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 _______ kJ

Answer must be in 3 sig fig!

Answer 1

Given reaction:

N2 + 3Cl2 → 2NCl3

Energy change = Bonds broken - bonds formed

ΔHrxn = Bond energy[ N≡N + 3*Cl-Cl] - Bond energy [6*N-Cl]

          = [945 + 3(240)] - [6(200)] = 465 kJ/mol

Now, atomic mass of Cl = 35.5 g/mol

Given mass of Cl = 85.3 g

# moles of Cl = 85.3/35.5 = 2.403 moles

ΔH corresponding to the above moles of Cl would be

= 2.403 moles * 465 kJ/ 1 mole = 1117.4 J

Ans: The reaction enthalpy for 85.3 g of chlorine = 1.12 kJ

Answer 2

Answer: 186

Explanation: I got this right on Edmentum.