Consider the equilibrium between COBr2, CO and Br₂. COBr₂(g) CO(g) + Br₂(g) K 1.96 at 381 K The reaction is allowed to reach equilibrium in a 20.0-L flask. At equilibrium, [COBr₂] = 2.10×10-2 M, [CO] = 0.203 M and [Br₂] = 0.203 M. (a) The equilibrium mixture is transferred to a 10.0-L flask. In which direction will the reaction proceed to reach equilibrium? to the left (b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 10.0-L flask. [COBr₂] = [Co] [Br₂] = ΣΣΣ

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Consider the equilibrium between COBr₂, CO and Br₂.
COBr₂(g) CO(g) + Br₂(g) K 1.96 at 381 K
The reaction is allowed to reach equilibrium in a 20.0-L flask. At equilibrium, [COBr₂] = 2.10×10-² M,
[CO] = 0.203 M and [Br₂] = 0.203 M.
(a) The equilibrium mixture is transferred to a 10.0-L flask. In which direction will the reaction proceed
to reach equilibrium?
to the left
(b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is
transferred to a 10.0-L flask.
[COBr₂] =
[CO]
[Br₂]
II
II
ΣΣΣ
Transcribed Image Text:Consider the equilibrium between COBr₂, CO and Br₂. COBr₂(g) CO(g) + Br₂(g) K 1.96 at 381 K The reaction is allowed to reach equilibrium in a 20.0-L flask. At equilibrium, [COBr₂] = 2.10×10-² M, [CO] = 0.203 M and [Br₂] = 0.203 M. (a) The equilibrium mixture is transferred to a 10.0-L flask. In which direction will the reaction proceed to reach equilibrium? to the left (b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 10.0-L flask. [COBr₂] = [CO] [Br₂] II II ΣΣΣ
Some PCI, is allowed to dissociate into PCI3 and Cl₂ at 523 K. At equilibrium, [PCI5] = 0.216 M, and
[PCI3] = [Cl₂] = 0.104 M. Additional PCI, is added so that [PCI5]new = 0.354 M and the system is allowed
to once again reach equilibrium.
PCI5(g) PCI3(g) + Cl₂(g)
(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?
[PCI5] =
[PCI3] =
[Cl₂]
=
K= 5.01x10-2 at 523 K
ΣΣΣ
Transcribed Image Text:Some PCI, is allowed to dissociate into PCI3 and Cl₂ at 523 K. At equilibrium, [PCI5] = 0.216 M, and [PCI3] = [Cl₂] = 0.104 M. Additional PCI, is added so that [PCI5]new = 0.354 M and the system is allowed to once again reach equilibrium. PCI5(g) PCI3(g) + Cl₂(g) (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? [PCI5] = [PCI3] = [Cl₂] = K= 5.01x10-2 at 523 K ΣΣΣ
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