Given that: 2no (g)→n2 (g)+o2 (g)n2 (g)+o2 (g)+cl2 (g)→2nocl (g) has an enthalpy change of −180.6 kj has an enthalpy change of 103.4 kj what is the enthalpy change, in kilojoules, for 2nocl (g)→2no (g)+cl2 (g)?

Explanation:

It is possible to answer this question knowing Hess's law that says you can sum half-reactions enthalpy cahnge to obtain enthalpy change of the total reaction. Using the reactions:

(1) 2NO(g) → N₂(g)+O₂(g) ΔH = -180,6 kJ

(2) N₂(g) + O₂(g) + Cl₂(g) → 2NOCl(g) ΔH = +103,4 kJ

The reverse reactions of (1) and (2) are:

N₂(g)+O₂(g) → 2NO(g) ΔH = +180,6 kJ

2NOCl(g) → N₂(g) + O₂(g) + Cl₂(g) ΔH = -103,4 kJ

The sum of these reactions is:

2NOCl(g) → 2NO(g) + Cl₂(g) ΔH = +180,6 kJ -103,4 kJ = 77,2 kJ

I hope it helps!

a electronegativity decreases beacuse the size of nucleus is bigger.