Question

Consider the following data: ${\Delta }_f{H}^{\circ }(N_2{H}_4,l)=50\text{kJ/mol},{\Delta }_f{H}^{\circ }(N{H}_3,g)=-46\text{kJ/mol}$, $B.E(N-H)=393\text{kJ/mol}$ and $B.E(H-H)=436\text{kJ/mol}$, ${\Delta }_{vap}H\left(N_2{H}_{4}l\right)=18\text{kJ/mol}$. Calculate the N-N bond energy in kJ/mol for $N_2{H}_4$.

• A. $190\text{kJ/mol}$
• B. $-190\text{kJ/mol}$
• C. $95\text{kJ/mol}$
• D. $-95\text{kJ/mol}$

Answer 1

The correct option is A $190\text{kJ/mol}$

Enthalpy of formation:
For one mole of product
$\Delta H=$ (Enthalpy required to break reactants into gaseous atoms) $-$ (Enthalpy released to form products from the gaseous atoms)

${\Delta }_f{H}_{N_2{H}_4\left(l\right)}=50$
$N_2\left(g\right)+2H_2\to N_2{H}_4\left(l\right)\left(1\right)$

$N_2{H}_4\left(l\right)\to N_2{H}_4\left(g\right)=\left(2\right)$
On addition:
$N_2\left(g\right)+2H_2\to N_2{H}_4\left(g\right)\Delta H=68kJ\left(3\right)$
For above expression enthalpy in term of bond energy is:
$68=(E_{N\equiv N}+2E_{H-H})-(E_{N-N}+4E_{N-H})$
$68=E_{N\equiv N}+2\times 436–(E_{N-N}+4\times 393)\left(4\right)$

And ${\Delta }_f{H}_{N{H}_3\left(g\right)}=-46kJ$
$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\to N{H}_3$
Enthalpy in terms of bond energy is:
$-46=\frac{1}{2}{E}_{N\equiv N}+\left(\frac{3}{2}{E}_{H-H}\right)-(3\times E_{N-H})$
$-46=\frac{1}{2}{E}_{N\equiv N}-525$
$E_{N\equiv N}=958kJ$
Substituting in equation 4
We get $E_{N-N}=190kJ/mol$