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
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1aba0b34-15dd-4381-a224-979828f87e27%2F5bde4a9d-8219-4161-adb6-9dc0e5dcf9d4%2Fxx32ez8_processed.jpeg&w=3840&q=75)
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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