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).

From Levine's Physical Chemistry, my textbook:

#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")#
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Answer 2

The entropy change in the surroundings can be calculated using the equation:

ΔS_surr = -ΔH_sys / T

Where: ΔS_surr = entropy change in the surroundings ΔH_sys = enthalpy change of the system T = temperature in Kelvin

First, you need to determine the enthalpy change of the system (ΔH_sys) for the reaction:

2NO2(g) -> N2O4(g)

Using standard enthalpies of formation: ΔH_sys = ΣΔHf(products) - ΣΔHf(reactants)

ΔH_sys = [0 kJ/mol - (-19.6 kJ/mol)] - [2(33.2 kJ/mol)]

ΔH_sys = -85.2 kJ/mol

Substitute ΔH_sys and the temperature (298 K) into the equation to find ΔS_surr.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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