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)#

1 Answer
Aug 2, 2017

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")#