84jmolk co2g21374jmolk c3h8g27030jmolk o2g20507jmolk

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 • • C3H8(g) + 5O2(g) ⇒ 3CO2(g) + 4H2O(g) Calculate entropy change in surrounding at 25°C given ∆ H°rxn = ­2044 kJ Determine ∆ Ssys, given: – – – – H2O(g) =188.84 J/mol•K...
View Full Document