Calculate the standard enthalpy of reaction, ΔrH°, for the reaction below using the thermochemical equations given. Submit your answer with no units and to one decimal place.

2 MnO₂ (s) + CO (g)  → Mn₂O₃ (s) + CO₂ (g) ΔrH° = ?

 Thermochemical Equation

ΔrH° (kJ/mol)

MnO₂ (s) + CO (g)  → MnO (s) + CO₂ (g)

-166

Mn₃O₄ (s) + CO (g)  → 3 MnO (s) + CO₂ (g)

-46.6

3 Mn₂O₃ (s) + CO (g)  → 2 Mn₃O₄ (s) + CO₂ (g)

-142.4

Standard enthalpy of reaction , ?rH° is -253.47 kJ/ mol

Step-by-step explanation

Eq 1 :

MnO2 (s) + CO (g) → MnO (s) + CO2 (g) ?rH° = -166

Eq 2 :

Mn3O4 (s) + CO (g) → 3 MnO (s) + CO2 (g) ?rH° = -46.6

Eq 3 :

3 Mn2O3 (s) + CO (g) → 2 Mn3O4 (s) + CO2 (g) ?rH° = -142.4

Rearranging the equations :

1. Multiply eq 1 by 2 .
2. Reverse eqn 2 and multiply by 2/3
3. Reverse eqn 3 and multiply by 1/3

Then , new equations become

Eq 1 :

2 MnO2 (s) + 2 CO (g) → 2 MnO (s) + 2 CO2 (g)

?rH° = - 332

Eq 2 :

2 MnO (s) + 2/3 CO2 (g) → 2/3 Mn3O4 (s) + 2/3 CO

?rH° = 31.07

Eq 3 :

2/3 Mn3O4 (s) + 1/3 CO2 (g) → 1 Mn2O3 (s) + 1/3 CO (g)

?rH° = 47.46

Net equation becomes :

2 MnO2 (s) + CO (g) →Mn2O3 (s) + CO2 (g)

?rH° = -332 + 31.07 + 47.46

?rH° = - 253.47 kJ/ mol