A) 2Fe(s) + O2(g) =>
B) 2FeO(s) ΔH = -544.0 kJ 4Fe(s) + 3O2(g) =>
C) 2Fe2O3(s) ΔH = -1648.4 kJ Fe3O4(s) =>
D) 3Fe(s) + 2O2(g) ΔH = +1118.4 kJ
Given the data above, determine the heat of reaction, ΔH, for the reaction below:
Reaction: Fe2O3(s) + FeO(s) => Fe3O4(s)

Respuesta :

Answer:  -22.2 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps. SGDSDGSDGSDGgsg

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(1)  [tex]2Fe(s)+O_2(g)\rightarrow 2FeO(s)[/tex]  [tex]\Delta H_1=-544.0kJ[/tex]

(2)  [tex]4Fe(s)+O_2(g)\rightarrow 2Fe_2O_3(s)[/tex]  [tex]\Delta H_2=-1648.4kJ[/tex]

(3)  [tex]Fe_3O_4(s)\rightarrow 3Fe(s)+2O_2(g)[/tex]  [tex]\Delta H=+1118.4kJ[/tex]

Reversing 1 ,2 and 3 and halving 1 and 2 and then adding we get net equation:

(4) [tex]Fe_2O_3(s)+FeO(s)\rightarrow Fe_3O_4(s)[/tex]  [tex]\Delta H_4=?[/tex]

[tex]\Delta H_4=\frac{-\Delta H_1}{2}+\frac{-\Delta H_2}{2}+(-\Delta H_3)=\frac{544.0}{2}+\frac{1648.4}{2}+(-1118.4)=-22.2kJ[/tex]

Therefore, the heat of reaction, ΔH, for the reaction is -22.2 kJ