Question

Bond dissociation enthalpies of ${H_2}(g)$ and ${N_2}(g)$ are 436.0 $kJmo{l^{ - 1}}$ and 941.8 $kJmo{l^{ - 1}}$ respectively and enthalpy of formation of $N{H_3}(g)$ is -46 $kJmo{l^{ - 1}}$. What is the enthalpy of atomization of $N{H_3}(g)$?
A) 390.3 $kJmo{l^{ - 1}}$
B) 1170.9 $kJmo{l^{ - 1}}$
C) 590 $kJmo{l^{ - 1}}$
D) 720 $kJmo{l^{ - 1}}$

Answer 1

Hint: First you need to write the balanced chemical reaction of ${N_2}(g)$ reacting with ${H_2}(g)$ to give $N{H_3}(g)$. Then, you must know the relation of enthalpy of formation of any substance with the bond dissociation enthalpies of the reactants and products in the gas phase. Also, enthalpy of atomisation is the enthalpy change in which total separation of all atoms occurs.

Complete step by step solution:
We are given:
Bond dissociation enthalpy of ${H_2}(g)$= ${\Delta _{bond}}{H_{({H_2})}}$= 436.0 $kJmo{l^{ - 1}}$
Bond dissociation enthalpy of ${N_2}(g)$= ${\Delta _{bond}}{H_{({N_2})}}$= 941.8 $kJmo{l^{ - 1}}$
Enthalpy of formation of $N{H_3}(g)$= -46 $kJmo{l^{ - 1}}$
To find: Enthalpy of atomisation of $N{H_3}(g)$
Now, the balanced chemical reaction of nitrogen gas,${N_2}$ with hydrogen gas, ${H_2}$ to give ammonia, $N{H_3}$ is as shown below:
\[{N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)\]
We know that enthalpy of formation is always given for 1 mole of a substance. So, enthalpy of formation for the above reaction that is, for 2 molecules of ammonia is:
\[{\Delta _f}{H_{(N{H_3})}}= 2 \times ( - 46)kJmo{l^{ - 1}}\]
Now, enthalpy of formation of any substance is related to the bond dissociation enthalpies of the reactants and products in the gas phase as:
\[\Delta H = \sum {{\text{bond enthalpy}}{{\text{s}}_{{\text{(Reactants)}}}} - } \sum {{\text{Bond enthalpy}}{{\text{s}}_{{\text{(Products)}}}}} \] ...................... Equation (1)
In our reaction, reactants are nitrogen and three moles of hydrogen gas and product is 2 molecules of ammonia gas. Now, structure of ammonia:

Let us consider the bond enthalpy of each N-H bond in ammonia be $x$ $kJmo{l^{ - 1}}$. Thus, overall bond enthalpy of ammonia is $3x$ $kJmo{l^{ - 1}}$.
Substituting the given bond enthalpy values and considered bond enthalpy value of $N{H_3}(g)$ in equation (1) as:
\[\Delta H = \sum {{\text{bond enthalpy}}{{\text{s}}_{{\text{(Reactants)}}}} - } \sum {{\text{Bond enthalpy}}{{\text{s}}_{{\text{(Products)}}}}}\]
\[{\Delta _f}{H_{(N{H_3})}} = [{\Delta _{bond}}{H_{({N_2})}} + 3 \times {\Delta _{bond}}{H_{({H_2})}}] - [2 \times {\Delta _{bond}}{H_{(N{H_3})}}]\]
\[2 \times ( - 46) = [941.8 + 3 \times 436.0] - [2 \times 3x]\]
\[\Rightarrow x = 390.3{\text{ }}kJmo{l^{ - 1}}\]
Now, enthalpy of atomisation is the enthalpy change in which total separation of all atoms occurs. We know from the structure of ammonia that 1 molecule of ammonia contains three nitrogen-hydrogen bonds.
Thus, enthalpy of atomisation of $N{H_3}(g)$, ${\Delta _a}{H_{(N{H_3})}}$:
\[{\Delta _a}{H_{(N{H_3})}}{\text{ = 3}} \times {\text{bond enthalpy of each N - H bond}}\]
\[{\Delta _a}{H_{(N{H_3})}} = 3 \times x\]
\[{\Delta _a}{H_{(N{H_3})}} = 3 \times 390.3\]
\[{\Delta _a}{H_{(N{H_3})}} = 1170.9{\text{ }}kJmo{l^{ - 1}}.\]

Thus, option B is the correct answer.

Note: Enthalpy of atomisation is generally defined as the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase whereas the bond dissociation enthalpy is the enthalpy change when one mole of covalent compounds is broken to form products in the gas phase.