when aluminum metal reacts with iron (III) oxide to form aluminum oxide and iron metal, 429.6 kJ of heat are given off for each mole of aluminum metal consumed, under constant pressure and standard conditions. What is the correct value for the standard enthalpy of reaction in the thermochemical reaction...
2Al + Fe2O3--> 2Fe + Al2O3
2019 ;-;
Well, that reaction sure sounds hot! It's like aluminum and iron (III) oxide are having a fiery party, giving off 429.6 kJ of heat per mole of aluminum consumed. So, the correct value for the standard enthalpy of reaction in this thermochemical reaction is -429.6 kJ/mol. Just remember, it's a chemically hot situation!
To determine the standard enthalpy of reaction for the thermochemical equation given, you need to use the concept of Hess's Law. Hess's Law states that the standard enthalpy change of a reaction is equal to the sum of the standard enthalpy changes of the individual steps of the reaction.
In this case, we'll consider the individual steps:
Step 1: Formation of iron (II) oxide (FeO) from iron (III) oxide (Fe2O3)
Fe2O3 + 3CO → 2FeO + 3CO2
Step 2: Formation of aluminum oxide (Al2O3) from aluminum (Al)
2Al + 3/2O2 → Al2O3
Step 3: Formation of iron (II) oxide (FeO) from iron (Fe)
Fe + 1/2O2 → FeO
Now, we'll use the given information to calculate the enthalpy change for each step:
Step 1: ΔH1 = 0 kJ (No heat is given off or absorbed)
Step 2: ΔH2 = unknown (To be determined)
Step 3: ΔH3 = -429.6 kJ (Heat is given off)
Given that the coefficients in the balanced thermochemical equation are:
2Al + Fe2O3 → 2Fe + Al2O3
We can write the overall reaction as the sum of the individual steps:
2Al + 3/2O2 + Fe → Al2O3 + 2FeO
Now, using Hess's Law, we can write the equation for the overall reaction:
ΔH = ΔH2 + ΔH3
Substituting the known values, we have:
ΔH = ΔH2 - 429.6 kJ
Therefore, the correct value for the standard enthalpy of reaction (ΔH) in the thermochemical reaction 2Al + Fe2O3 → 2Fe + Al2O3 is -ΔH2 = -(-429.6 kJ) = +429.6 kJ.
To determine the standard enthalpy of reaction for the given thermochemical reaction, you need to use the concept of Hess's law.
Hess's law states that the standard enthalpy change of a reaction can be calculated by considering the standard enthalpy changes of other reactions. In this case, you can use the standard enthalpy of formation values to calculate the enthalpy change.
The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.
First, you need to determine the standard enthalpy of formation for each of the compounds involved in the reaction: aluminum oxide (Al2O3) and iron (Fe). These values can be found in reference tables or online databases.
The standard enthalpy of formation for aluminum oxide (Al2O3) is -1675 kJ/mol, and for iron (Fe) is 0 kJ/mol (since it is in its standard state).
Next, you can use these values to calculate the standard enthalpy of reaction by summing up the enthalpies of formation for the products and subtracting the enthalpies of formation for the reactants:
ΔH = Σ(ΔHf,products) - Σ(ΔHf,reactants)
For the given reaction: 2Al + Fe2O3 → 2Fe + Al2O3
ΔH = [2 × ΔHf(Fe) + ΔHf(Al2O3)] - [2 × ΔHf(Al) + ΔHf(Fe2O3)]
Substituting the respective values:
ΔH = [2 × 0 kJ/mol + (-1675 kJ/mol)] - [2 × 0 kJ/mol + 0 kJ/mol]
Simplifying the equation:
ΔH = (-2 × 1675 kJ/mol) - (0 kJ/mol) = -3350 kJ/mol
Therefore, the correct value for the standard enthalpy of reaction in the thermochemical reaction 2Al + Fe2O3 → 2Fe + Al2O3 is -3350 kJ/mol.