Calculate the enthalpy change (delta H) in kJ for the following reaction.

2Al(s) + Fe2O3 (s) ==> 2 Fe(s) + Al2O3(s)

use enthalpy changes for the combustion of aluminum and iron.

2Al(s) + 3/2 O2(g) ==> Al2O3(s) Delta H: -1669.8 kJ

2Fe(s) + 3/2 O2(g) ==> Fe2O3 (s) Delta H: -824.2 kJ

Well, well, well! Looks like we have a reaction on our hands! Let's calculate the enthalpy change, shall we?

First, let's look at the balanced equations and enthalpy changes provided:

2Al(s) + 3/2 O2(g) ==> Al2O3(s) ΔH = -1669.8 kJ
2Fe(s) + 3/2 O2(g) ==> Fe2O3(s) ΔH = -824.2 kJ

From the given information, we can clearly see that we need to reverse the second equation and then add the two equations together to cancel out the common stoichiometry to isolate the desired reaction:

2Al(s) + 3/2 O2(g) ==> Al2O3(s) ΔH = -1669.8 kJ
-2Fe(s) - 3/2 O2(g) ==> -Fe2O3(s) ΔH = +824.2 kJ
------------------------------
2Al(s) - 2Fe(s) + 3/2 O2(g) ==> Al2O3(s) + Fe2O3(s) ΣΔH = -1669.8 kJ + 824.2 kJ = -845.6 kJ

And there you have it! The enthalpy change (ΔH) for the given reaction is -845.6 kJ. Keep in mind that this is a general calculation and may not reflect the actual conditions you encounter. Stay cool and keep clowning around!

To calculate the enthalpy change (ΔH) for the reaction, we need to use the enthalpy changes for the combustion of aluminum and iron, which are given.

The reaction can be divided into two steps:
1. Combustion of aluminum:
2Al(s) + 3/2 O2(g) → Al2O3(s) ΔH: -1669.8 kJ (Given)

2. Combustion of iron:
2Fe(s) + 3/2 O2(g) → Fe2O3(s) ΔH: -824.2 kJ (Given)

The values for the reactions involve 2 moles of aluminum and iron, so the enthalpy change for the overall reaction will be twice the given values. Let's multiply the reactions by the appropriate coefficients and add them together to get the overall reaction:

2 × (2Al(s) + 3/2 O2(g) → Al2O3(s))
= 4Al(s) + 3O2(g) → 2Al2O3(s)

2 × (2Fe(s) + 3/2 O2(g) → Fe2O3(s))
= 4Fe(s) + 3O2(g) → 2Fe2O3(s)

Now, add the two reactions together to get the overall reaction:

4Al(s) + 3O2(g) + 4Fe(s) + 3O2(g) → 2Al2O3(s) + 2Fe2O3(s)

The ΔH for the overall reaction will be the sum of the enthalpy changes for the individual reactions. Let's calculate it:

ΔH for overall reaction = 2 × ΔH for the combustion of aluminum + 2 × ΔH for the combustion of iron

= 2 × (-1669.8 kJ) + 2 × (-824.2 kJ)

= -3339.6 kJ - 1648.4 kJ

= -4988 kJ

Therefore, the enthalpy change (ΔH) for the reaction is -4988 kJ.

To calculate the enthalpy change (ΔH) for the given reaction, you need to use the enthalpy changes of the combustion reactions for aluminum (Al) and iron (Fe).

The given reaction shows that 2 moles of aluminum (Al) react with 1 mole of iron(III) oxide (Fe2O3) to produce 2 moles of iron (Fe) and 1 mole of aluminum oxide (Al2O3).

The enthalpy change for the combustion of aluminum (Al) is given as -1669.8 kJ. This means that when 2 moles of aluminum react with 3/2 moles of oxygen gas (O2), it releases 1669.8 kJ of energy.

Similarly, the enthalpy change for the combustion of iron (Fe) is given as -824.2 kJ. This means that when 2 moles of iron react with 3/2 moles of oxygen gas (O2), it releases 824.2 kJ of energy.

To calculate the enthalpy change for the reaction given, you can use Hess's Law, which states that the enthalpy change for a reaction is the sum of the enthalpy changes for the individual reactions that make up the overall reaction.

Since you need 2 moles of aluminum (Al) in the given reaction, you need to double the enthalpy change for the combustion of aluminum. Similarly, since you need 2 moles of iron (Fe), you also need to double the enthalpy change for the combustion of iron.

Calculating the enthalpy change for the given reaction:

2[(2Al(s) + 3/2 O2(g) ==> Al2O3(s)] (multiply by 2)

2[(-1669.8 kJ)] = -3339.6 kJ (enthalpy change for the combustion of aluminum)

2[(2Fe(s) + 3/2 O2(g) ==> Fe2O3(s)] (multiply by 2)

2[(-824.2 kJ)] = -1648.4 kJ (enthalpy change for the combustion of iron)

Now, subtract the enthalpy change for the combustion of iron from the enthalpy change for the combustion of aluminum:

-3339.6 kJ - (-1648.4 kJ) = -1691.2 kJ

Therefore, the enthalpy change (ΔH) for the given reaction is -1691.2 kJ.

415 kj