Question

What is the enthalpy change for this reaction?

Answer 1

boyy that question is too difficult 

Answer 2

`\Delta _rH=-906.04kJ/mol`

Step-by-step 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):

`\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`

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:

`\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`

Best regards.