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Some SbCl, is allowed to dissociate into SbCl3 and Cl₂ at 521 K. At equilibrium, [SbCl5] = 0.267 M, and [SbCl3] = [Cl₂] = 8.17x10-2 M. Additional SbCl3 is added so that [SbCl3]new = 0.154 M and the system is allowed to once again reach equilibrium. SbCl5(g) SbCl3(g) + Cl₂(g) K = 2.50x102 at 521 K A (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations of reactants and products after the system reaches equilibrium? [SbCl5] = [SbCl3] = [Cl₂] ΣΣΣ

Some SbCl, is allowed to dissociate into SbCl3 and Cl₂ at 521 K. At equilibrium, [SbCl5] = 0.267 M, and [SbCl3] = [Cl₂] = 8.17x10-2 M. Additional SbCl3 is added so that [SbCl3]new = 0.154 M and the system is allowed to once again reach equilibrium. SbCl5(g) SbCl3(g) + Cl₂(g) K = 2.50x102 at 521 K A (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations of reactants and products after the system reaches equilibrium? [SbCl5] = [SbCl3] = [Cl₂] ΣΣΣ

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter15: Principles Of Chemical Reactivity: Equilibria
Section: Chapter Questions
Problem 3PS: Kc = 5.6 1012 at 500 K for the dissociation of iodine molecules to iodine atoms. I2(g) 2 I(g) A...
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Some SbCl, is allowed to dissociate into SbCl3 and Cl₂ at 521 K. At equilibrium, [SbCl5] = 0.267 M, and [SbCl3] = [Cl₂] = 8.17×102 M. Additional SbCl3 is added so that [SbC13]new = 0.154 M and the system is allowed to once again reach equilibrium. SbCl5(g) SbCl3(g) + Cl₂(g) K = 2.50x10-2 at 521 K (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations of reactants and products after the system reaches equilibrium? [SbCls] = [SbCl3] = [Cl₂] 11 ΣΣΣ Autor
Transcribed Image Text:Some SbCl, is allowed to dissociate into SbCl3 and Cl₂ at 521 K. At equilibrium, [SbCl5] = 0.267 M, and [SbCl3] = [Cl₂] = 8.17×102 M. Additional SbCl3 is added so that [SbC13]new = 0.154 M and the system is allowed to once again reach equilibrium. SbCl5(g) SbCl3(g) + Cl₂(g) K = 2.50x10-2 at 521 K (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations of reactants and products after the system reaches equilibrium? [SbCls] = [SbCl3] = [Cl₂] 11 ΣΣΣ Autor
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