# What is the DeltaH_f^@ of liquid water ? Show calculation with bond energies

Mar 21, 2017

281.2kJ/mol

#### Explanation:

$\Delta H = \text{Bond energies of reactant - Bond energies of product}$

$\Delta {H}^{o} f$ of something is the $\Delta H$ of the reaction when the substance is formed . It can be gaseous or liquid. But if we use bond energies to calculate it only tell us the $\Delta {H}^{o} f$ under standard conditions.
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 $\Delta {H}^{o} f$ of water vapour is different from $\Delta {H}^{o} f$ of water(liquid). But if you use bond energy you will only get to know the $\Delta {H}^{o} f$ under standard conditions which means we can't find the $\Delta {H}^{o} f$ water because water is formed under non-standard conditions. But if you know $\Delta {H}^{o} f$ of water vapour you can find the $\Delta {H}^{o} f$ of liquid water.

So let us calculate the $\Delta {H}^{o} f$ of water vapour

$\Delta H = \text{Bond energies of reactant - Bond energies of product}$

The reaction to form water vapor is

${H}_{2} + \frac{1}{2} {O}_{2} = {H}_{2} O$

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 $\Delta {H}^{o} f$ of water vapor is

685.5kJ/mol - 926kJ/mol = -240.5kJ/mol

Convert -240.5kJ/mol to 240.5kJ/mol because $\Delta {H}^{o} f$ is different from $\Delta H$ and tells us the amount of energy released and is positive for exothermic reactions unlike $\Delta H$.

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 $\Delta {H}^{o} f$ of gaseous water

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.