Le Châtelier’s Principle. Concentration Pressure and volume Temperature Catalysts.
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Transcript of Le Châtelier’s Principle. Concentration Pressure and volume Temperature Catalysts.
Le Châtelier’s Principle
Concentration Pressure and volume Temperature Catalysts
Consider: CO2(g) + H2O(g) ↔ H2CO3(g)
If more CO2(g) is added, what will happen?◦ Le Châtelier's Principle: “system will react to relieve
the stress” Since CO2(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 H2CO3(g) ◦ also can say:
eq’m shifts to the product side the forward rxn is more favoured than before
Change of concentration:◦ [CO2(g)] ↑: what is being added. Only some is used
up in forward rxn◦ [H2O(g)] ↓: forward rxn is favoured, more is used up
◦ [H2CO3(g)] ↑: forward rxn is favoured, more is produced
If more H2CO3(g) is added, what will happen?◦ Since H2CO3(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 CO2(g) and H2O(g)
[CO2(g)] ↑ : reverse rxn is favoured [H2O(g)] ↑ : reverse rxn is favoured [H2CO3(g)] ↑ : is what is being added, some is used
in reverse rxn
If some H2O(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
[CO2(g)] ↑ : reverse rxn is favoured [H2O(g)] ↓ : is being removed, some is made
in reverse rxn [H2CO3(g)] ↓ : reverse rxn is favoured
Note: The value of Keq does not change when changes in concentration cause a shift in equilibrium.
Gas Flash Back Standard Temperature and Pressure (STP) 0 °C
(273.15 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 = P1 + P2 + P3 …
Consider: N2O4 (g ) ↔ 2 NO2 (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
◦ H2 (g ) + Cl2 (g ) ↔ 2 HCl (g )
Note: The Keq 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) + O2 (g) ↔ 2 CO2 (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. COCl2 (g) ↔ CO(g) + Cl2 (g)
b. PCl3 (g) + Cl2 (g) ↔ PCl5 (g)
c. H2 (g) + I2 (g) ↔ 2 HI(g)
d. C(s) + 2 H2 (g) ↔ CH4 (g)
e. C(s) + H2O (g) ↔ CO (g) + H2 (g)
Shifts to the left
No change
Shifts to the right
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 Keq will change
Consider: N2O4(g) ↔ 2 NO2(g) H = + 58.0 kJ
The forward rxn is endothermic. By adding more heat, eq’m will shift to use up the additional heat, favouring the forward direction.
N2O4(g) + HEAT ↔ 2 NO2(g)
Consider: H2 (g) + I2 (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 Keq will change with the change of temperature,◦ Keq will ↑ if rxn favours the product
◦ Keq 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
Stress Shifts to: Forms more: Value of Keq
add catalyst none none no change
[ ] ↑i.of reactantii.of product
i.Rightii.Left
i.Productii.Reactant
i.No changeii.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 changeii.No Change
Temp ↑i.Exothermicii.Endothermic
i.Leftii.Right
i.Reactantii.Product
i.Decreaseii.Increase
Temp ↓i.Exothermicii.Endothermic
i. Rightii. Left
i. Product ii. Reactant
i. Increaseii. Decrease