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Calculate ΔH for
2NOCl(g) → N₂(g) + O₂(g) + Cl₂(g)
given the following reactions:
1/2 N₂(g) + 1/2 Cl₂(g) → NO(g) ΔH = 90.3kJ
NO(g) + 1/2 Cl₂(g) → NOCl(g) ΔH = -38.6kJ

Answer :

The enthalpy of the reaction,

2NOCl(g) → N₂(g) + O₂(g) + Cl₂(g) is -103.3kJ.

According to Hess's law of constant heat summation, the heat evolved or absorbed in a given chemical equation is the same whatever the process occurs in one step or several steps .

According to the given law, the chemical equation can be treated as ordinary algebraic expression and it can be added or subtracted to produce the required equation. That means the enthalpy change of the complete reaction is the sum of the enthalpy changes of the intermediate reactions.

Given,

The final reaction is,

2NOCl(g) → N₂(g) + O₂(g) + Cl₂(g)

The intermediate balanced chemical reaction can be,

1/2 N₂(g) + 1/2 O₂(g) + 1/2 Cl₂(g) → NO(g), ΔH = 90.3kJ

NO(g) + 1/2 Cl₂(g) → NOCl(g), ΔH = -38.6kJ

First we will reverse the reaction 1 and 2 both and multiply by 2 then adding both the equation, we get :

2NOCl(g)→ 2NO + Cl₂(g) ∆H°1 = 2 × 38.6 = 77.2kJ

2 NO → N₂(g) + O₂(g) + Cl₂(g) ∆H°2 = 2 × (-90.3) = -180. 6 kJ

The expression for final enthalpy is,

∆H = ∆H°1 + ∆H°2

∆H = -180. 6 + 77.2

∆H = -103.3

Thus, we calculated that the enthalpy of the reaction,

2NOCl(g) → N₂(g) + O₂(g) + Cl₂(g) is -103.3kJ.

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