Exercise 3 | Thermochemistry

Given that:

C (s) + O2 (g) → CO2 (g)                     ∆H1 = -393.5 kJ

2 Fe(s) + 3/2 O2 (g) → Fe2O3 (s)       ∆H2 = -824.2 kJ


Calculate the enthalpy change for 2 Fe2O(s) + 3 C (s) → 4 Fe (s)  + 3 CO2 (g)

   2 Fe2O(s) → 4 Fe (s) + 3 O2 (g)                              ∆H = -2 ∆H2

+ 3 C (s) + 3 O2 (g) → 3 CO2 (g)                                   ∆H = 3 ∆H1

= 2 Fe2O(s) + 3 C (s) → 4 Fe (s)  + 3 CO2 (g)          ∆H3

 

So, ∆H3 = 3 ∆H1 - 2 ∆H2 = 467.9 kJ