chemistry hess law

Question

Consider the following equations:

2AI(s)+3/2O2(g)--->AI2O3(s) deltaH= -1670kJ

Mn(s)+O2(g)--->MnO2(s) deltaH= -520kJ

what is the standard enthalpy change, in kJ, of the reaction below?

4AI(s)+3MnO2(s)--->2AI2O3(s)+3Mn(s)

Answer 1

Hess' Law: rearrange the given equations in order to obtain the target equation

4AI(s) + 3MnO₂(s) ---> 2AI₂O (s) + 3Mn(s) ... TARGET EQUATION

Eq.1: 2AI(s) + 3/2 O₂(g) ---> AI₂O₃(s) ... ∆H= -1670kJ

Eq.2: Mn(s) + O₂(g) ---> MnO₂(s) ... ∆H= -520kJ

Procedure:

Multiply Eq.1 by 2: 4AI(s) + 3O₂(g) ---> 2AI₂O₃(s) ... ∆H= -3340kJ

Reverse Eq.2 and multiply by 3: 3MnO₂(s) --> 3Mn(s) + 3O₂(g) ... ∆H = +1560 kJ

Add them together to get:

4AI(s) + 3O₂(g) + 3MnO₂(s) --> 2AI₂O₃(s) + 3Mn(s) + 3O₂(g)

Combine and cancel like terms to end up with:

4AI(s) + 3MnO₂(s) --> 2AI₂O₃(s) + 3Mn(s) ... TARGET EQUATION

To find the ∆H, just add the ∆H values of the above equations:

∆H = -3340 kJ + 1560 kJ = -1780 kJ