## delta S(system) vs delta S(surroundings)

$\Delta S = \frac{q_{rev}}{T}$

Megan Purl 1E
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Joined: Fri Sep 29, 2017 7:07 am

### delta S(system) vs delta S(surroundings)

Under what conditions would delta S(sys) = -delta S(surr)?

Niyanta Joshi 1F
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Joined: Thu Jul 13, 2017 3:00 am

### Re: delta S(system) vs delta S(surroundings)

delta S (universe)= delta S (sys) + delta S (surr)
When the reaction is at equilibrium and it is a reversible reaction, the total delta S of the universe is zero.
Hence, delta S (sys) = - delta S (surr) for these conditions.

Veritas Kim 2L
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### Re: delta S(system) vs delta S(surroundings)

You can always assume that delta S(system)= -delta S(surrounding) for a reversible expansion at equilibrium because the heat that flows to the surrounding is being released by the system.

AtreyiMitra2L
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### Re: delta S(system) vs delta S(surroundings)

This can only occur in isothermal, reversible reactions. It would not be considered a violation of the second law of thermodynamics (law that says the entropy is always increasing) because these never actually occur in nature.

Rohan Chaudhari- 1K
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### Re: delta S(system) vs delta S(surroundings)

Delta S of the universe must equal 0, which happens in reversible reactions at equilibriums.