Calculate the enthalpy change (delta H) in kJ for the following reaction.
2Al(s) + Fe2O3 (s) ==> 2 Fe(s) + Al2O3(s)
use enthalpy changes for the combustion of aluminum and iron.
2Al(s) + 3/2 O2(g) ==> Al2O3(s) Delta H: -1669.8 kJ
2Fe(s) + 3/2 O2(g) ==> Fe2O3 (s) Delta H: -824.2 kJ
Delta H = -1669.8 kJ + 824.2 kJ = -845.6 kJ
To calculate the enthalpy change (ΔH) for the given reaction, you need to use Hess's law. Hess's law states that if a reaction can be expressed as a sum of several other reactions, then the enthalpy change for the original reaction is the sum of the enthalpy changes of those reactions.
In this case, the reaction can be expressed as a combination of the combustion reactions of aluminum and iron:
2Al(s) + 3/2 O2(g) ==> Al2O3(s) ΔH = -1669.8 kJ
2Fe(s) + 3/2 O2(g) ==> Fe2O3(s) ΔH = -824.2 kJ
To calculate the enthalpy change for the desired reaction, you need to manipulate the given reactions to match the desired reaction, both in terms of reactants and products.
The given reactions have 2 moles of aluminum (Al) on the left side and 2 moles of aluminum oxide (Al2O3) on the right side. The desired reaction also has 2 moles of aluminum (Al) on the left side and 1 mole of aluminum oxide (Al2O3) on the right side.
To achieve this, multiply the second reaction by 2:
4Fe(s) + 3O2(g) ==> 2Fe2O3(s) (multiply by 2)
Now, the desired reaction can be obtained by combining the two modified reactions:
2Al(s) + Fe2O3(s) ==> 2Fe(s) + Al2O3(s) (desired reaction)
Combining the two modified reactions also combines their respective enthalpy changes.
The ΔH value for the desired reaction is the sum of the ΔH values of the modified reactions:
ΔH = -1669.8 kJ + (-824.2 kJ)
ΔH = -2494 kJ
Therefore, the enthalpy change (ΔH) for the given reaction is -2494 kJ.
To calculate the enthalpy change (ΔH) for the given reaction, we need to use Hess's Law, which states that if a reaction can be expressed as the sum of two or more chemical equations, then the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.
Given equations:
1) 2Al(s) + 3/2 O2(g) ==> Al2O3(s) ΔH = -1669.8 kJ
2) 2Fe(s) + 3/2 O2(g) ==> Fe2O3(s) ΔH = -824.2 kJ
We need to manipulate the given equations to match the desired reaction:
1) Reverse the first equation:
Al2O3(s) ==> 2Al(s) + 3/2 O2(g)
Multiply ΔH by -1 to reverse its sign:
ΔH1 = 1669.8 kJ
2) Multiply the second equation by 2 to get 2Fe(s) in the desired reaction:
4Fe(s) + 3O2(g) ==> 2Fe2O3(s)
Multiply ΔH by 2 to account for the change in stoichiometric coefficients:
ΔH2 = -1648.4 kJ
Now, we can add the two manipulated equations together to get the desired reaction:
ΔH total = ΔH1 + ΔH2
ΔH total = 1669.8 kJ + (-1648.4 kJ)
ΔH total = 21.4 kJ
Therefore, the enthalpy change (ΔH) for the given reaction is 21.4 kJ.