Calculate ΔH for the reaction

Question

Calculate ΔH for the reaction:

2N2 (g) + 6H2O (g) → 3O2 (g) + 4NH3 (g) 

Answer 1

This question relies on the idea of state functions and Hess's Law. In short, Hess's Law states the ΔG, ΔH, or ΔS for a particular reaction may be determined by summing the ΔG, ΔH, or S values for any cumulative series of subreactions for that particular reaction. Note that in the case of entropy, you can sum the entropy values for each reactant and product in the reaction; we can explain this using the third law of thermodynamics, but it's not quite the discussion here!

Supposing that this was all of the information you were provided, namely that you were not given a series of subreactions, then you would also need to reference a chart (likely in your textbook). I referenced the chart in the following link:

https://sistemas.eel.usp.br/docentes/arquivos/5817712/TDQ%20I/R-standard_enthalpy_of_formation.pdf

ΔHº_rxn = ΣΔHº_{formation products} - ΣΔHº_{formation reactants}

ΔHº_rxn = [3O_2(g) + 4NH_3(g)] - [2N_2(g) + 6H₂O_(g)]

=> ΔHº_formation O_2(g) = 0 because it is an element in its most stable form

=> ΔHº_formation N_2(g) = 0 because it is an element in its most stable form

ΔHº_rxn = [4NH_3(g)] - [6H₂O_(g)]

<=> [4*(-46.2 KJ*mol^-1)] - [6*(-241.8 KJ*mol^-1)]

ΔHº_rxn = 1266 kJ*mol^-1

I hope this helps!

Cheers!

Answer 2

Find a table of heats of formation in the back of your text: The enthalpy of reactions is sum of ν_iΔH_f,i where ν_i are the stoichiometric coefficients (defined as negative for reactants)

4ΔH_f,NH3(g) - 6ΔH_f,H2O(g)

Elements in their natural state have ΔH_f = 0 and be careful to pick the right form of water.