Solution: Using the following data, calculate the standard heat of formation of ICl(g) in kJ/mol:Cl2 (g) → 2Cl (g)          ΔH° = 242.3 kJI2 (g) → 2I (g)               ΔH° = 151.0 kJICl (g) → I (g) + Cl (g)    ΔH° = 211.3 kJI2 (s) → I2 (g)               ΔH° = 62.8 kJ

Problem

Using the following data, calculate the standard heat of formation of ICl(g) in kJ/mol:

Cl2 (g) → 2Cl (g)          ΔH° = 242.3 kJ

I2 (g) → 2I (g)               ΔH° = 151.0 kJ

ICl (g) → I (g) + Cl (g)    ΔH° = 211.3 kJ

I(s) → I2 (g)               ΔH° = 62.8 kJ


Solution

Formation of ICl(g) from it elements:

Natural element form of iodine  I2(s)

Natural element form of chlorine  Cl2(g) 


In a formation equation, one mole of the product should be forming:

Formation equation of ICl(g): ½ I2(s) + ½ Cl2(g)  ICl(g)

Manipulate reaction such that when you add them you arrive to the formation equation of ICl(g):

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