Question

Given the following bond enthalpies:
B.E $(N\equiv N)=942KJ/mol$; B.E $(H-H)=436KJ/mol$
B.E $(N-N)=163KJ/mol$; B.E $(N-H)=390KJ/mol$
Determine enthalpy change for the following polymerisation reaction per mole of $N_2(g)$ consumed
$n{N}_2(g)+n{H}_2(g)\to -(-NH-NH-{)}_n-$

Answer 1

Hint:Bond enthalpy is the strength or energy associated with a bond between atoms or molecules. It is also determined as the energy required to break the bonds between atoms. For a given reaction the bond enthalpy is calculated as the sum of the bond energies of the bonds of the reactants subtracted by the bond energies of the bonds of the products.
Formula used:
 $\mathrm{\Delta }H=\mathrm{\Delta }H(reac\tan t)-\mathrm{\Delta }H(product)$

Complete step by step answer:
Bond enthalpy is referred to as the strength or energy in a chemical bond. It is a quantity which also defines the extension and stability of a reaction. Bond energy is the energy which is required to break a bond between atoms or molecules. Thus, this energy is an endothermic reaction and hence, will always be positive whereas the reaction enthalpy can be negative or positive because it involves breaking and formation of bonds. Bond enthalpy is also determined as the energy required to break the bond because the amount of energy or strength the bond holds will also be the energy required to break that bond between the molecules.

Bond enthalpy is calculated as the sum of the bond energies of the bonds of the reactants subtracted by the bond energies of the bonds of the products. The energy of the reactant’s bonds is known as energy in and the energy of the product’s bond is known as energy out.
Thus, bond enthalpy = energy in – energy out. Bond enthalpy is represented by $\mathrm{\Delta }H$
$\therefore \mathrm{\Delta }H=\mathrm{\Delta }H(energyin)-\mathrm{\Delta }H(energyout)$ where energy in is the bonds broken and energy out is bonds formed.
For the given question we know that for per mole
$N_2+H_2\to -(NH-NH)-$
 $\Rightarrow \mathrm{\Delta }H(N\equiv N)+\mathrm{\Delta }H(H-H)=\mathrm{\Delta }H(2(N-N)+2(N-H))$
$\Rightarrow \mathrm{\Delta }H(N\equiv N)+\mathrm{\Delta }H(H-H)-\mathrm{\Delta }H(2(N-N)-2(N-H))$
Now substitute the values given
$\Rightarrow \mathrm{\Delta }H=942+436-2(163)-2(390)$
$\Rightarrow \mathrm{\Delta }H=272KJ/mol$

Thus, the enthalpy change is $272KJ/mol$.

Note:
Remember that the bonds on the reactant side are being broken and on the product side, bonds are formed. The atoms which are bonded together may or may not have equally distributed electrons. Thus, this separation results in electric charge along the bond which leads to a molecule or an atom bonded to another to acquire dipole-dipole moment. This is referred to as bond polarity.