Question

In solid ammonia, each $N{{H}_{3}}$ molecule has six other $N{{H}_{3}}$ molecules as nearest neighbors. $\Delta H$ of sublimation of $N{{H}_{3}}$ at the melting point in $30.8kJmo{{l}^{-1}}$ and the estimated $\Delta H$ of sublimation in the absence of hydrogen bonding is $14.4kJmo{{l}^{-1}}$. What is the strength of the hydrogen bond in solid ammonia $kJmo{{l}^{-1}}$? Give your answer in the nearest integer number.

Answer 1

We know that the basic case of ammonia with bonding is given by the amount of hydrogen bonding is limited by the fact that each nitrogen has one lone pair. So, on an average each ammonia molecule forms one hydrogen bond with its lone pair electron.

Complete step-by-step answer: Given: First of all we know that solid ammonia, which contains $N{{H}_{3}}$ molecules, has an electronegative atom $N$ capable of hydrogen bonding. Hydrogen bonding thus strengthens the interaction between the ammonia ($N{{H}_{3}}$) molecules. Now, if we look into the data given
Here we have given solid ammonia and for each $N{{H}_{3}}$ molecule we have six another$N{{H}_{3}}$ molecules which are as nearest as a neighboring molecule.
$\Rightarrow$So the sublimation of $N{{H}_{3}}$at melting point is given by; $\Delta H$ of sublimation of $N{{H}_{3}}$ at melting point $=30.8kJmo{{l}^{-1}}$
$\Rightarrow$Similarly for absence of sublimation of $Hbonding$ is given by; $\Delta H$of sublimation in absence of $Hbonding$ $=14.4kJmo{{l}^{-1}}$
Therefore, by subtracting both the $\Delta H$we can acquire the total strength of all Hydrogen bonds
$=30.8-14.4=16.4\dfrac{kJ}{mol}$
$\Rightarrow 16.4kJmo{{l}^{-1}}$
Also as we earlier saw that are six nearest neighboring molecules with solid ammonia even though each one of hydrogen bond involves every single time a two molecules this sums up the effective neighboring which is $3$
So, the strength of hydrogen bonds in solid ammonia can be given by dividing total strength divide by effective neighbors.
$=\dfrac{16.4}{3}\dfrac{kJ}{mol}=5.4666\dfrac{kJ}{mol}$
Here $5.4666kJmo{{l}^{-1}}$ with roundup can be written as $5.5kJmo{{l}^{-1}}$
$\Rightarrow 5.5kJmo{{l}^{-1}}$
Therefore, the strength of hydrogen bonds in solid ammonia is $5.5kJmo{{l}^{-1}}$ .
Additional Information: The extent of a hydrogen bonding depends upon electronegativity and number of hydrogen atoms available for bonding. Among nitrogen, fluorine, and oxygen, increasing order of electronegativity are $N\text{ }<\text{ }O\text{ }<\text{ }F$ Hence expected order of extent of hydrogen bonding is $HF\text{ }>\text{ }{{H}_{2}}O\text{ }>\text{ }N{{H}_{3}}$ But, actual order is ${{H}_{2}}O\text{ }>\text{ }HF\text{ }>\text{ }N{{H}_{3}}$

Note: Note that we should understand the fact that sublimation requires all the forces to be broken in solid to convert it into gas. So, higher value of enthalpy of sublimation of ammonia can be attributed to hydrogen bonding.