Question

Given: N2 + 3Cl2 → 2NCl3, ΔH = 464 kJ/mol Use the given bond energies and the periodic table to calculate the energy change in the reaction. The ΔH when 85.3 grams of chlorine reacts in the given reaction is kilojoules.

Answer 1

The given chemical reaction is:

$N_{2}+3Cl_{2}-->2NCl_{3}$

Δ$H^{0}_{reaction}=$∑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 = $85.3 g Cl_{2}* \frac{1 mol Cl_{2}} {70.91 g Cl_{2} }$

                             = 1.20 mol $Cl_{2}$

Heat change when 1.20 mol chlorine reacts

                             = $1.20 mol * \frac{467kJ}{mol} =560.4 kJ$