Welcome to the Entropy debriefing

Answer the questions below, and click on the Check buttons when you are ready.

1. The barium hydroxide and ammonium chloride reaction has positive values for both DSº and DHº, implying that at some temperatures it is product-favored and at others reactant-favored. Calculate the temperature at which the reaction switches from one to the other.

Ba(OH)2 • 8 H2O (s) + 2 NH4Cl (s) BaCl2 • 2 H2O (s) + 2 NH3 (aq) + 8 H2O (l)

Correct! This demonstration would be ineffective for teaching a class of chemistry students in outer space!
Remember to include units with your answer.
Maybe you are confusing degrees Celsius with Kelvins.
While numerically correct, your answer has the incorrect number of significant digits. Try again.
That is incorrect. Try again.
The correct answer is shown below. This demonstration would be ineffective for teaching a class of chemistry students in outer space!

DHº = 63 kJ      DSº = 368 J/K      (from the previous page)
DSuniv = 0 = DSsyst + DSsurr
0 = DSsyst - DHsyst/T
DH/T = DS
T = DH/DS
T = 63 kJ/368 J/K • 1000 J/1 kJ = 170 K = -100 ºC

2. Chemical cold packs used in athletic medicine are designed to lower the temperature at the injury site when ammonium nitrate is dissolved in water:

NH4NO3 (s) NH4NO3 (aq)

a. What is DHº for the reaction?

b. What is DSº for the reaction?

c. Is the reaction product-favored at 25 ºC?

Yes

d. Is it product-favored at -40 ºC?

Yes
No No
Correct! Dissolving ammonium nitrate in water is an endothermic process, which produces the cold needed to reduce swelling in injuries.
Remember to include units with your answer.
Your answer is off by a factor of 1000. Perhaps you have mixed up joules and kilojoules.
While numerically correct, your answer has the incorrect number of significant digits. Try again.
That is incorrect. Try again.
The solution is shown below. Dissolving ammonium chloride in water is an endothermic process, which produces the cold needed to reduce swelling in injuries.

DHº = 1 mol • DHºf(NH4NO3(aq)) - 1 mol • DHºf(NH4NO3(s))
DHº = 1 mol • (-339.87 kJ/mol) - 1 mol •(-365.56 kJ/mol)
DHº = -339.87 kJ + 365.56 kJ = 25.69 kJ

Correct! Dissolving ammonium nitrate in water has a positive entropy change, as you might expect from a solid's entering a solution.
Remember, the units for entropy are J/K.
Your answer is off by a factor of 1000. Perhaps you have mixed up joules and kilojoules.
While numerically correct, your answer has the incorrect number of significant digits. Try again.
That is incorrect. Try again.
The solution is shown below. Dissolving ammonium nitrate in water has a positive entropy change, as you might expect from a solid's entering a solution.

DSº = 1 mol • (Sº(NH4NO3(aq)) - 1 mol • (Sº(NH4NO3(s))
DSº = 1 mol • (259.8 J/K) - 1 mol • (151.08 J/K)
DSº = 259.8 J/K - 151.08 J/K = 108.7 J/K

Correct! At 25 ºC, the entropy change of the universe is positive, so dissolving ammonium nitrate in water is a product-favored reaction at this temperature.
The solution is shown below. The entropy change of the universe is positive, so dissolving ammonium nitrate in water is a product-favored reaction at this temperature.

DSuniv = DSsyst + DSsurr
DSuniv = DS - DH/T = 108.7 J/K - 25.69 kJ/(273.15 + 25.0)K • 1000 J/kJ
DSuniv = 22.5 J/K > 0

Correct! At -40 ºC, the entropy change of the universe is negative, so dissolving ammonium nitrate in water is a reactant-favored process at this temperature. It is probably cold enough to cool the injury without a cold pack anyway!
The solution is shown below. The entropy change of the universe is negative, so dissolving ammonium nitrate in water is a reactant-favored process at this temperature. It is probably cold enough to cool the injury without a cold pack anyway!

DSuniv = DSsyst + DSsurr
DSuniv = DS - DH/T = 108.7 J/K - 25.69 kJ/(273.15 - 40.0)K • 1000 J/kJ
DSuniv = -1.5 J/K < 0

Congratulations!

You have completed the Entropy module.

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