Question

How do i set up this problem? answer is provided but struggling with how to actually solve it.

11. Calculate AS and AGº for the following reaction: C3H3(g) +502(g) → 3C02(g) + 4H2O(1) Using: Alf" C3H8, gas) = -103.85 kJ/mol; AH,^[ O2, gas)= 0 kJ/mol; AH [ CO2, gas]=-393.51 kJ/mol; AH?°[ H20, liquid]=-285.83 kJ/mol Smº[ C3Hs, gas) = 270.2 J/(K-mol); Smº[ O2, gas)= 205.14 J/(K-mol); Smº[ CO2, gas)= 213.74 J/(K-mol); Smº[ H2O, liquid)= 69.91 J/(K-moll (Ans ASⓇ = -375.0J/K and AGO= -2,108 kJ)

Answer 1

(1) son: - - C3 HB (g) + 502(g) 300, (g) + 4 Hou! Step1: Calculation IgH = 3x - 393.51KI mott) + 4x(-285.83K [ sx (0 kJ molt) + (-103.85 KJ mat) = - 1180.53 - 1143.32) KJ molt -[(-103.85) KJ mothy 1-2323.85 + 103.85 KJ molt -2, 220 kJ mott step 2: Calculation of Aso : A5° = {Centropy change of products) Centropy change of reactants = (4x(69.91J K molt) + 3x( 213.74 Jkt met T- Tsx (205: 14 J Khanott) + 270,2 J K mer (279.64 + 641.22 ) Jkt moty (6025.1 + 270.2) Jkt mort 7 = 920,86 J K t mort) - (1295.9 J K 1 moth) 7 - 375.04 T t molt 45° = - 375.0 J K t molt

Step 3 : I Calculation of Go : - AGO = 3HO - TOSO = (-2,220000 I mott) - (298K) (-375.0 Jkt moth) = -2, 2200 oogmott) + (11,750 I mott) – – 2108.250 kJ molt OGO = -2108.0 KJ molt I
I I have used the unit of entropy and Gibbs free energy in terms of Per mole . So do not get confused.