Question

Consider the following data:

$\Delta f{H}^0(N_2{H}_4,l)=50kJ/mol$

$\Delta f{H}^0(N{H}_3,g)=-46kJ/mol$

$B.E.(N-H)=393kJ/mol$ and $B.E.(H-H)=436kJ/mol$, also ${\Delta }_{vap}H(N_2{H}_4,l)=18kJ/mol$

The N- N bond energy in $N_2{H}_4$ is

• A. 226 kJ/mol
• B. 154 kJ/mol
• C. 190 kJ/mol
• D. 45.45 kJ/mol

Answer 1

The correct option is C

190 kJ/mol

$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\to N{H}_3$

$\left(g\right)\left(g\right)\left(g\right)$

Let BF of N≡ N s x

$-46=\frac{x}{2}+\frac{3}{2}\times 436-3\times 393$

$\therefore x=958$
$N_2{H}_{4\left(I\right)}\to N_{2\left(g\right)}+2H_{2\left(g\right)}$
$\Delta H=-50kJ/mol$

$\Delta H=\left[\Delta H_{vap}\right(N_2{H}_4,l)+4\times BE(N-H)+BE(N-N)+BE(N-N\left)\right]-[B.Eof(N\equiv N\left)\right]+2BE(H-H)-50=(18+4\times 393+y)-[958+2\times 436]$

$50=(1590+y)-\left(1830\right)$

$BEof(N-N)$ (or)

$Y=190KJ/mol$