LeChatelier's Principle

What you can expect to learn from this page:

Be able to identify, using LeChatelier's principle, how a chemical equation at equilibrium will react as to restore equilibrium due to a change in pressure, temperature, or concentration.

Background Summary:

Before starting this model you should already have studied: rates of reaction, collision theory, catalysts, activation energy, equilibrium, and equilibrium constants. By understanding LeChatelier's Principle you can easily how a equilibrium will shift towards products or reactants from a change in temperature, pressure, or concentration. For example LeChatelier's Principle is applied in industries to help manipulate reactions in order to yield greater products or reactant, this will be explained more in depth later.
Quick Review of Topics:
Reaction Rates:Simply the speed at which a reaction occurs. The rate of a reaction can be increased or decreased by raising or lowering temperature, pressure, or concentration.
Collision Theory: The theory that when two molecules collide that need to have a necessary amount of energy, and by oriented in a certain position for a reaction to occur.
Catalysts:A substance that is added to a reaction, and provides an alternative route for the chemicals to react with a lower activation energy.
Activation Energy: The necessary energy that must be present in a system for a reaction to occur.
Equilibrium: When reactants are converted into products at the same rate that products are converted into reactants. It does not mean that there are an equal number of reactant and product molecules, just that they are reacting at equal rates.
Equilibrium Constants: The ratio of product concentrations to reactant concentrations when equilibrium is achieved is a constant, meaning it obviously cannot change (with the exception that when there is a change in temperature because the constant is dependent upon temperature). Other equation for the equilibrium constant equation works as follows:
2A+ 4B<=>1C + 3D
is plugged into this equation:
Kc=[C]¹ [D]³/[A]² [B]⁴

Lesson:

Part 1: Concentration

Assume that the reaction below is at equilibrium.
CH₄(g) + O₂(g) <=> H₂CO(g) + H2O(g)
H₂O is added. How will the reaction shift?
A.) Left-Correct
B.) Right-Incorrect, if there are more H₂0 molecules on the products side of the equation then equilibrium will not be balanced because there are more particles on the right side so the reaction now is more favored to move left to balance out. If the equation were to move further right it would imbalance equilibrium even more and LeChatelier's principle states that an equation will adjust to return to equilibrium.
C.) No Change-Incorrect, if there are more H₂0 molecules on the products side of the equation then equilibrium will not be balanced because there are more particles on the right side so the reaction now is more favored to move left to balance out.

Part 2: Temperature

Assume that the reaction below is at equilibrium and is exothermic.
2H₂(g) + O₂(g) <=> 2H₂O(g)
The temperature of the system is increased. How will the reaction shift?
A.) Right-Incorrect, increasing the temperature of a reaction will cause the equilibrium to shift in the direction that will remove energy. Since the reaction is exothermic, energy is released with products. To remove energy, the reaction will shift away from products (to the left), not the right.
B.) Left- Correct
C.) No Change-Incorrect, increasing the temperature of a reaction will cause the equilibrium to shift in the direction that will remove energy. Since the reaction is exothermic, energy is released with products. To remove energy, the reaction will shift away from products (to the left).

Part 3: Pressure

Assume that the reaction below is at equilibrium.
I₂(g) + 2HCl(g) <=> H₂(g) + 2ICl(g)
The volume is increased. How will the reaction shift?
A.) Left-Incorrect, there are an even number of moles of molecules on each side of the equation, meaning they all both have the same odds of a collision occurring. Therefore, by increasing the volume the pressure is increased, and because the pressure is increased the odds of a collision is increased, but equally on both sides. The equation would obviously not shift left because that would throw off equilibrium.
B.) Right-Incorrect, there are an even number of moles of molecules on each side of the equation, meaning they all both have the same odds of a collision occurring. Therefore, by increasing the volume the pressure is increased, and because the pressure is increased the odds of a collision is increased, but equally on both sides. The equation would obviously not shift right because that would throw off equilibrium.
C.) No Change-Correct

Homework (if any):

To assess your knowledge of LeChatelier's Principle go to this link below. It will automatically generate questions that will require you to use LeChatelier's Principle. Once you have successfully answered 10 questions out of 13 you know how to use LeChatelier's Principle effectively.

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