Question

Calculate the enthalpy change for the reaction below given the standard heat of combustion for ammonia, NH3, is -226 kJ/mol. 4NH3 (g) + 5O2(g) --> 4NO(g) + 6H2O(g)

Answer 1

The given equation is: 4NH3(g) + 5O2(g) ----> 4NO(g) + 6 H2O(g)

and the given data is, Standard heat of combustion for ammonia, = -226 kJ/mol

The Standard heat of combustion of any substance = -(Standard heat of formation of that substance), because in on process substance is being made(formation) and in the other substance the substance is being decomposed/destroyed (combustion).

Therefore, Standard heat of formation of NH3= 226 kJ/mol

Standard heat of combustion for O2 = Standard heat of O2 = 0 kJ/mol , Since it is present in its pure form and no energy is used in its formation.

Standard heat of formation of NO(g) = 90.25 kJ/mol

Standard heat of formation of H2O(g) = −241.8 kJ/mol

Accorfing to Hess's law,

= {[4(-90.25)] + [6(−241.8)]} - {[4(226)] + [5(0)]} kJ/mol

= {361 - 1450.8} - {904}   kJ/mol

        

Therefore, the enthalpy change for the above reaction is -1993.8 kJ/mol, negative sign shows its an exothermic reaction.