Answer:
[tex]\Delta _rH=-906.04kJ/mol[/tex]
Explanation:
Hello,
In this case of thermochemistry, the first step is to know all the involved species' enthalpies of formation as shown below (extracted from the NIST database):
[tex]\Delta _fH_{NH_3}^{(G)}=-45.9kJ/mol\\ \Delta _fH_{O_2}^{(G)}=0kJ/mol\\\Delta _fH_{H_2O}^{(G)}=241.8kJ/mol\\\Delta _fH_{NO}^{(G)}=90.29kJ/mol[/tex]
Now, for this particular chemical reaction and taking into account the involved stoichiometric coefficient, the feasible expression to compute the enthalpy change for that reaction is:
[tex]\Delta _rH=6\Delta _fH_{H_2O}^{(G)}+4\Delta _fH_{NO}^{(G)}-\Delta _fH_{NH_3}^{(G)}\\\Delta _rH=6(-241.8kJ/mol)+4(90.29kJ/mol)-4(-45.9kJ/mol)\\\Delta _rH=-906.04kJ/mol[/tex]
Best regards.
