Question

The enthalpy of formation of $NH_3$ is $- 4 6\, kJ \, mol^{-1}$. The enthalpy change for the reaction : $2 NH _{3}(g) \longrightarrow N _{2}(g)+3 H _{2}(g)$ is

• A. #ERROR!
• B. #ERROR!
• C. #ERROR!
• D. #ERROR!

Answer 1

Answer: (A)

#ERROR!

Answer 2

$2 NH _{3}(g) \longrightarrow N _{2}(g)+3 H _{2}(g)$ $\Delta H_{r}=-\left(2 \times\right.$ enthalpy of formation of $\left.NH _{3}\right)$ $=-(2 \times-46)=92\, kJ$