Hess's Law Problem. Please Help

Question

Given the following data:

Fe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO2 (g) ΔHO = -23.4 kJ

3 Fe2O3 (s) + CO (g) → 2 Fe3O4 (s) + CO2 (g) ΔHO = -48.3 kJ

Fe3O4 (s) + CO (g) → 3 FeO (s) + CO2 (g) ΔHO = +21.8 kJ

Calculate ΔHO for the reaction:

FeO (s) + CO (g) → Fe (s) + CO2 (g)

Answer 1

Hello John, you have to combine the three given equations to get the required equation. 

 

Thus,

 

Take Rxn 1 and times by 3:       3 Fe2O3 (s) + 9 CO (g) → 6 Fe (s) + 9 CO2 (g)         ΔHO = -23.4 x 3 kJ

 

reverse Rxn 2:                              Fe3O4 (s) + CO2 (g)  → 3 Fe2O3 (s) + CO (g)      ΔHO = 48.3 kJ  (note sign change )

 

reverse Rxn 3 and times by 2:     6 FeO (s) + 2 CO2 (g) →  2 Fe3O4 (s) + 2 CO (g)        ΔHO = - 21.8 x 2 kJ

 

If you add the three equation you will get:

                                 6 FeO (s) + 6 CO (g) → 6 Fe (s) + 6 CO2 (g)    ΔH = (-23.4 x 3 )kJ + 48.3 kJ + (- 21.8 x 2) kJ 

 

Thus for

 

                                  FeO (s) + CO (g) → Fe (s) +  CO2 (g)       ΔH = (-23.4 x 3 )kJ + 48.3 kJ + (- 21.8 x 2) kJ 

                                                                                                                             6

 

I will let you to finish the answer.

 

      

 

 

Answer 2

The goal is the reaction   FeO + CO → Fe + CO2  It is easier to work toward the triple of this, then divide ΔH0 by 3

That is: work towards      3 FeO + 3 CO → 3Fe + 3 CO2

 

Since  FeO is wanted on the LHS, start with the reverse of the third reaction:

 3 FeO + CO2   →  Fe3O4 +CO                       ΔH0 = -21.8

 Next,  to balance out the unwanted   Fe3O4 on the RHS,   use the reverse of one half the second equation

  Fe3O4 +(1/2) CO2  → (3/2) Fe2O3 + (1/2) CO        ΔHo  = 48.3 /2 = 24.15

 

Combining these two equations and noticing the cancellation of the Fe3O4  results in

   3 FeO +(3/2) CO2  →  (3/2) Fe2O3 +(3/2) CO       ΔH0 = 2.35

 

Next, to balance out the unwanted (3/2) Fe2O3 on the RHS use (3/2) times the first equation

 

 (3/2) Fe2O3 + (9/2) CO → 3 Fe + (9/2) CO2         ΔH0 = (3/2) (-23.4) = - 35.1

 

Combining with the equation just previous and noticing the cancellation of the (3/2) Fe2O3  gives

 

 3FeO + (3/2) CO2 + (9/2) CO →  3 Fe + (3/2) CO +(9/2) CO2    ΔH0 =  -32.75

 

Combining the coefficients of the CO and CO2  results in

 

  3FeO +3 CO  →  3 Fe + 3 CO2            ΔH0 =  - 32.75

 

Finally dividing through by 3 results in

 

  FeO + CO  →  Fe + CO2      ΔH0 = - 10.92  kj