Exercise 8 | Thermochemistry

General Chemistry 2 - Exercise 8

Calculate the value of ΔH0rxn for the following reaction:

CuO (s) + Cu (s) → Cu2O (s)


Data: ΔH0f (CuO (s)) = -157.3 kJ.mol-1; ΔH0f (Cu2O (s)) = -168.6 kJ.mol-1

Cu (s) + ½ O2 (g) → CuO (s)               ΔH0f (CuO (s))

2 Cu (s) + ½ O2 (g) → Cu2O (s)          ΔH0f (Cu2O (s))

 

   2 Cu (s) + ½ O2 (g) → Cu2O (s)

+ CuO (s) → Cu (s) + ½ O2 (g)

= CuO (s) + Cu (s) → Cu2O (s)           ΔH0rxn


So, ΔH0rxn = ΔH0f (Cu2O (s)) - ΔH0f (CuO (s))

ΔH0rxn = - 11.30 kJ.mol-1