Given that:
2Al(s)+(3/2)O2(g)--->Al2O3(s)
change in H(rxn)= -1601 kJ/mol
2Fe(s)+(3/2)O2(g)--->Fe2O3(s)
change in H(rxn)= -821 kJ/mol
Calculate the standard enthalpy change for the following reaction:
2Al(s)+Fe2O3(s)--->2Fe(s)+Al2O3(s)
_______kJ
Thank you so much :)
To calculate the standard enthalpy change for the reaction:
2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s)
We need to use the Hess's Law, which states that the enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps.
First, we can write the individual reactions and their corresponding enthalpy changes:
1. 2Al(s) + (3/2)O2(g) ---> Al2O3(s) (change in H1 = -1601 kJ/mol)
2. 2Fe(s) + (3/2)O2(g) ---> Fe2O3(s) (change in H2 = -821 kJ/mol)
We can observe that the overall reaction is the sum of these two reactions, where the reactant Al2O3(s) in the first reaction is the product in the second reaction.
To calculate the overall enthalpy change, we sum the individual enthalpy changes while paying attention to their stoichiometric coefficients:
(change in H1) + (change in H2)
(-1601 kJ/mol) + (-821 kJ/mol)
= -2422 kJ/mol
Therefore, the standard enthalpy change for the reaction:
2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s)
is -2422 kJ/mol.
To calculate the standard enthalpy change for the given reaction, we will use Hess's law. Hess's law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps.
First, let's write the given reactions in terms of the desired reaction:
1. Reverse the second equation:
Fe2O3(s) --> 2Fe(s) + (3/2)O2(g)
2. Multiply the first equation by 2 to balance the number of Al atoms:
4Al(s) + 3O2(g) --> 2Al2O3(s)
Now, let's add the two equations together to obtain the desired reaction:
4Al(s) + 3O2(g) + Fe2O3(s) --> 2Al2O3(s) + 2Fe(s) + (3/2)O2(g)
By canceling out the common molecules on both sides of the equation, we get:
4Al(s) + Fe2O3(s) --> 2Al2O3(s) + 2Fe(s)
Now, we can add the enthalpy changes of the individual reactions to calculate the standard enthalpy change of the desired reaction:
ΔH(rxn) = ΔH1 + ΔH2
ΔH1: The enthalpy change of the first reaction (2Al(s) + (3/2)O2(g) --> Al2O3(s))
ΔH1 = -1601 kJ/mol
ΔH2: The enthalpy change of the second reaction (Fe2O3(s) --> 2Fe(s) + (3/2)O2(g))
ΔH2 = -821 kJ/mol
Now, we substitute the enthalpy changes into the equation:
ΔH(rxn) = -1601 kJ/mol + (-821 kJ/mol)
ΔH(rxn) = -2422 kJ/mol
Therefore, the standard enthalpy change for the reaction 2Al(s) + Fe2O3(s) --> 2Fe(s) + Al2O3(s) is -2422 kJ.
Use equation 1 as is.
Reverse equation 2 (change the sign) and add to equation 1. Add the delta Hs to arrive at the total delta H for the new reaction.