# How do you calculate the enthalpy change in the following reaction 4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O?

Jun 1, 2017

Well first you need some data............I eventually get $\Delta {H}_{\text{rxn}} = - 1028.6 \cdot k J \cdot m o {l}^{-} 1$.

#### Explanation:

And I quote $\Delta {H}_{f}^{\circ}$ values for the given substances. I use this site, and perhaps you should check that I have got the values right.......

$N {H}_{3} \left(g\right)$, $\Delta {H}_{f}^{\circ} = - 80.8 \cdot k J \cdot m o {l}^{-} 1$

$N O \left(g\right)$, $\Delta {H}_{f}^{\circ} = + 90.29 \cdot k J \cdot m o {l}^{-} 1$

${H}_{2} O \left(l\right)$, $\Delta {H}_{f}^{\circ} = - 285.8 \cdot k J \cdot m o {l}^{-} 1$

And of course $\Delta {H}_{f}^{\circ}$ ${O}_{2} \left(g\right)$ $=$ $0$, by definition for an element in its standard state under standard conditions.

And DeltaH_"reaction"^@=SigmaDelta_f^@("products")-SigmaDelta_f^@("reactants")

And we do this summation for the given reaction:

$4 N {H}_{3} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 4 N O \left(g\right) + 6 {H}_{2} O \left(l\right)$

$\Delta {H}_{\text{rxn}}^{\circ} = \left\{\left(6 \times - 285.5 + 4 \times 90.29\right) - \left(4 \times - 80.8\right)\right\} \cdot k J \cdot m o {l}^{-} 1$

$= - 1028.6 \cdot k J \cdot m o {l}^{-} 1$ (i.e. per mole of reaction as WRITTEN.....).

The enthalpy is strongly negative, i.e. exothermic, because the reaction forms water..........a thermodynamic sink.......