Given the standard enthalpy changes for the following two reactions: (1) 2Fe(s) + O2(g)2FeO(s) ΔH° = -544.0 kJ (2) 2Zn(s) + O2(g)2ZnO(s)ΔH° = -696.6 kJ what is the standard enthalpy change for the reaction:

-76.3 kJ

Explanation:

Here is the complete question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g) → 2FeO(s)ΔH° = -544.0 kJ

(2) 2Zn(s) + O₂(g) → 2ZnO(s)ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:

(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)ΔH° = ?

Solution

Since (1) 2Fe(s) + O₂(g) → 2FeO(s)ΔH° = -544.0 kJ

reversing the reaction, we have

2FeO(s) → 2Fe(s) + O₂(g) ΔH° = +544.0 kJ (4)

Adding reactions (2) and (3), we have

2FeO(s) → 2Fe(s) + O₂(g) ΔH° = +544.0 kJ (4)

2Zn(s) + O₂(g) → 2ZnO(s)ΔH° = -696.6 kJ  (2)

This gives

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)ΔH° =

The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)

= +544.0 kJ - 696.6 kJ)

= -152.6 kJ

Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)

we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).

So, ΔH° = -152.6 kJ/2 = -76.3 kJ

So, the standard enthalpy change for the reaction

FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ

your answer would be “dry”