Question

INCLASS EXERCISE ON CHEMICAL EQUILIBRIUM (FINAL) DUE FRIDAY 11/08/2019 NAME Given these equilibrium reactions: 1. X Saw + 2024 <------- 2 SO200) Ka = 1.86 x 10105 2. 1/8 SW + 3/2021) <---.....→ S03e) K = 1.77 x 105 Calculate the equilibrium constant kas for the reaction below, which is important in the formation of acid-rain air pollution. 2 SO260 + 02 -----------------→2 503)

Answer 1

1.15 (s) + 2O₂(g) 250, (g) Ka=1.86810105 2. 15 (s) + 3/20(g) So (8); K 62 = 177x1053 Taking reverse of the equations & C2 = 1 3.3.250,(9) $ (9 +20,49); kós = 7186xldos Multiplying equation 2 by 2; we got: 53 4. 1 S₂ (3) +3 0₂ (9) 250 (9) ; K= 177 x 10 Adding equations 3 and 4; we get: - 250,9+0,6) 2016. d kiss kexko 177x1053 TR = 9.52x10-53