Question

Based on the values of B.E. given, ${\Delta }_f{H}^0$ of $N_2{H}_4\left(g\right)$ is:
$N-N=159kJmo{l}^{-1};H-H=436kJmo{l}^{-1}$
$N\equiv N=941kJmo{l}^{-1};{N-H=398kJmol}^{-1}$

• 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

$62{\text{kJ mol}}^{-1}$

$N_2+2H_2\to N_2{H}_4$

$\Rightarrow △H$ ( bond breaking ) $=△H(N\equiv N)+2△H(H-H)$

$=941+2\left(436\right)$

$=1813$

$\Rightarrow △H$ ( bond formation ) $=-[△H(N-N)+4△H(N-H)]$

$=-[159+4\left(398\right)]$

$=-1751$

$\Rightarrow △H=△H$ ( bond formation ) + $△H$ ( bond breaking )

$=-1751+1813$

$=62kJ/mol$

Hence, the answer is $62kJmo{l}^{-1}.$