Question

Using the provided table and the equation below, determine the heat of formation for CuBr2. CuCl2(s) + Br2(I) → CuBr2 (s) + CI2(g) ΔH = 64.1 kJ/mol 

Answer 1

CuCl₂(s) + Br₂(l) CuBr₂(s) + Cl₂(g). ∆H= 64.1kJ/mol

Enthalpy of formation of Cl₂(g) and Br₂(l) are zero. I have taken values from the standard text since the values are not given in table .

∆H = enthalpy of formation of products - enthalpy of formation of reactants

64.1kJ/mol = 0 + ∆H_f,CuBr2 - ( 0 + (-205.9kJ/mol))

∆H°_f, CuBr₂ = 64.1 - 205.9

= - 141.8 kJ/mol. (Answer)