# EXAM 2 ANSWER - Δ U cycle = 0 Δ H cycle = 0 18 8.21 kJ b...

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16) C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) Δ H = -1410.94 kJ/mol b) yes because it is exothermic and thus releases energy c) -1291 kJ/mol d) yes because Δ G c is negative 17) *you had to find temperature using PV=nRT for each point in order to calculate q w 1 2 = -2 kJ or -2.3 kJ q 1 2 = +5.67 kJ w 2 3 = 0 q 2 3 = -3.40 kJ w 3 1 = +1.57 kJ q 3 1 = -1.57 kJ w total = -0.7 kJ q total = +0.70 kJ
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Unformatted text preview: Δ U cycle = 0 Δ H cycle = 0 18) 8.21 kJ b) 92.7 J/K or 0.0927 kJ/K c) The values are different which is expected. The Δ S vap at the lower temperature is larger because at this temperature the increase in thermal disorder is greater than at a higher temperature where the molecules have more thermal disorder. d) Δ S total = 0...
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