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Q.
Enthalpy of formation of $NH_3$ is $- X \,kJ$ and $ΔH_{H-H}, ΔH_{N-H}$ are respectively $Y \,kJ\, mol^{-1}$ and $Z\, kJ\, mol^{-1}$. The value of
$ΔH_{N ≡ N}$ is
Thermodynamics
Solution:
$\frac{1}{2} N _{2}( g )+\frac{3}{2} H _{2}( g ) \longrightarrow NH _{3}( g ) ; \Delta H _{ f }=- x \,kJ$
$\Delta H _{ f }=\Delta H =( B .D . E )_{\text {reactants }}-( B .D .E )_{\text {products }}$
$=\frac{1}{2} \Delta H _{ N _{2}}+\frac{3}{2} \Delta H _{ H _{2}}-3 \Delta_{ N - H }$
$=\frac{\Delta H_{N \equiv N}}{2}+\frac{3}{2} X Y-3 X Z$
We have,
$-X=\frac{\Delta H_{N \equiv N}}{2}+\frac{3 Y}{2}-3 Z$
or, $\Delta H_{ N \equiv N }=\left(3 Z-\frac{3 Y}{2}-X\right) \times 2$
$=6 Z-3 Y-2 X$