Question

The enthalpy of formation of $N{H}_3$(g) is $-46.2$ kJ mol${}^{-1}$. The heat of the following reaction is:

$2N{H}_3\left(g\right)\to N_2\left(g\right)+3H_2\left(g\right)$

• A. $-46.2$ kJ
• B. $+46.2$ kJ
  
• C. $-92.4$ kJ
  
• D. $+92.4$ kJ
  

Answer 1

The correct option is D $+92.4$ kJ

$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\left(g\right)\to N{H}_3\left(g\right);\Delta H=-46.2kJmo{l}^{-1}$
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right);\Delta H=-46.2\times 2$ $=-92.4kJ$

So, for the reverse reaction sign will be opposite i.e. $\Delta H=92.4kJ$