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If the temperature of the surroundings is -244.32 °C, calculate the entropy change (in J/K) for...

If the temperature of the surroundings is -244.32 °C, calculate the entropy change (in J/K) for the system (ΔSsys), surroundings (ΔSsur) and universe (ΔSuniverse) when 40.4 g of liquid oxygen (O2) freezes. Report your answers to two decimal places. Tfus(°C) -218.79 Tvap(°C) -182.96 ΔH°fus (kJ/mol) 0.44 ΔH°vap (kJ/mol) 6.82

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Answer #1

Heat of fusion, ΔHfus=0.44 kJ/mol=440 J/mol

Temperature of fusion, Tfus=-218.79 ℃ =54.21 K

Surrounding temperature, Ts=-244.32℃ =28.68 K

Heat released by liquid oxygen in freezing=moles of oxyge*heat of fusion

                                                                =(40.4/32)*440=555.5 J

ΔSsys=-dQ/T= -555.5/54.21= -10.247 J/K.

ΔSsurr=dQ/T=555.5/28.68 = 19.369 J/K

ΔSuniverse=ΔSsys +ΔSsurr=-10.247+19.369=9.122

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