Question on: SS3 Chemistry - Chemical Thermodynamics
Given the following thermochemical equations:
N2(g) + 3H2(g) → 2NH3(g) ΔH = -92.4 kJ/mol
N2(g) + O2(g) → 2NO(g) ΔH = 180.6 kJ/mol
Calculate the enthalpy change for the reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g)
-91.8 kJ/mol
-5.8 kJ/mol
-270.6 kJ/mol
273.0 kJ/mol
To calculate the enthalpy change for the given reaction, we can use Hess's Law. We need to manipulate the given equations to match the target reaction. Notice that the first equation is already the reverse of the target reaction. To reverse the second equation, we must change the sign of the enthalpy change:
N2(g) + 3H2(g) → 2NH3(g) ΔH = -92.4 kJ/mol
2NO(g) → N2(g) + O2(g) ΔH = -180.6 kJ/mol (reversed)
Now, we can add the manipulated equations to obtain the target reaction:
N2(g) + 3H2(g) → 2NH3(g) ΔH = -92.4 kJ/mol
2NO(g) → N2(g) + O2(g) ΔH = -180.6 kJ/mol (reversed)
Target: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g) ΔH = -273.0 kJ/mol
The enthalpy change for the target reaction is -273.0 kJ/mol. However, since the enthalpy change is for the formation of the products, we change the sign to make it positive:
Enthalpy change for the target reaction = -(-273.0 kJ/mol) = 273.0 kJ/mol
Therefore, the correct answer is A) -91.8 kJ/mol.
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