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)?

Jun 2, 2017

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

Explanation:

And here, $\Delta {H}_{\text{rxn}}^{\circ} = \ldots \ldots \ldots \ldots \ldots \ldots$

$\left(4 \times 90.3 - 6 \times 241.8 - \left\{4 \times - 45.9\right\}\right) \cdot k J \cdot m o {l}^{-} 1$

$= - 906 \cdot k J \cdot m o {l}^{-} 1$.

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

ANd of course, $\Delta {H}_{f}^{\circ} = 0$ for an element (here dixoygen) in its standard state (they even spoon-feed you this fact).