Question

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`]

Answer 1

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).

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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")`