When solid ammonium nitrate (NH_4NO_3) dissolves in water, the temperature of th
ID: 523211 • Letter: W
Question
When solid ammonium nitrate (NH_4NO_3) dissolves in water, the temperature of the water can drop dramatically. Taking NH_4NO_3 as the system, predict the signs of the entropy change of the system (delta S_sys) and surroundings (delta S_sur). a. delta S_sys is less than 0 and delta S_surr is less than 0. b. delta S_sys equals 0 and delta S_surr is less than 0. c. delta S_sys is greater than 0 and delta S_surr is less than 0. d. delta S_sys is less than 0 and delta S_surr is greater than 0. e. delta S_sys is greater than 0 and delta S_surr is greater than 0.Explanation / Answer
Answer:
Second Law of Thermodynamics be used to predict whether a reaction is product- or reactant-favored? The entropy change of the universe can be broken up into two parts, the entropy change of the system and the entropy change of the surroundings:
DSuniv = DSsyst + DSsurr
DSsyst, the entropy change of the system, represents the change in order of the molecules of the system, similar to what was discussed in Entropy 2. It can be calculated using absolute entropies as has been described on the previous page.
It may seem unlikely that the entropy change of the surroundings can be calculated just from what is known about the system. But, the only way the system affects the surroundings is by a transfer of heat. Since the surroundings are so much bigger than the system, its temperature is certain to stay constant. Thus, equation (1) can be used to calculate DSsurr:
DSsurr = qsurr/T = -DHsyst/T
Why does DHsyst have a negative sign? Any heat lost by the system is gained by the surroundings, and conversely, any heat gained by the system is lost by the surroundings. So, qsurr and DHsyst must always have opposite signs, which is why DHsyst is given a negative sign. DHsyst can be calculated in the way described in the Thermochemical Equations module.
Now that DSsyst and DSsurr are known, DSuniv can be determined. If it is greater than zero, the reaction is product-favored. If it is less than zero, the reaction is reactant-favored.
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