# 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 $J {K}^{-} 1$]

Apr 8, 2017

I got $+ \text{191.9 J/K}$.

It must be positive for the surroundings, because $\Delta {S}_{s y s}^{\circ}$ is negative for a gas reaction with $\Delta {n}_{g a s} < 0$.

Furthermore, we have to have $\Delta {S}_{s u r r}^{\circ} > 0$ according to the second law of thermodynamics, because only then would $\Delta {S}_{u n i v} \ge 0$ for a reversible ($\Delta {S}_{u n i v} = 0$) OR irreversible process ($\Delta {S}_{u n i v} > 0$) in a thermodynamically-closed system (energy transfer, no mass transfer).

From my textbook (Physical Chemistry, Levine):

${\Delta}_{f} {H}_{298 , N {O}_{2} \left(g\right)}^{\circ} = \text{33.18 kJ/mol}$

${\Delta}_{f} {H}_{298 , {N}_{2} {O}_{4} \left(g\right)}^{\circ} = \text{9.16 kJ/mol}$

By the state property of $\Delta H$, one can find the $\Delta {H}_{\text{rxn}}^{\circ}$ for the reaction

$2 {\text{NO"_2(g) rightleftharpoons "N"_2"O}}_{4} \left(g\right)$

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

$\Delta {H}_{\text{rxn}}^{\circ} = {\sum}_{P} {n}_{P} {\Delta}_{f} {H}_{298 , P}^{\circ} - {\sum}_{R} {n}_{R} {\Delta}_{f} {H}_{298 , R}^{\circ}$

$= \left(\text{1 mol" cdot "9.16 kJ/mol") - ("2 mols"cdot"33.18 kJ/mol}\right)$

$= - \text{57.20 kJ}$

This appears to be an exothermic process with respect to the system (the molecules). So, at $\text{298 K}$, and constant pressure, we have:

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

$= - \text{57.20 kJ"/("298 K") xx "1000 J"/"1 kJ" = -"191.9 J/K}$

From the system, heat is released into the surroundings, and thus, the surroundings collect heat and $\Delta {S}_{\text{surr}}^{\circ} > 0$. Since $\Delta {S}_{\text{univ}}$ for a reversible process is zero, we have that:

$\Delta {S}_{\text{univ"^@ = 0 = DeltaS_"sys"^@ + DeltaS_"surr}}^{\circ}$

$\implies \textcolor{b l u e}{\Delta {S}_{\text{surr"^@) = -DeltaS_"sys"^@ = color(blue)(+"191.9 J/K}}}$