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Unformatted text preview: univ = ∆ Ssys + ∆ Ssurr
Therefore, for water freezing:
– ∆ Ssys is negative (exothermic)
– ∆ Ssurr is positive (endothermic) ∀ ∆ S is large at low temperatures (closer to freezing point): has very little E, so big difference made
∀ ∆ S is small at high temps (further away from fp): already has more E due to higher temp ⇒ little difference
– IOW, magnitude of ∆ S depends on temp (at above 0°C) 18 Quantifying entropy changes in surrounding
• ∆ Hsys ∝ ∆ Ssurr
∀ ∆ Ssurr ∝ 1/T
∀ ∆ Ssurr ∝ ∆ Hsys/T ∀ ∴ exothermic processes have tendency to be more spontaneous at low temps than high – They increase entropy of surrounding more if T is small
• As temp increases (greater E in surrounding), ∆ H decreases, producing a smaller ∆ Ssurr 19 Problem
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• C3H8(g) + 5O2(g) ⇒ 3CO2(g) + 4H2O(g)
Calculate entropy change in surrounding at 25°C given ∆ H°rxn = 2044 kJ
Determine ∆ Ssys, given:
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– H2O(g) =188.84 J/mol•K...
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This note was uploaded on 01/17/2014 for the course PHY 53890 taught by Professor Appling during the Fall '13 term at University of Texas.
 Fall '13
 Appling

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