What is the enthaphly change when 2 moles of fe(111)oxide is reduced by 3 moles of aluminium ♢Hf Al2O3= -1676kJ mol-1 ♢Hf (Fe2O3)= -825kJmol-1
To calculate the enthalpy change (∆H) when 2 moles of Fe2O3 is reduced by 3 moles of aluminum (Al), you need to use the enthalpy of formation (∆Hf) values of Al2O3 and Fe2O3.
Given:
∆Hf Al2O3 = -1676 kJ/mol
∆Hf Fe2O3 = -825 kJ/mol
The reaction equation is:
2 Fe2O3 + 3 Al -> 3 Fe + Al2O3
To find the ∆H of the overall reaction, you can use Hess's Law, which states that the ∆H of a reaction is equal to the sum of the ∆H of the individual reactions.
The reaction can be broken down into two steps:
1. The formation of Al2O3:
Al + (3/2)O2 -> Al2O3
The ∆H for this reaction can be obtained from the given ∆Hf of Al2O3:
∆H1 = ∆Hf Al2O3 = -1676 kJ/mol
2. The formation of Fe from Fe2O3:
Fe2O3 -> 2 Fe + (3/2) O2
The ∆H for this reaction can be obtained by reversing the sign of the ∆Hf of Fe2O3:
∆H2 = -(-825 kJ/mol) = 825 kJ/mol
Now, we can calculate the overall ∆H of the reaction by adding ∆H1 and ∆H2:
∆Hoverall = ∆H1 + ∆H2
∆Hoverall = -1676 kJ/mol + 825 kJ/mol
∆Hoverall = -851 kJ/mol
Therefore, the enthalpy change when 2 moles of Fe2O3 is reduced by 3 moles of aluminum is -851 kJ/mol.