Given,
NH3(g)+3Cl2(g)→NCl3(g)+3HCl; ΔH1
N2(g)+3H2(g)→2NH3(g); ΔH2
H2(g) + Cl2(g)→ 2HCl(g); ΔH3
The heat of formation of NCl3 in terms of ΔH1,ΔH2 and ΔH3 is:
ΔH1+ΔH22−32ΔH3
NH3(g)+3Cl2(g)→NCl3(g)+3HCl; ΔH1...(i)
N2(g)+3H2(g)→2NH3(g); ΔH2...(ii)
H2(g)+Cl2(g)→2HCl(g); ΔH3...(iii)
12N2(g)+32Cl2(g)→NCl3(g); ΔH4...(iv)
Equation (iv) can be obtained by:
(i)+(ii)×12+(iii)×(−32)
ΔH4=ΔH1+ΔH22−32ΔH3