3 Given: SO₂Cl₂(g) = SO₂(g) + Cl₂(g) Kc = 119 at 298 K In this reaction at equilibrium, the concentration of SO₂Cl2 is 0
Posted: Mon Jul 11, 2022 1:39 pm
- 3 Given So Cl G So G Cl G Kc 119 At 298 K In This Reaction At Equilibrium The Concentration Of So Cl2 Is 0 1 (21.05 KiB) Viewed 26 times
- 3 Given So Cl G So G Cl G Kc 119 At 298 K In This Reaction At Equilibrium The Concentration Of So Cl2 Is 0 2 (31.97 KiB) Viewed 26 times
- 3 Given So Cl G So G Cl G Kc 119 At 298 K In This Reaction At Equilibrium The Concentration Of So Cl2 Is 0 3 (32.79 KiB) Viewed 26 times
Question 2 (1 point) Given: SO₂Cl₂(g) = SO₂(g) + Cl₂(g) Kc = 119 at 298 K Given that this reaction starts with 0.2 M of the SO₂Cl2, what is the expression for K at equilibrium? Hints: Considering the magnitude of K and starting concentrations, predict the direction of reaction. Create an ICE table and fill in in term of x. Write the Kc in terms of the species first. Then, sub in the things you know! x²/(0.2+x) (0.2+x)/x² 2x/(0.2-x) (0.2+x)/2x (0.2-x)/2x (0.2-x)/x² x²/(0.2-x) 2x/(0.2+x)
Question 3 (1 point) Given: SO₂Cl₂(g) SO₂(g) + Cl₂(g) Kc 119 at 298 K Given that the reaction starts with 3 M of each of the products the SO2 and Cl2, what is the concentration of the reactant, SO₂Cl2(g), at equilibrium? Hints: Considering the magnitude of K and starting concentrations, predict the direction of reaction. Create an ICE table and fill in in term of x. Write the Kc in terms of the species first. Then, sub in the things you know! Can you think of a way to simplify and avoid the quadratric? 06M 03M 0.08 M 0.03 M