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Given the following thermochemical equations N2... - SS3 Chemistry Chemical Thermodynamics Question

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)

A

-91.8 kJ/mol

 

B

-5.8 kJ/mol

 

C

-270.6 kJ/mol

 

D

273.0 kJ/mol

correct option: a

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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