Question

What is the $\Delta H_f^ \circ$ of liquid water? Show calculation with bond energies.

Answer 1

The enthalpy change for the formation of one mole of substance from its constituent elements in their standard states is known as the standard heat of formation, $\Delta H_f^ \circ$. As a result, in their standard states, elements have $\Delta H_f^ \circ = 0$.

Complete answer:
Oxygen and hydrogen combine to form water vapour (gaseous form) under normal conditions, but hydrogen and water combine to form liquid water under unusual conditions. As a result, $\Delta H_f^ \circ$ of water vapour differs from $\Delta H_f^ \circ$ of water (liquid). However, if we use bond energy, we will only be able to find out the $\Delta H_f^ \circ$ under normal conditions, which means we won't be able to find the $\Delta H_f^ \circ$ of water because water is formed under non-standard conditions.
We can find the $\Delta H_f^ \circ$of liquid water if we know the $\Delta H_f^ \circ$of water vapour.
So let's first find out the $\Delta H_f^ \circ$of water vapour.
$\Delta H =$Reactant bond energies $-$ product bond energies
The equation to form water vapor is
${H_2} + \dfrac{1}{2}{O_2} = {H_2}O$
The product of the equation is water
Bonds in water are $2O - H$ bonds
$O - H$ bond energy $= 463kJ{(mol)^{ - 1}}$
Thus energy required to break water
$463kJ{(mol)^{ - 1}} \times 2 = 926kJ{(mol)^{ - 1}}$
We have multiplied $463$ with $2$ as water has $2O - H$ bonds.
The reactants of the equation are Oxygen and Hydrogen:
Bonds in oxygen $= O = O$
$O = O$ bond energy $= 499kJ{(mol)^{ - 1}}$
Thus energy required to break water
$499kJ{(mol)^{ - 1}} \times \dfrac{1}{2} = 249.5kJ{(mol)^{ - 1}}$
We have multiplied $499$ with $\dfrac{1}{2}$ as $\dfrac{1}{2}$ is the coefficient of oxygen in the equation.
Bonds in hydrogen $= H - H$
$H - H$ bond energy $= 436kJ{(mol)^{ - 1}}$
Thus energy required to break hydrogen
$436kJ{(mol)^{ - 1}} \times 1 = 436kJ{(mol)^{ - 1}}$
We have multiplied $436$ with $1$ as $1$ is the coefficient of hydrogen in the equation.
Thus, the total bond energy of the reactants are
$249.5kJ{(mol)^{ - 1}} + 436kJ{(mol)^{ - 1}} = 685.5kJ{(mol)^{ - 1}}$
Hence, $\Delta H_f^ \circ$of water vapor is
$\Delta H =$ Reactant bond energies $-$ product bond energies
$= 685.5kJ{(mol)^{ - 1}} - 926kJ{(mol)^{ - 1}} = - 240.5kJ{(mol)^{ - 1}}$
We convert $- 240.5kJ{(mol)^{ - 1}}$ to $240.5kJ{(mol)^{ - 1}}$ because $\Delta H_f^ \circ$ is different from $\Delta H$ and tells us the amount of energy liberated and is positive for exothermic reactions unlike $\Delta H$.
To calculate $\Delta H_f^ \circ$ of liquid water
We know that the energy required to evaporate $1mol$ of water is $40.7kJ{(mol)^{ - 1}}$
So, we add it to the $\Delta H_f^ \circ$ of gaseous water,
$240.5 + 40.7 = 281.2kJ{(mol)^{ - 1}}$
Therefore, $281.2kJ{(mol)^{ - 1}}$ is the $\Delta H_f^ \circ$ of liquid water.

Note:
Real-life examples of enthalpy include refrigerator compressors and chemical hand warmers. Under constant pressure, both the vaporisation of refrigerants in the compressor and the reaction of iron oxidation in a hand warmer produce a change in heat content.