Solveeit Logo
Login

Question

Question: Given are the bond enthalpies: \({{\varepsilon }_{\left( N=N \right)}}=945kJmo{{l}^{-1}}\) , \({{\va...

Given are the bond enthalpies: ε(N=N)=945kJmol−1{{\varepsilon }_{\left( N=N \right)}}=945kJmo{{l}^{-1}} , ε(H−H)=436kJmol−1{{\varepsilon }_{\left( H-H \right)}}=436kJmo{{l}^{-1}} and ε(N−H)=391kJmol−1{{\varepsilon }_{\left( N-H \right)}}=391kJmo{{l}^{-1}} . The enthalpy change of the reaction
N2(g)+3H2(g)→2NH3(g){{N}_{2\left( g \right)}}+3{{H}_{2\left( g \right)}}\to 2N{{H}_{3\left( g \right)}} is
(A) −93kJmol−1-93kJmo{{l}^{-1}}
(B) −89kJmol−1-89kJmo{{l}^{-1}}
(C) −105kJmol−1-105kJmo{{l}^{-1}}
(D) 105kJmol−1105kJmo{{l}^{-1}}

Explanation

Solution

The general knowledge about bond dissociation energy, enthalpy of formation and Hess’s law, we can solve the given illustration easily.
We have a relationship that relates bond enthalpy and enthalpy of formation; we can solve by that method too.

Complete step by step solution:
Let us firstly understand the given terms;
Bond dissociation enthalpy (or simply bond enthalpy/energy)-
It is a measure of strength of a chemical bond. It can also be defined as the standard enthalpy change when the bond (A-B) is cleaved by homolysis to give fragments of A and B.
Enthalpy of formation (or reaction)-
It is the change in the enthalpy of reaction during the formation of 1 mole of the substance from its constituent elements, with all the contributing substances in their standard states.
Hess’s law-
The law states that the change in enthalpy for a reaction is the same whether the reaction takes place in one step or a series of steps regardless of path by which reaction occurs.
Now, we can describe the required equations in an illustration and solve it as follows,
Given that,
Bond enthalpy of H2{{H}_{2}} = 436 kJ/mol
Bond enthalpy of N2{{N}_{2}} = 945 kJ/mol
Bond enthalpy of N−HN-H = 391 kJ/mol
The reaction is described as follows;
N2(g)+3H2(g)→2NH3(g){{N}_{2\left( g \right)}}+3{{H}_{2\left( g \right)}}\to 2N{{H}_{3\left( g \right)}}
This can be divided into simpler steps as,
Step (I)- Dissociation of hydrogen and nitrogen molecules into their respective atoms.
3H2→6H N2→2N \begin{aligned} & 3{{H}_{2}}\to 6H \\\ & {{N}_{2}}\to 2N \\\ \end{aligned}
Step (II)- Combination of these atoms to form NH3N{{H}_{3}} .
6H+2N→2NH36H+2N\to 2N{{H}_{3}}
Since, the first step is an endothermic process thus, ΔH1>0\Delta {{H}_{1}}>0 . Whereas, the next step releases the energy thus, ΔH2<0\Delta {{H}_{2}}<0 .
So,
ΔH1=ΔHN−N+(3×ΔHH−H) ΔH1=945+(3×436)=2253kJ/mol \begin{aligned} & \Delta {{H}_{1}}=\Delta {{H}_{N-N}}+\left( 3\times \Delta {{H}_{H-H}} \right) \\\ & \Delta {{H}_{1}}=945+\left( 3\times 436 \right)=2253kJ/mol \\\ \end{aligned}
And,
ΔH2=−2×3×ΔHN−H ΔH2=−2346kJ/mol \begin{aligned} & \Delta {{H}_{2}}=-2\times 3\times \Delta {{H}_{N-H}} \\\ & \Delta {{H}_{2}}=-2346kJ/mol \\\ \end{aligned}
Thus, applying Hess’s law we get,
ΔH′=ΔH1+ΔH2 ΔH′=2253−2346=−93kJ/mol \begin{aligned} & \Delta H'=\Delta {{H}_{1}}+\Delta {{H}_{2}} \\\ & \Delta H'=2253-2346=-93kJ/mol \\\ \end{aligned}

Therefore option (A) is correct.

Note: An alternative method to solve the same illustration can be the direct relationship of bond dissociation enthalpy and enthalpy of reaction as stated,
ΔHreaction=∑(B.E)reactan⁡t−∑(B.E)product\Delta {{H}_{reaction}}=\sum{{{\left( B.E \right)}_{reac\tan t}}-\sum{{{\left( B.E \right)}_{product}}}} .
Solving by this method will give the same answer.