The given chemical reaction is:
[tex]N_{2}+3Cl_{2}-->2NCl_{3}[/tex]
Δ[tex]H^{0}_{reaction}=[/tex]∑BE(reactants)-∑BE(products)
= {(941 kJ/mol) + (3 * 242 kJ/mol)} -[{2*(3*200 kJ/mol)}]
= 467 kJ/mol
Calculating the change in heat when 85.3 g chlorine reacts in the above reaction:
Moles of chlorine = [tex]85.3 g Cl_{2}* \frac{1 mol Cl_{2}} {70.91 g Cl_{2} }[/tex]
= 1.20 mol [tex]Cl_{2}[/tex]
Heat change when 1.20 mol chlorine reacts
= [tex]1.20 mol * \frac{467kJ}{mol} =560.4 kJ[/tex]
