How do you calculate the standard enthalpy of reaction, #DeltaH_(rxn)^@# for the following reaction from the given standard heats formation (#DeltaH_f^@#) values: #4NH_3(g) + 5O_2(g) - 4NO(g) + 6H_2O(g)#?

Question

Mateo Aguilar · Accepted Answer

#DeltaH_"rxn"^@=DeltaH_f^@("products")-DeltaH_f^@("reactants")#

And here, #DeltaH_"rxn"^@=..................#

#(4xx90.3-6xx241.8-{4xx-45.9})*kJ*mol^-1#

#=-906*kJ*mol^-1#.

The formation of water clearly dominates the enthalpy change, it is a thermodynamic sink.

ANd of course, #DeltaH_f^@=0# for an element (here dixoygen) in its standard state (they even spoon-feed you this fact).

Owen Ambrose · Answer

undefined $$ \Delta H_{\text{rxn}}^{\circ} = \sum (\Delta H_{f}^{\circ} \text{ of products}) - \sum (\Delta H_{f}^{\circ} \text{ of reactants}) $$

$$ \Delta H_{\text{rxn}}^{\circ} = [4 \Delta H_{f}^{\circ}(\text{NO}) + 6 \Delta H_{f}^{\circ}(\text{H}_2\text{O})] - [4 \Delta H_{f}^{\circ}(\text{NH}_3) + 5 \Delta H_{f}^{\circ}(\text{O}_2)] $$

How do you calculate the standard enthalpy of reaction, #DeltaH_(rxn)^@# for the following reaction from the given standard heats formation (#DeltaH_f^@#) values: #4NH_3(g) + 5O_2(g) -&gt; 4NO(g) + 6H_2O(g)#?

How do you calculate the standard enthalpy of reaction, #DeltaH_(rxn)^@# for the following reaction from the given standard heats formation (#DeltaH_f^@#) values: #4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O(g)#?