Question

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)`

Answer 1

I got `"71.8 kJ"`, at `0^@ "C"` and a constant atmospheric pressure of `"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,

`DeltaH_(rxn)^@ = DeltaE_(rxn)^@ + PDeltaV_(rxn)`.

To find `DeltaH_(rxn)^@`, you first need to find the change in the system volume, `DeltaV_(rxn)`. You gained `~~ 50%` more gas, so how much did the volume change?

Well, if you know that `"28.2 L"` of `"NO"(g)` is produced at `0^@ "C"` and `"1 bar"` (adjust your standard pressure accordingly!), then for ideal gases...

`barV ~~ "22.711 L/mol"`

The mol ratios are:

`"2 mols NO" = "1 mol O"_2`

`"1 mol NO" = "1 mol NO"_2`

As such, there are:

`n_(NO) = 28.2 cancel"L NO" xx "mol"/(22.711 cancel"L") = "1.242 mols NO"`

`=> n_(O_2) = "0.621 mols O"_2`

`=> n_(NO_2) = "1.242 mols NO"_2`

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

`DeltaV_(rxn) = V_(NO) + V_(O_2) - V_(NO_2)`

`= n_(NO)barV_(NO) + n_(O_2)barV_(O_2) - n_(NO_2)barV_(NO_2)`

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

`DeltaV_(rxn) = (n_(NO) + n_(O_2) - n_(NO_2))barV`

`= barVDeltan`

`= "22.711 L/mol" xx (1.242 + 0.621 - 1.242) "mols"`

`~~` `"14.104 L"`

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

`color(blue)(DeltaH_(rxn)^@) = DeltaE_(rxn)^@ + PbarVDeltan`

`=` `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))`

`= "70.4 kJ" + "1.41 kJ"`

`=` `color(blue)("71.8 kJ")`