1. Le Chatelier Concentration
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2 Le Chatelier: Volume
3 Le Chatelier: Temperature
4. Le Chatelier: Endo or Exothermic
5. Le Chatelier: Summary True/False
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Due Jun 22 at 11:55 PM
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Consider the following system at equilibrium where AH 108 kJ/mol, and K-1.29-102, at 600 K
COCI₂ (g) CO(g) + Cl₂ (g)
When 0.16 moles of COCI₂ (g) are removed from the equilibrium system at constant temperature:
The value of K
increases
The value of Qc decreases
remains the same
The reaction must
Orun in the forward direction to restablish equilibrium.
Orun in the reverse direction to restablish equilibrium.
O remain the same. It is already at equilibrium.
The concentration of CO will
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HW & LeChatelier's Principle
1 Le Chatelier Concentration
Preparation
Question
Question
Question
2. Le Chatelier Volume
X LOWL CHT17 UAB S2023
Progress:
05 groups
Due Jun 22 at 11:55 PM
Finish Assignment
3. Le Chatelier Temperature
4 Le Chatelier Endo or Exothermic 1 pts
5 Le Chateller Summary True/False
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[Review Topics]
Factors that determine the equilibrium conditions of a system are:
1) Concentration,
2) Temperature (K changes), and
3) Volume (Pressure) for gaseous systems.
Le Chatelier's Principle:
A change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner
as to reduce or counteract the effect of the change.
References]
Concentration
When a system is at equilibrium, Q=K. Changing the concentration of reactants or products that are part of the equilibrium constant expression
cause the reactant quotient, Q. to change. Q will no longer be equal to K and the system will seek to restore itself to a new equilibrium that offsets
the change.
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For example: Adding reactant to an equilibrium system.
2 NO₂(g) N₂O4(8) K = [N₂O4]/[NO₂)²
Adding NO₂(g) will cause the reaction to run in the forward direction to consume the added reactant. Since NO₂ is in the denominator, Q is less
than K, and more product needs to be produced to reestablish equilibrium. Therefore the concentration of N₂O(g) will increase.
2 H₂O(l) 2 H₂(g) + O₂(g) K = [H₂)² [0₂]
Adding H₂O(l) will have no effect on the equilibrium because the reactant is a liquid. Notice that the concentration of H₂O(l) is not a part of the
equilibrium constant expression. Since the concentration of product gases do not change when H₂O(l) is added, Q is still equal to K and there is
no driving force for the system to change. As long as some liquid is present to establish the equilibrium, adding more will have no effect on the
concentration of products.
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