Question

Based on the values of bond energies $(B.E.)$ given, ${\Delta }_f{H}^{\circ }$ of $N_2{H}_4\left(g\right)$ is:
Given : $N-N=159{\text{kJ mol}}^{-1};H-H=436{\text{kJ mol}}^{-1}N\equiv N=941{\text{kJ mol}}^{-1};N-H=398{\text{kJ mol}}^{-1}$

• A. $711{\text{kJ mol}}^{-1}$
• B. $62{\text{kJ mol}}^{-1}$
• C. $-98{\text{kJ mol}}^{-1}$
• D. $-711{\text{kJ mol}}^{-1}$

Answer 1

The correct option is B $62{\text{kJ mol}}^{-1}$

The required reaction is
$N_2\left(g\right)+2H_2\left(g\right)\to N_2{H}_4\left(g\right)$
$\therefore {\Delta }_{f}H\left(N_2{H}_4\right(g\left)\right)=(B.E{.}_{N\equiv N}+2\times B.E{.}_{H-H})-(B.E{.}_{N-N}+4\times B.E{.}_{H-H})=(941+2\times 436)-(159+4\times 398)=1813-1751=62{\text{kJ mol}}^{-1}$