IB Chemistry on Le Chatelier's Principle, Equilibrium Constant and Dynamic Equilibrium

Factors affecting the position of Equilibrium

Le Chatelier’s Principle

• A system in dynamic equilibrium is disturbed, the position of equilibrium will shift so as to cancel

out the effect of change and a new equilibrium can be established again

Effect of Concentration on the position of equilibrium

Increase SCN- or Fe3+ Conc

•Equilibrium shift to right →

•Formation of complex ion Fe(SCN)2+ (red blood)

Fe3+ + SCN- ↔ Fe(SCN)+2

(yellow) (red Blood)

Increase Concentration • Rate of reaction increase ↑ • Rate constant - no change • Kc, equilibrium constant - no changes • Position of equilibrium shifted to a side to decrease concentration again ↓

Decrease Fe3+ Conc

• By adding OH- will shift equilibrium to left ←

•Fe(SCN)2+ breakdown to form more Fe3+ (yellow)

Decrease SCN- Conc

• By adding Ag+ will shift equilibrium to left

• Fe(SCN)2+ breakdown to form more SCN- (yellow)

• Increase in Conc ↑ - position of equilibrium shift to right/left - Conc will be Reduced ↓ • Decrease in Conc ↓ – position of equilibrium shift to right/left - Conc will be Increased ↑

Click to view video

http://www.youtube.com/watch?v=ZOYyCTvLa9E&feature=plcp&context=C41f458bVDvjVQa1PpcFNfJOeCPY-GOhxMAjp0kIsgiOqrhy9kk8c=

http://www.youtube.com/watch?v=ZOYyCTvLa9E&feature=plcp&context=C41f458bVDvjVQa1PpcFNfJOeCPY-GOhxMAjp0kIsgiOqrhy9kk8c=






Decrease H+ Conc

• By adding OH-

•Equilibrium shift to left ←

•Formation of CrO42- (yellow)


Increase H+ Conc

• By adding H+

• Shift equilibrium to right →

• Formation of Cr2O72- (orange)

2CrO42- + 2H+ ↔ Cr2O7

2- + H2O (yellow) (orange)


Click to view video

http://www.youtube.com/watch?v=zP9qEiaL4kQ&feature=relmfu






Decrease CI- Conc

•By adding Ag+ to form AgCI

•Equilibrium shift to right →

•Formation of Co(H2O)62+ (pink)


Increase CI- Conc

• By adding HCI

• Shift equilibrium to left ←

• Formation of CoCl42- (blue)

CoCl42- + 6H2O ↔ Co(H2O)6

2+ + 4CI –

(blue) (pink)

Increase H2O Conc

• By adding H2O

• Shift equilibrium to right →

• Formation of Co(H2O)62+ (pink)


Click to view video

http://www.youtube.com/watch?v=cWr3UDo-WeU&feature=plcp&context=C493411fVDvjVQa1PpcFNfJOeCPY-GOj7US1LiEwFKrWfbtfSRbZg=

http://www.youtube.com/watch?v=GS9kIj9n-BU&feature=related





Effect of Pressure on the position of equilibrium

Increasing Pressure ↑

• By reducing Volume

• Equilibrium shift to left ←

• Less molecules on left side

•Pressure drops ↓

• Formation of N2O4 (colourless)

Increase Pressure • Rate of reaction increases • Rate constant unchanged • Position of equilibrium shift to reduce pressure • Kc, equilibrium constant unchanged

Increase in pressure ↑ - favour reaction with a decrease in pressure ↓ Decrease in pressure ↓ - favour reaction with an increase in pressure ↑

N2O4 (g) ↔ 2NO2(g)

(colourless) (brown)

Decreasing Pressure ↓

• By Increasing Volume

• Equilibrium shift to right →

• More molecules on right side

•Pressure increase ↑

• Formation of NO2 (brown)

Increase pressure ↑ – collision more frequent - shift equilibrium to left - to reduce number of molecules - pressure decrease again ↓

Decrease pressure ↓ – collision less frequent – shift equilibrium to right – to increase number of molecules – pressure increase again ↑

Click to view video


Effect of Pressure on the position of equilibrium


• By reducing Volume





Increase in pressure ↑ - favour reaction with a decrease in pressure ↓ Decrease in pressure ↓ - favour reaction with an increase in pressure ↑

N2O4 (g) ↔ 2NO2(g)



• By Increasing Volume





N2(g) + 3H2(g) ↔ 2NH3(g)

( 4 vol/moles ) (2 vol/moles)





• Formation of NH3 (product)





• Formation of H2 and N2 (reactants)

Click to view video





Effect of Temperature on the position of equilibrium

Decrease Temp ↓

• By Cooling it down

• Favours exothermic reaction


• To increase Temp

• Formation of Co(H2O)62+ (pink)

Increase Temp ↑

• By Heating it up

• Favours endothermic reaction


• To reduce Temp

• Formation of CoCl42- (blue)

CoCl42- + 6H2O ↔ Co(H2O)6

2+ + 4CI – ΔH = -ve (exothermic) (blue) (pink)

