Calculate the ∆H for the following reaction: 4Al + 3MnO2🡪 2Al2O3 + 3Mn

Question

You are given these two equations:

4Al  + 3O2 🡪 2Al2O3  ∆H = -3352 kJ

Mn  + O2 🡪  MnO2  ∆H = -521 kJ

Answer 1

To solve this, we need to manipulate these two equations in a way so that that they can be added together in order to yield the resulting equation that we are looking for. To start, let's lay out all of the equations as we have them:

Equation 1: 4 Al + 3 O2 → 2 Al₂O₃ ΔH = -3352 kJ

Equation 2: Mn + O₂ → MnO₂ ΔH = -521 kJ

Final Equation: 4Al + 3MnO₂ → 2 Al₂O₃ + 3Mn ΔH = ???? kJ

In order to get to the final equation, we have to possibly multiply or divide equations one and two by themselves, reverse them, or add them together to get the result. To start, we know that we have to have 4 Al on the reactant side of the final equation. Equation one already gives us this, so let's keep it as is.

Equation 1: 4 Al + 3 O2 → 2 Al₂O₃ ΔH = -3352 kJ

Next, wwe need 3MnO₂ on the reactant side of the final equation. Equation two gives us one MnO₂ on the product side of the equation, therefore we must reverse the equation and the multiply the entire equation by three to get the desired reactant:

Equation 2: Mn + O₂ → MnO₂ ΔH = -521 kJ

Reversed equation 2: MnO₂ → Mn + O₂ ΔH = +521 kJ

Reversed and tripled equation 2: 3 MnO₂ → 3 Mn + 3 O₂ ΔH = +1563 kJ

When we reverse a chemical equation and the products become the reactants and vice versa, the ΔH much also flip signs. When we multiply a chemical equation by a number, we must multiply the ΔH by that same number. Now that we have done this, we can add equation 1 and the modified equation 2 in order to obtain the desired final equation:

Equation 1: 4 Al + 3 O2 → 2 Al₂O₃ ΔH = -3352 kJ

Modified equation 2: 3 MnO₂ → 3 Mn + 3 O₂ ΔH = +1563 kJ

Addition of equation 1 + 2: 4 Al + 3 MnO₂ → 2 Al₂O₃ + 3 Mn ΔH = +1563 kJ - 3352 kJ = -1789 kJ

When adding two equations together, we add the ΔH of each equation as well in order to obtain the ΔH of the added equation. Looking at the addition of the two equations, we were able to obtain the final equation that we were looking for. Therefore, the ΔH that we are looking for is equal to the ΔH for that equation:

Final answer: ΔH = -1789 kJ

Answer 2

Hey again Priscilla,

Yes, I agree, this is a more difficult problem. This one involves flipping reactions and multiplying by integers. My strategy on this problem would be to write the given equations on top of each other, then ask myself: what species do not appear in the final equation? In this example, the species is O₂. Therefore, let's focus on cancelling this term out.

I also notice that O₂ appears on the left side for both given reactions. In order to cancel this term, we will need to flip one of the reactions. I see that Al, one of our desired species in the final reaction, is on the left where it needs to be. Therefore, let's flip the second reaction:

4Al + 3O₂ -> 2Al₂O₃

MnO₂ -> Mn + O₂

It is very important to stop and take note here that we will flip the sign of the delta H for this specific reaction when adding up the final enthalpy.

Next, I see that the O₂ doesn't quite cancel out. There are 3 O₂ on the left and 1 O₂ on the right. Therefore, let's multiple the second equation by 3, this will allow us to cancel the O₂:

4Al + 3O₂ -> 2Al₂O₃

3MnO₂ -> 3Mn + 3O₂

Again, remember that we will multiple the enthalpy change in the second reaction by 3 (in addition to flipping the sign). When we add these reactions together and cancel out the O₂, we get the final reaction:

4Al + 3MnO₂ -> 2Al₂O₃ + 3Mn

The enthalpy then adds up as: -3352 kJ + -3(-521 kJ) = -1789 kJ.

These problems are best solved by trying things out until you can cancel out the necessary species, just keep with it!

Thank you,

Jacob P.