SCH 4U - Le Chatelier's Principle

Introduction

In industry, we always seek to maximize product yield.  For instance, Haber wished to increase the yield of ammonia, so he did experiments with the process under a variety of conditions.

Le Chatelier’s Principle - Henri Louis Le Chatelier (1850-1936)

Le Chatelier's Principle states, ‘If a system at equilibrium is disturbed by a change in T, P or [  ] of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.’

Basically, this means that a reaction at equilibrium will always attempt to undo any action that you have visited upon it.  It does this by shifting the reaction the direction that will help to erase your meddling.  

Le Chatelier dit, "La chimie est amusante!"

Change in Reactant or Product Concentration

‘If a chemical system is at equilibrium and a reactant or product is added, the reaction will shift to reestablish equilibrium by consuming part of the added substance.  Conversely, removing a substance will result in the reaction shifting to form more of the substance.’

            N₂(g) + 3H₂(g) ↔ 2NH₃(g)

• ADD N₂ or H₂ - equilibrium shifts right (→), toward products in an attempt to use up the added reactant - of course, because of the reaction's response, this means that both reactants will get used and NH₃ will be produced
• ADD NH₃ - equilibrium shifts left (←), toward reactants in an attempt to use up the added product - of course, because of the reaction's response, this means that both reactants will be produced and NH₃ will be used
• REMOVE N₂ or H₂ - equilibrium shifts left (←), toward reactants, in an attempt to replace the removed reactant - of course, because of the reaction's response, this means that both reactants will be produced and NH₃ will be used
• REMOVE NH₃ - equilibrium shifts right (→), toward products, toward products in an attempt to replace the removed product - of course, because of the reaction's response, this means that both reactants will get used and NH₃ will be produced

Effect of Temperature Changes

‘When heat is added to a system, the equilibrium shifts in the direction that absorbs the heat.’

Co(H₂O)₆²⁺(aq) + 4Cl^-(aq) ↔ CoCl₄^2-(aq) + 6H₂O(l)     ΔH > 0 (endothermic)

pink coloured                           blue coloured

*note becasue this reaction is endothermic, the heat term would be placed on the left side of the equation to give: 

 

Co(H₂O)₆²⁺(aq) + 4Cl^-(aq) + HEAT ↔ CoCl₄^2-(aq) + 6H₂O(l)

 

and although we know that heat isn't material, we can treat it as if it is.  So, in this case, heat would behave just like the reactants did up above.

• AT ROOM TEMP – violet solution (because there are equivalent amounts of Co(H₂O)₆²⁺ & CoCl₄^2-)
  
• INCREASE TEMP - blue solution, because reaction shifts right (→), toward products (toward CoCl₄^2-) to decrease the system's temp
• DECREASE TEMP - pink solution, because reaction shifts left (←), toward reactants(toward Co(H₂O)₆²⁺) to increase the system's temp

 

Check out this video (watch from 8:00-10:15) to see the above in action.

Effects of Volume and Pressure Changes

‘Reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas.’

Before we begin, recall that pressure (P) and volume (V) are inversely proportional, meaning that as one increases, the other decreases.  

1N₂O₄(g) ↔ 2NO₂(g)

• INCREASE P/DECREASE V - equilibrium shifts left (←), toward reactants because there are fewer moles of gas on the reactant side of the equation, so this will lower the system's pressure
• DECREASE P/INCREASE V - equilibrium shifts right (→), toward products because there are more moles of gas on the product side of the equation, so this will raise the system's pressure

H₂(g) + I₂(g) ↔ 2HI(g)

• INCREASE/DECREASE P or V - equilibrium does not shift (⎯) because both sides have equivalent numbers of moles of gas.

TryIt! 

For the rxn at equilibrium,  

BaCO₃(s) + 2H⁺(aq) + heat ↔ Ba²⁺(aq) + H₂O(l) + CO₂(g), 

which way will the reaction shift in response to the following actions?  Answers at the end of the lesson.

(a) addition of hydrogen ion

(b) removal of water

(c) increase in temperature

(d) decrease in pressure

(e) addition of carbon dioxide

 

 

Effect of Catalysts

Recall that catalysts provide an alternate reaction pathway, with a lower activation energy.  But, remember, the catalyst lowers the E_a of both the forward and reverse reactions.

‘A catalyst increases the rate at which equilibrium is reached, but it does not change the composition of the equilibrium mixture.’

Effect of Adding Inert Gases

They equally decrease the probability of substances colliding for the reactant and product molecules, thus, inert gases will not affect the system. 

‘Addition of an inert gas decreases the rate at which equilibrium is reached, but it does not change the composition of the equilibrium mixture.’

Want some further explanation?  Check out Professor Dave.

 

Want to see equilibrium in action?  Check out this video.

 

 

TryIt! Answers:

(a) →

(b) ⎯  water is a liquid (which can't change [  ]), so it has no effect on the eq.

(c) →    

(d) → the product side has the gas (the only state affected by pressure changes)

(e) ←

 

Homework #17, 19-21, 23 (we will be ignoring all the graphical questions in this section - I think we could all use a bit of a break 😊).