Question

Using the provided table and the equation below, determine the heat of formation (in kJ/mol) for CuBr₂.
CuCl₂ (s) + Br₂(l) → CuBr₂ (s) + Cl₂ (g) ∆H° = 64.1 kJ/mol

CuCl2= -205.9 kJ/mol
Br2= 0 kJ/mol
Cl2= 0 kJ/mol

Answer 1

The heat of formation of CuBr₂ (s) = -141.8 kJ/mol

The heat of reaction ΔH = ∑ΔHproducts - ∑ΔHreactants

ΔH = ΔHCuBr₂ (s) + ΔHCl₂ (g) - [ΔHCuCl₂ (s) + ΔHBr₂(l)]

where ΔHCuBr₂ (s) = heat of formation of CuBr₂ (s),  ΔHCl₂ (g) = heat of formation of Cl₂ (g) = 0 kJ/mol, ΔHCuCl₂ (s) = heat of formation of CuCl₂ (s) = -205.9 kJ/mol, ΔHBr₂(l) = heat of formation of Br₂(l) = 0 kJ/mol and ΔH = heat of reaction = 64.1 kJ/mol

Substituting the values of the variables into the equation, we have

ΔH = ΔHCuBr₂ (s) + ΔHCl₂ (g) - [ΔHCuCl₂ (s) + ΔHBr₂(l)]

64.1 kJ/mol = ΔHCuBr₂ (s) + 0 kJ/mol - [-205.9 kJ/mol + 0 kJ/mol]

64.1 kJ/mol = ΔHCuBr₂ (s) + 0 kJ/mol + 205.9 kJ/mol - 0 kJ/mol

64.1 kJ/mol = ΔHCuBr₂ (s) + 205.9 kJ/mol

ΔHCuBr₂ (s) = 64.1 kJ/mol - 205.9 kJ/mol

ΔHCuBr₂ (s) = -141.8 kJ/mol

So, the heat of formation of CuBr₂ (s) = -141.8 kJ/mol

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