Increase in Temp ↑ – Favours endothermic reaction – Absorb heat to reduce Temp again ↓ Decrease in Temp ↓ – Favours exothermic reaction – Release heat to increase Temp again ↑

Increase Temperature • Rate of reaction increases • Rate constant also changes • Rate of forward and Rate of reverse increases but to different extend • Position of equilibrium shift to endothermic to decrease Temp • Kc, equilibrium constant change

Click to view video

http://www.youtube.com/watch?v=cWr3UDo-WeU&feature=plcp&context=C493411fVDvjVQa1PpcFNfJOeCPY-GOj7US1LiEwFKrWfbtfSRbZg=

http://www.youtube.com/watch?v=BGfYf8OQzuk&feature=relmfu





Effect of Temperature on the position of equilibrium

Decrease Temp ↓

• By Cooling it down ↓

• Favours exothermic reaction


• To increase Temp ↑


Increase Temp ↑

• By Heating it up ↑

• Favours endothermic reaction


• To reduce Temp ↓


Increase in Temp ↑ – Favours endothermic reaction – to absorb heat to reduce Temp again ↓ Decrease in Temp ↓ – Favours exothermic reaction – to release heat to increase Temp again ↑

Increase Temperature • Rate of reaction increases • Rate constant also changes • Rate of forward and Rate of reverse increases but to different extend • Position of equilibrium shift to endothermic to decrease Temp • Kc, equilibrium constant change

N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1


Click to view video


Effect of Temperature on equilibrium constant, Kc

DecreaseTemp ↓ – equilibrium shift to left - exothermic side to increase Temp - more reactant A produced

Rate of reverse kr > kf

Kc = kf/kr

Kc decrease ↓ because ratio of kf/kr decrease ↓ or ratio of product /reactant decrease ↓

Increase Temp ↑ – equilibrium shift to right, endothermic side to decrease Temp ↓ - more product B produced

Rate of forward kf > Rate of reverse kr

Kc = kf/kr or [conc product]/[conc reactant]

Kc increase ↑ because ratio of kf/kr increase ↑ or ratio of product /reactant increase ↑

A ↔ B ΔH = +ve Rate forward = kf

Rate reverse = kr

Arrhenius Equation show the relationship between temperature and rate constant

• Temperature will affect the rate constant for forward and reverse

For endothermic reaction Increase Temp ↑- position equilibrium shift to right – endothermic side – to absorb heat – Temp decrease ↓

Increase Temp ↑ – forward rate, kf > reverse rate, kr - Kc = kf/kr = Kc increase ↑

Increase Temp ↑- more product form, less reactants – Kc = ratio of product/reactants – Kc increases ↑

N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1

Conclusion :

Endothermic Reaction – Temp increase ↑ – Kc increase ↑ – Product increase ↑


Effect of Temperature on equilibrium constant, Kc

A ↔ B ΔH = -ve Rate forward = kf

Rate reverse = kr

H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1

Arrhenius Equation show the relationship between temperature and rate constant

• Temperature will affect the rate constant for forward and reverse

Increase Temp ↑ – equilibrium shift to left, endothermic side to decrease Temp ↓ - more reactant A produced

Rate of reverse kr > Rate of forward kf

Kc = kf/kr or [conc product]/[conc reactant]

Kc decrease ↓ because ratio of kf/kr decrease ↓ or ratio of product /reactant decrease ↓

DecreaseTemp ↓– equilibrium shift to right - exothermic side to increase Temp ↑ - more products B produced

Rate of forward kf > kr

Kc = kf/kr

Kc increase ↑ because ratio of kf/kr increase ↑ or ratio of product /reactant increase ↑

For exothermic reaction Increase Temp ↑- position equilibrium shift to left – endothermic side – to absorb heat – Temp decrease ↓

Increase Temp ↑ – reverse rate, kr > forward rate, kf - Kc = kf/kr = Kc decrease ↓

Increase Temp ↑- more reactant form, less product – Kc = ratio of product/reactants – Kc decreases ↓

Conclusion :

Exothermic Reaction – Temp increase ↑ – Kc decrease ↓ – Product decrease ↓

Catalyst • Provide an alternative pathway with lower activation energy • Increase the rate of forward and reverse to the same extent/factor • Position of equilibrium and Kc remains unchange • Catalyst shorten the time it takes to reach equilibrium


Effect of Catalyst on equilibrium constant, Kc

Without catalyst, takes long reaching equilibrium With catalyst, reaching equilibrium fast

Effect of catalyst on Rate, Rate constant and Kc on ammonia production

N2(g) + 3H2(g) ↔ 2NH3(g) ΔH = - 92kJmol-1

Adding catalyst

• Rate increase ↑

• Rate constant increase ↑

• Equilibrium constant Kc – No change

• Amount of product and reactants remain

the unchanged

N2O4 (g) ↔ 2NO2 (g)

Rf (rate forward) = Rr (rate reverse) kf [N2O4] = kr [NO2]2

Kc = (NO2)2 (N2O4)

