bond energy = 463kJ/mol
Thus energy required to break water
463kJ/mol * 2 = 926kJ/mol
Reactants are
Oxygen
Bonds in oxygen = O=O
O=O bond energy = 499
Thus energy required to break water
499kJ/mol * 1/2 = 249.5
Hydrogen
Bonds in hydrogen = H-H
H-H bond energy = 436
Thus energy required to break hydrogen
436kJ * 1 = 436kJ/mol
Thus total bond energy of reactants
(249.5kJ/mol + 436kJ/mol = 685.5kJ/mol)#
Therefore
685.5kJ/mol - 926kJ/mol = -240.5kJ/mol#
Convert -240.5kJ/mol to 240.5kJ/mol because
But How to calculate DeltaH of liquid water
It is very simple. Just check the energy required to evaporate 1mol of water or that is 40.7kJ/mol
Just add it to the
240.5 + 40.7 = 281.2kJ/mol
Does it make sense?
Yes because when water vapor is formed 40.7kJ is used to make 1mol of water steam but when liquid is formed that energy is not used and the reaction is more exothermic.