# 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?

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[-192 #J K^-1# ]

[-192

##### 1 Answer

I got

It must be positive for the surroundings, because

Furthermore, we have to have

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

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

by recognizing that the stoichiometric coefficients are

#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

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

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