Le Châtelier’s Principle

 Concentration  Pressure and volume  Temperature  Catalysts

 Consider: CO 2(g) +H 2 O (g) ↔ H 2 CO 3(g)  If more CO 2(g) is added, what will happen? ◦ Le Châtelier's Principle: “system will react to relieve the stress”  Since CO 2(g) is on the reactant side ◦ rate of the forward rxn will ↑ to "use up" the additional reactant ◦ the eq’m shifts to the right, producing more H 2 CO 3(g) ◦ also can say:  eq’m shifts to the product side  the forward rxn is more favoured than before

 Change of concentration: ◦ [CO 2(g) ]↑: what is being added. Only some is used up in forward rxn ◦ [H 2 O (g) ]↓: forward rxn is favoured, more is used up ◦ [H 2 CO 3(g) ] ↑: forward rxn is favoured, more is produced

 If more H 2 CO 3(g) is added, what will happen? ◦ Since H 2 CO 3(g) is on the product side  rate of the reverse rxn will ↑  the eq’m shifts to the left  eq’m shifts to the reactant side  the reverse rxn is favoured ◦ producing more CO 2(g) and H 2 O (g)  [CO 2(g) ]↑ : reverse rxn is favoured  [H 2 O (g) ]↑ : reverse rxn is favoured  [H 2 CO 3(g) ] ↑ : is what is being added, some is used in reverse rxn

 If some H 2 O (g) is removed: what will happen? ◦ rate of the reverse rxn will ↑, to “fill the void” ◦ the eq’m to shifts to the left ◦ eq’m shifts to the reactant side ◦ the reverse rxn is favoured  [CO 2(g) ]↑ : reverse rxn is favoured  [H 2 O (g) ]↓ : is being removed, some is made in reverse rxn  [H 2 CO 3(g) ] ↓ : reverse rxn is favoured

 Note: The value of K eq does not change when changes in concentration cause a shift in equilibrium.

Gas Flash Back  Standard Temperature and Pressure (STP) 0 °C ( K) 1 atm STP 1 mol = 22.4 L “molar volume”  the volume of a gas is determined by the space between the molecules, not the type of gas  as V ↓, P ↑ and vice versa  as V ↓ the concentration of the gas ↑  Total Pressure = P 1 + P 2 + P 3 …

 Consider: N 2 O 4 (g ) ↔2 NO 2 (g) ◦ Since the product has 2 times number of moles as reactant, it exerts twice as much pressure  If ↑ pressure, the stress can be reduced by favouring the side with the fewest moles of gas, which has less pressure. ◦ eq’m shifts to the left  If ↓ pressure, stress can be reduced by favouring the side with the most moles of gas, which has more pressure. ◦ eq’m shifts to the right  If number of moles of reactant = moles of product, a change of pressure has no effect ◦ H 2 (g ) + Cl 2 (g ) ↔2 HCl (g )

 Note: The K eq does not change  Remember: only count the number of moles of gases. ◦ solids, liquids, aqueous solution are not be affected by changing volume or pressure  Pressure can also be changed by adjusting levels of an inert gas  2 CO (g) + O 2 (g) ↔ 2 CO 2 (g)  the addition of Ne (g) ↑ the pressure, shifting the eq’m to the right.

 Predict the effect on eq’m when pressure is increased a.COCl 2 (g) ↔ CO (g) + Cl 2 (g) b.PCl 3 (g) + Cl 2 (g) ↔ PCl 5 (g) c.H 2 (g) + I 2 (g) ↔ 2 HI (g) d.C (s) + 2 H 2 (g) ↔ CH 4 (g) e.C (s) + H 2 O (g) ↔ CO (g) + H 2 (g) Shifts to the left No change Shifts to the right Shifts to the left

Temperature Flash Back  Exothermic: heat released, heat is a product,  H is negative  Endothermic: heat absorbed, heat is a reactant,  H is positive  When temperature is the stress that affect a system at eq’m, ↑ of temp favours reaction that absorbs heat (i.e. endothermic rxn)  the value of K eq will change

 Consider: N 2 O 4(g) ↔2 NO 2(g)  H = kJ  The forward rxn is endothermic. By adding more heat, eq’m will shift to use up the additional heat, favouring the forward direction.  N 2 O 4(g) + HEAT ↔ 2 NO 2(g)

 Consider: H 2 (g) + I 2 (g) ↔ 2 HI (g) + HEAT  The forward rxn is exothermic. By removing heat (making the system colder), eq’m will shift to “replace” the heat that has been removed, favouring the forward direction.  The K eq will change with the change of temperature, ◦ K eq will ↑ if rxn favours the product ◦ K eq will ↓ if rxn favours the reactant

Catalysts (or inhibitors)  speeds up both the forward and the reverse rxns, ◦ no uneven change.  will not affect the position of the eq’m

StressShifts to:Forms more:Value of K eq add catalyst none no change [ ] ↑ i.of reactant ii.of product i.Right ii.Left i.Product ii.Reactant i.No change ii.No change Pressure (Gases) i.↑ ii.↓ i.side with less mole (g) ii.side with more mole (g) i.side with less mole (g) ii.side with more mole (g) i.No change ii.No Change Temp ↑ i.Exothermic ii.Endothermic i.Left ii.Right i.Reactant ii.Product i.Decrease ii.Increase Temp ↓ i.Exothermic ii.Endothermic i.Right ii.Left i.Product ii.Reactant i.Increase ii.Decrease