• Kc is a ratio of rate constant and is temperature dependent

• Kc is a ratio of products conc to reactants conc

• Magnitude of Kc indicate how far/extend of the reaction proceeds towards a product at a given temperature

Small Kc : N2(g) + O2(g) ↔ 2NO(g) Kc = 1 x 10 -30 small ↓

Kc = (NO)2

(N2)1(O2)

1 Kc small → low product ↓, more reactants ↑, close to no reaction at all

Large Kc : 2COg) + O2 ↔ 2CO2(g) Kc = 2.2 x 10 22 high ↑

Kc = (CO2)2

(CO)2(O2)1 Kc large → high product ↑, small reactants ↓ , close to completion

Intermediate Kc : 2HI(g) ↔H2(g) + I2(g) Kc = 0.02

Kc = (H2)(I2)

(HI)2 Kc intermediate → significant amount of reactants and products

Equilibrium Law

Kc = (C)c(D)d

(A)a(B)b

At equilibrium:

N2O4 (g) ↔ 2NO2 (g)

Rf (rate forward) = Rr (rate reverse) kf [N2O4] = kr [NO2]2

kf/kr = (NO2)2 (N2O4)

Kc = kf / kr

aA + bB ↔ cC and dD

When a reversible reaction achieved dynamic equilibrium - aA + bB ↔ cC and dD

• Equilibrium constant Kc = ratio of molar conc of product (raised to power of their respective stoichiometry coefficient) to

molar conc of reactants (raised to power of their respective stoichiometry coefficient)

• Kc is constant at constant temperature

Equilibrium Law applies to system at equilibrium

• Kc is constant at constant temperature

• Kc unaffected by Concentration, Pressure and Catalyst

• Addition of reactants and products will only shift position of equilibrium to right or left changing to a new equilibrium

concentration but Kc remains the same

• Magnitude of Kc indicate the extend or how far the reaction forms product/reactants but not how fast

• Kc High – High product but rate can be very slow

• Kc Low – Low product but rate can be very fast

Reaction between H2 + I2 ↔ 2HI Kc = 46.4 at 730K Kc = [HI]2 = 46.4 [H2][I2]

Effect of Concentration on equilibrium constant, Kc • Increase in Conc ↑ - position of equilibrium shift to right/left - Conc will be Reduced ↓ • Decrease in Conc ↓ – position of equilibrium shift to right/left - Conc will be Increased ↑

Reaction contain different Initial Conc of H2 and I2 and HI but at equilibium Kc remains the same regardless of initial conc

H2 + I2 ↔ 2HI Initial Conc Equilibrium Conc

Kc = [HI]2 = 46.4 [H2][I2]

Equilibrium Law

Kc unchanged

How dynamic equilibrium is shifted when H2 is added ?

N2(g) + 3H2(g) ↔ 2NH3 (g)

• When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc • Rate of forward and reverse will increase • New equilibrium concentration will be achieved when rate of forward kf = rate of reverse kr • More products NH3 and less reactants N2 but Kc value remains unchanged

H2 added New equilibrium Conc

At equilibrium

Rate forward kf = Rate reverse kr

At equilibrium again

Rate forward kf = Rate reverse kr

Equilibrium shifted to right

Rate forward kf > Rate reverse kr

Kc = 4.07 Qc = 2.24 Kc = 4.07

How dynamic equilibrium is shifted when H2 is added ?

N2(g) + 3H2(g) ↔ 2NH3 (g)

• When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc • Rate of forward and reverse will increase • New equilibrium concentration will be achieved when rate of forward kf = rate of reverse kr • More products NH3 and less reactants N2 but Kc value remains unchanged

Equilibrium Conc for H2 = 0.82M

Equilibrium Conc for N2 = 0.20M

Equilibrium Conc for NH3 = 0.67M

N2(g) + 3H2(g) ↔ 2NH3 (g)

• Kc = (NH3)2 (N2)(H2)3

• Kc = (0.67)2

(0.20)(0.82)3

• Kc = 4.07

N2(g) + 3H2(g) ↔ 2NH3 (g)

• Qc = (NH3)2 (N2)(H2)3

• Qc = (0.67)2

(0.20)(1.00)3

• Qc = 2.24

New Conc for H2 = 1.00M

Equilibrium Conc for N2 = 0.20M

Equilibrium Conc for NH3 = 0.67M

New Equilibrium Conc for H2 = 0.90M

New Equilibrium Conc for N2 = 0.19M

New Equilibrium Conc for NH3 = 0.75M

Qc < Kc

2.24 < 4.07

• Kc must remain constant

• Shift to the right

• Increase products and decrease reactants

• New equilibrium Conc is achieved

• Qc = Kc again

N2(g) + 3H2(g) ↔ 2NH3 (g)

• Kc = (NH3)2 (N2)(H2)3

• Kc = (0.75)2

(0.19)(0.90)3

• Kc = 4.07

H2 added New equilibrium Conc

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IB Chemistry on Le Chatelier's Principle, Equilibrium Constant and Dynamic Equilibrium