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
see above.
To calculate the standard enthalpy change for the given reaction, we can 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 the individual reactions.
In this case, we need to determine the enthalpy change for the desired reaction by combining the enthalpy changes of the given reactions. We will use the stoichiometric coefficients of the reactions to manipulate them in order to match the desired reaction.
Given reactions:
1. 2Al(s) + (3/2)O2(g) ---> Al2O3(s) (enthalpy change = -1601 kJ/mol)
2. 2Fe(s) + (3/2)O2(g) ---> Fe2O3(s) (enthalpy change = -821 kJ/mol)
To match the desired reaction:
2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s)
We can reverse reaction 2 and multiply it by 2 so that it produces 2Fe(s):
2Fe2O3(s) ---> 4Fe(s) + 3O2(g)
Now, reverse reaction 1 and multiply it by 2 to produce 2Al2O3(s):
4Al2O3(s) ---> 8Al(s) + 6O2(g)
Combine the two manipulated reactions:
2Fe2O3(s) + 4Al(s) ---> 4Fe(s) + 2Al2O3(s)
To calculate the enthalpy change for this desired reaction, we can sum the enthalpy changes of the manipulated reactions:
[2Fe2O3(s) + 4Al(s) ---> 4Fe(s) + 2Al2O3(s)] (enthalpy change = ?)
= [2Fe2O3(s) ---> 4Fe(s) + 3O2(g)] + [4Al2O3(s) ---> 8Al(s) + 6O2(g)]
Now, let's add their enthalpy changes:
(-821 kJ/mol) + (-1601 kJ/mol) + 6(0 kJ/mol) + 3(0 kJ/mol)
Simplifying:
-821 kJ/mol + -1601 kJ/mol
= -2422 kJ/mol
Therefore, the standard enthalpy change for the given reaction:
2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s)
is -2422 kJ/mol.
To calculate the standard enthalpy change for the reaction:
2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s)
We can use the Hess's Law and the given enthalpy changes for the reactions:
Reaction 1: 2Al(s) + (3/2)O2(g) ---> Al2O3(s)
Change in H(rxn) = -1601 kJ/mol
Reaction 2: 2Fe(s) + (3/2)O2(g) ---> Fe2O3(s)
Change in H(rxn) = -821 kJ/mol
We need to manipulate the given reactions to get the desired overall reaction.
2Al(s) + Fe2O3(s) <--- 2Fe(s) + Al2O3(s)
We can reverse reaction 1 and multiply it by 2:
2Al2O3(s) ---> 4Al(s) + 3O2(g)
Change in H1 = - 2(1601 kJ/mol) = - 3202 kJ/mol
We can multiply reaction 2 by 2:
4Fe(s) + 3O2(g) ---> 2Fe2O3(s)
Change in H2 = 2(-821 kJ/mol) = - 1642 kJ/mol
Now, we can add the manipulated reactions together to get the overall reaction:
2Al2O3(s) + 4Fe(s) + 3O2(g) ---> 4Al(s) + 2Fe2O3(s)
And add the enthalpy changes:
Change in H overall = Change in H1 + Change in H2
= - 3202 kJ/mol + (- 1642 kJ/mol)
= - 4844 kJ/mol
Therefore, the standard enthalpy change for the reaction 2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s) is -4844 kJ/mol.