The enthalpy of vaporization of water at 100.0°C is 40.55 kJ/mol. What is the en

Question

The enthalpy of vaporization of water at 100.0°C is 40.55 kJ/mol. What is the entropy change in the surroundings when one mole of water vapor condenses at 100.0°C in a large room maintained at a temperature of 15.00°C? (answer in J/K) The enthalpy and entropy of fusion of ice at 0.00°C are 6.00 kJ/mol and 21.5 J/mol K. respectively. What is the entropy change of the universe when one mole of ice melts in a large room maintained at 30.00°C? Assume the final temperature of the water is 0.000. (answer in J/K)

Explanation / Answer

Q1.

-Qsystem = Qsurroundings

Qsystem = n*HRn = 1*40.55 = 40.55 kJ = 40550 J

Qsurroundings = -Qsystem = -40550 J

then...

T surr = 15°C +273 = 288 K

Ssurroundings = Qsurr / T = -40550 /288 = -140.798 J/K

Q2

fussion --> energy is gained by ssystem, lost by surroundings

Ssystem = n*Sfusion = 21.5 J/K

Ssurroundings = Qsurroundings/Tsurr = -Qsystem/Tsurr = -6000/(30+273) = -19.80 J/K

Suniverse = Ssystem + Ssurr = 21.5 - 19.80 = 1.7 J/K

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