Question

1. Suppose the following reaction: 2 SO, (g) + O2(g) → 2 SO, (g) is carried out at 25°C and 1.00 atm in a constant pressure cylinder fitted with a movable piston. Initially, 0.030 mol SO, and 0.030 mol O2 are present in the cylinder. The reaction is then initiated with a catalyst. A)(1 pt) Calculate the volume of the cylinder before the reaction is initiated. B)(1 pt) What is the limiting reactant? C ipt) Assuming that the reaction goes to completion and that the temperature and pressure remain constant, what is the final volume of the cylinder (including any excess reactant)? D) (1 pt) How much work takes place, and is the work done on or by the system? EX(1 pt) What is the standard state change in enthalpy for the reaction? Standard enthalpies of formation: SO, (g) = -296.83 kJ/mol SO; (g) = -395.72 kJ/mol

Answer 1

PA NO: 2502 (g) + O, (q) - 250 At 25°C, latm, 250₂ to 0.03 0.03 250₂ for calculation Limiting Reagent (LR) Divide the moles by their respective coefficients For so, = 0.03 For o = 0.03. = 0.015 = 0.03 (B) SO, is limiting Reagent 250 At to At t= t 200₂ to 0.03. 0.03 o 10.03-0.03 0.03 0.015 0.03 Total moles after reaction 0.03+0.015 = 0.045 Before reaction - Applying PV=nRT (A) x V = (0.061* 0.0821x(298) v = 1.46L Teachers Sicheure

Ence 10 For final volume PV=nRT IX V = 0.0458o.o8L1 x 298 2 Work done = POV = 1x (V7-V) 1x (H.1-1.46) . -0.36 atm L Wat do since volume contradim took place, work is done on the - System V SO₂ = SO₃ = O2 = -296.83 kJ/mul - 395.72 kJ/mul 0 (in native state) Ahr = - = 2 = sn (Prodult) - in reatant 2x (SOJ - 2 x Sb -395.72 + 296.83) -197.78 kJ/mol Teacher's Signature
feel free to ask any doubt in comment section