Find the enthalpy for : 4Fe + 3O2 = 2Fe2O3
I got the following informations:
Fe + 3H2O = Fe(OH)3 + 3/2H2 - Enthalpy is 160.9 kj
H2 + 1/2O2 = H2O - Enthalpy is -285.8 kj
Fe2O3 + 3H2O = 2Fe(OH)3 - Enthalpy is 288.6
I try using Hess Law but cannot solve it. Can anyone help me find the enthalpy? Thanks
I think you multiply equation 1 by 4.
Reverse equation 3 and multiply by 2.
Add in equation 2 multiplied by 6.
To find the enthalpy for the reaction 4Fe + 3O2 = 2Fe2O3, you can use Hess's Law by manipulating the given reactions and their corresponding enthalpies.
First, let's multiply the first equation (Fe + 3H2O = Fe(OH)3 + 3/2H2) by 2 to get the same number of Fe(OH)3 as in the desired equation:
2Fe + 6H2O = 2Fe(OH)3 + 3H2
Then, multiply the second equation (H2 + 1/2O2 = H2O) by 3:
3H2 + 3/2O2 = 3H2O
Finally, multiply the third equation (Fe2O3 + 3H2O = 2Fe(OH)3) by 2:
2Fe2O3 + 6H2O = 4Fe(OH)3
Now, let's combine these manipulated equations to form the desired equation:
4Fe + 6H2O + 3H2 + 3/2O2 = 2Fe2O3 + 2Fe(OH)3 + 3H2O
Next, adjust the equation by canceling out the common species on both sides:
4Fe + 4H2O + 3/2O2 = 2Fe2O3 + 2Fe(OH)3
Now, we can calculate the enthalpy change for the reaction using Hess's Law:
Enthalpy change = Σ Enthalpy of products - Σ Enthalpy of reactants
Enthalpy change = (2 × 288.6) + (2 × 160.9) - (4 × 0) - (4 × 285.8)
Calculating the values:
Enthalpy change = 577.2 + 321.8 - 0 - 1143.2
Enthalpy change = -244.2 kJ
Therefore, the enthalpy change for the reaction 4Fe + 3O2 = 2Fe2O3 is -244.2 kJ.
To find the enthalpy change for the reaction 4Fe + 3O2 = 2Fe2O3, you can use Hess's Law by combining the given equations in a way that cancels out the intermediates. Here's the step-by-step process:
1. Start with the given equation: 4Fe + 3O2 = 2Fe2O3
2. Let's consider the first equation: Fe + 3H2O = Fe(OH)3 + 3/2H2, which has an enthalpy change of +160.9 kJ.
3. Multiply this equation by 2 to match the stoichiometry of Fe2O3 in the target equation. This gives us: 2Fe + 6H2O = 2Fe(OH)3 + 3H2, with an enthalpy change of 2 × (+160.9 kJ) = +321.8 kJ.
4. Next, consider the second equation: H2 + 1/2O2 = H2O, with an enthalpy change of -285.8 kJ.
5. Multiply this equation by 3 to match the stoichiometry of O2 in the target equation. This gives us: 3H2 + 3/2O2 = 3H2O, with an enthalpy change of 3 × (-285.8 kJ) = -857.4 kJ.
6. Now, sum the modified equations obtained in steps 3 and 5 to eliminate the intermediates: (2Fe + 6H2O) + (3H2 + 3/2O2) = (2Fe(OH)3 + 3H2O) + 3H2O.
This simplifies to: 2Fe + 3/2O2 = 2Fe(OH)3 + 6H2O, with an enthalpy change of +321.8 kJ -857.4 kJ = -535.6 kJ.
7. Finally, consider the third equation: Fe2O3 + 3H2O = 2Fe(OH)3, with an enthalpy change of +288.6 kJ.
8. Since this equation already contains the desired product and reactant, it can be directly used without modification.
9. Now, subtract the equation obtained in step 6 from the equation in step 8 to get the target equation: (Fe2O3 + 3H2O) - (2Fe + 3/2O2) = (2Fe(OH)3) + 6H2O - (2Fe(OH)3 + 6H2O).
Simplifying gives us: Fe2O3 - 2Fe + 3/2O2 = 0, with an enthalpy change of +288.6 kJ -(-535.6 kJ) = +824.2 kJ.
Therefore, the enthalpy change for the given reaction 4Fe + 3O2 = 2Fe2O3 is +824.2 kJ.