# What is the DeltaH_(rxn)^@ for the following reaction if 70.4 kJ of energy is absorbed when 28.2L of NO (g) is produced at STP? 2 NO_2 (g) -> 2 NO (g) + O_2 (g)

Aug 2, 2017

I got $\text{71.8 kJ}$, at ${0}^{\circ} \text{C}$ and a constant atmospheric pressure of $\text{1 bar}$.

This should be larger than the change in internal energy of the reaction, as the change in enthalpy is a bit higher due to changes in system volume.

Well, at constant atmospheric pressure,

$\Delta {H}_{r x n}^{\circ} = \Delta {E}_{r x n}^{\circ} + P \Delta {V}_{r x n}$.

To find $\Delta {H}_{r x n}^{\circ}$, you first need to find the change in the system volume, $\Delta {V}_{r x n}$. You gained ~~ 50% more gas, so how much did the volume change?

Well, if you know that $\text{28.2 L}$ of $\text{NO} \left(g\right)$ is produced at ${0}^{\circ} \text{C}$ and $\text{1 bar}$ (adjust your standard pressure accordingly!), then for ideal gases...

$\overline{V} \approx \text{22.711 L/mol}$

The mol ratios are:

${\text{2 mols NO" = "1 mol O}}_{2}$

${\text{1 mol NO" = "1 mol NO}}_{2}$

As such, there are:

${n}_{N O} = 28.2 \cancel{\text{L NO" xx "mol"/(22.711 cancel"L") = "1.242 mols NO}}$

$\implies {n}_{{O}_{2}} = {\text{0.621 mols O}}_{2}$

$\implies {n}_{N {O}_{2}} = {\text{1.242 mols NO}}_{2}$

This gives a change in volume governed by the change in mols:

$\Delta {V}_{r x n} = {V}_{N O} + {V}_{{O}_{2}} - {V}_{N {O}_{2}}$

$= {n}_{N O} {\overline{V}}_{N O} + {n}_{{O}_{2}} {\overline{V}}_{{O}_{2}} - {n}_{N {O}_{2}} {\overline{V}}_{N {O}_{2}}$

Since we have assumed all of these gases are ideal, we have:

$\Delta {V}_{r x n} = \left({n}_{N O} + {n}_{{O}_{2}} - {n}_{N {O}_{2}}\right) \overline{V}$

$= \overline{V} \Delta n$

$= \text{22.711 L/mol" xx (1.242 + 0.621 - 1.242) "mols}$

$\approx$ $\text{14.104 L}$

Thus, the change in enthalpy of reaction at "STP" is:

$\textcolor{b l u e}{\Delta {H}_{r x n}^{\circ}} = \Delta {E}_{r x n}^{\circ} + P \overline{V} \Delta n$

$=$ overbrace("70.4 kJ")^(DeltaE_(rxn)^@) + overbrace(cancel"1 bar")^(P)overbrace((14.104 cancel"L" cdot (8.314472 xx 10^(-3) "kJ")/(0.083145 cancel("L"cdot"bar"))))^(DeltaV_(rxn))

$= \text{70.4 kJ" + "1.41 kJ}$

$=$ $\textcolor{b l u e}{\text{71.8 kJ}}$