How do you calculate the entropy change in the surroundings when 1.00 mol #N_2O_4(g)# is formed from #NO_2(g)# under standard conditions at 298 K?

[-192 #J K^-1#]

1 Answer
Apr 8, 2017

I got #+"191.9 J/K"#.

It must be positive for the surroundings, because #DeltaS_(sys)^@# is negative for a gas reaction with #Deltan_(gas) < 0#.

Furthermore, we have to have #DeltaS_(surr)^@ > 0# according to the second law of thermodynamics, because only then would #DeltaS_(univ) >= 0# for a reversible (#DeltaS_(univ) = 0#) OR irreversible process (#DeltaS_(univ) > 0#) in a thermodynamically-closed system (energy transfer, no mass transfer).


From my textbook (Physical Chemistry, Levine):

#Delta_fH_(298,NO_2(g))^@ = "33.18 kJ/mol"#

#Delta_fH_(298,N_2O_4(g))^@ = "9.16 kJ/mol"#

By the state property of #DeltaH#, one can find the #DeltaH_"rxn"^@# for the reaction

#2"NO"_2(g) rightleftharpoons "N"_2"O"_4(g)#

by recognizing that the stoichiometric coefficients are #2# and #1#, respectively.

#DeltaH_"rxn"^@ = sum_P n_P Delta_fH_(298,P)^@ - sum_R n_R Delta_fH_(298,R)^@#

#= ("1 mol" cdot "9.16 kJ/mol") - ("2 mols"cdot"33.18 kJ/mol")#

#= -"57.20 kJ"#

This appears to be an exothermic process with respect to the system (the molecules). So, at #"298 K"#, and constant pressure, we have:

#DeltaS_"sys"^@ = (q_"rev")/T = (DeltaH_("sys")^@)/T#

#= -"57.20 kJ"/("298 K") xx "1000 J"/"1 kJ" = -"191.9 J/K"#

From the system, heat is released into the surroundings, and thus, the surroundings collect heat and #DeltaS_("surr")^@ > 0#. Since #DeltaS_"univ"# for a reversible process is zero, we have that:

#DeltaS_"univ"^@ = 0 = DeltaS_"sys"^@ + DeltaS_"surr"^@#

#=> color(blue)(DeltaS_"surr"^@) = -DeltaS_"sys"^@ = color(blue)(+"191.9 J/K")#