Question

For the reaction:

Answer 1

Hey there!

Given the reaction :

CH₃I + Cl⁻ → CH₃Cl + I⁻  

Change in enthalpy for the reaction :

ΔrH = (Bond energies of breaking) - (Bond energies of formation)

Bond energy of breaking = CH₃ - I = 215 kJ/mol

Bond energy of formation =  CH₃ - Cl = 330 kJ/mol

ΔrH = 215 kJ/mol - 330 kJ/mol = -115 kJ/mol

4.184 Kj/mol = 1 Kcal/mol

-115 kJ/mol = -115/4.184 = -27.49 Kcal/mol