Question

1a. The equilibrium for the formation of HOCl in the gas phase from chlorine dioxide (C120) is given below. What are the expected equilibrium pressures if a cylinder has initial pressures as follows: PC120 = PH20 = 0.45 atm; Phoci = 0.65 atm ? (10 points) Cl20 (g) + H2O (g) = 2 HOCl2 (g) Kp=0.085 at 395 °C 1b. Does the total pressure in the cylinder increase, decrease or remain the same as the reactions goes from its original condition to equilibrium ? Provide a brief statement that supports your prediction. (6 points)

Answer 1

1..] Op cleocg) + H2O(g) = 2 HOW. Lgs (reaction quotient) = Proch can 0.65 - Suro. Puro (6.45) - 2.08 0.085 Lence reaction will nove Qe > Kp backwards. = Sütial P: Equib. p : cd₂O(g) + 0.45 0.45+X Hocal 045 0-45+X 2HOCO, (g) 0-65 O-GS- 2x = 0.08s ors 0-65- 0.65-2x12 (0.45+x)" 0-45+2) 0-65 -2% 0. 292 0-450 + x 0-65-2x = 0.132 + 0.292 % 0.650 - 0.132 = 2.292 Ł X = 0.518 0.226 2.292 ON_ atry ... Palea = And (BH2o) eq = 0.450 + = 0458 +0.226= 0.676 (sandeq = 0.676 atm And I Croslead O 65 - 22 = 0.65 -0.452 = 0.198 atom Hence, / faster = (Brno)eq = Phounen = 0.676 atra 0.676 atra 0.198 athu There is no change in number of moles of gas in reactant and product side. Hence, total pressure in cylinder will remain some gf a moles of Hoca decompose, then 1 mole each of cho and ho will form. Hence, no net change in moles of gases Ex - 2