Answer:
2 NO (g) → N2 (g) + O2 (g)
2 NOCl (g) → 2 NO (g) + Cl2 (g)
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2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)
ΔH = [90.3 kJ x 2 x -1] + [-38.6 kJ x -1 x 2] = -103.4 kJ
The ΔH for the reaction is -103.4 kJ
Formal Charge is calculated as, Lewis structure is attached below,
Formal Charge = [# of valence electrons] - [electrons in lone pairs + 1/2 the number of bonding electrons]
# of valence electrons of Cl = 7
electrons in lone pairs = 4
number of bonding electrons = 6
Formal Charge = [7] - [4 + 6/2]
Formal Charge = [7] - [4 + 3]
Formal Charge = [7] - [7]
Formal Charge = 0
Result: Formal charge on Cl in ClF₃ is
zero.
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