Le Chatelier’s principle and more...

Feature of Equilibrium State ExplanationEquilibrium is dynamic The reaction has not stopped

but the forward and reverse reaction are still occurring

Equilibrium is achieved in a closed system

A closed system prevents exchange of matter with the surroundings

The concentrations of reactants and products remain constant

They are being produced and degraded at an equal rate

The macroscopic properties do not change

Color and density are properties that depend upon concentrations

Equilibrium can be reached from either direction

The same equilibrium mixture will result under the same conditions, regardless of the starting point.

Quick review of equilibrium

Catalyst

• the same process still has to happen, catalysts just help out by lowering the activation energy

• increase the RATE of a reaction….and therefore the decrease the time in which equilibrium is reached – they speed up the forward and reverse reactions

equally• therefore decreases the time required for the system to

achieve equilibrium• less time equals $$$ when making chemicals

http://www.google.com/url?sa=i&rct=j&q=physical+equilibrium+system&source=images&cd=&cad=rja&docid=CaNC2M60DIaISM&tbnid=9vbrFrIwD4ihWM:&ved=0CAUQjRw&url=http://www.chem.queensu.ca/people/faculty/mombourquette/firstyrchem/equilibrium/&ei=dWlKUbD1JMbliwK2kYGQDw&bvm=bv.44158598,d.cGE&psig=AFQjCNHuA1x2NZ_RUMbgZPuavfQuh2rZNg&ust=1363917107771225

Le Chatelier’s principle and more...

– Nice video- 20 minutes– Another good one- 15 minutes– states when a system in chemical equilibrium is

disturbed by a change, the system shifts in a way that tends to counteract this change of variable

– a change imposed on an equilibrium system is called a stress• a stress usually involves a change in the

temperature, pressure, or concentration• the equilibrium always responds in such a way

so as to counteract the stress

Stress 1. Temperature change

• this is the ONLY STRESS THAT WOULD ACTUALLY CHANGE KC

• increasing temperature– favors the “cold side”/endothermic/the reaction that

needs heat– adding heat favors the reaction to the left since it needs

+92 kJ– Kc decreases

DH = + 92 kJ N2 (g) + 3H2 (g) 2NH3 (g) DH = - 92 kJ

Haber Process again

McGraw Hill Flash animation

• decreasing temperature– favors the “hot side”/exothermic• the reaction is already giving off heat it

doesn’t need so cooling down is good• Kc increases

DH = + 92 kJ N2 (g) + 3H2 (g) 2NH3 (g) DH = - 92 kJ

Stress 2. Pressure change

• an increase in pressure causes the equilibrium to shift in the direction that has the fewer number of moles– results in a decrease in N2 and H2 and an increase in NH3

• an decrease in pressure causes the equilibrium to shift in the direction that has the most number of moles– results in a an increase in N2 and H2 and an decrease in NH3

• does NOT affect the equilibrium constant Kc

N2 (g) + 3H2 (g) 2NH3 (g) DH = - 92 kJ

Haber Process again

McGraw Hill Flash animation

Stress 3. Concentration change

• the equilibrium responds in such a way so as to diminish the increase or equalize the ratio– increasing concentration of reactants shifts the reaction

to the right (forward, more product)– increasing concentration of products shifts the reaction

to the left (reverse, more reactants)• does NOT affect the equilibrium constant Kc

N2 (g) + 3H2 (g) 2NH3 (g) DH = - 92 kJ

Haber Process again

Kc =[NH3]2

[N2] [H2]3

McGraw Hill Flash animation

Practice Problem• Predict the effect of the following changes on the

reaction in which SO3 decomposes to form SO2 and O2. 2 SO3(g) 2 SO2 (g) + O2 (g) Ho = 197.78 kJ

• increasing the temperature of the reaction – shifts right

• increasing the pressure on the reaction– shifts left

• adding more O2 when the reaction is at equilibrium – shifts left

• removing O2 from the system when the reaction is at equilibrium– shifts right

backwards Contact Process

Le Chatelier’s Principle – Summary

Change Effect on Equilibrium Change in Kc?

Increase concentration

Shifts to opposite side No

Decrease concentration

Shifts to same side No

Increase pressure Shifts to side with least moles of gas

No

Decrease pressure Shifts to side with most moles of gas

No

Increase temperature

Shifts in endothermic direction

Yes

Decrease temperature

Shifts in exothermic direction

Yes

Add a catalyst No change No

Le Châtelier’s Principle

• Changes in Concentration

Change Shifts the Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left

aA + bB cC + dD

AddAddRemove Remove

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s Principle

• Changes in Concentration

N2 (g) + 3H2 (g) 2NH3 (g)

AddNH3

14.5

Le Châtelier’s Principle

• Changes in Volume and Pressure(Only a factor with gases)

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas

Decrease pressure Side with most moles of gas

Decrease volume

Increase volume Side with most moles of gas

Side with fewest moles of gas

• For each of the following reactions, predict how the equilibrium will shift as the temperature is increase

N2(g) + O2(g) 2NO(g) ∆H = +kJ mol-1

2SO2(g) + O2(g) 2SO3(g) ∆H = -kJ mol-1

Right

left

Effect of: Position of Equilibrium

Value of Kc

Concentration Changes No changePressure Changes if

reaction involves a change in the number of gas molecules

No change

Temperature Changes Changes (depends on if exothermic or endothermic)

Catalyst No change No change

Haber Process (don’t think you need to know this anymore/I will not test you on it)

• N2 (g) + 3H2 (g) 2NH3 (g) DH = - 92 kJ– good catalyst is iron– optimum temp is 450°C• good for kinetic theory (molecules moving fast,

more collisions with more energy)• bad for equilibrium, reaction is exothermic

– optimum pressure 250 atm• good for kinetics (more collisions)• favors the products (only 2 mol vs. 4 in the

reactancts)

Contact Process (don’t think you need to know this anymore/I will not test you on it)

2 SO2(g) + O2(g) 2 SO3(g) Ho = - 197.78 kJ – good catalyst is V2O5

– optimum temp is 450°C• good for kinetic theory (molecules moving fast,

more collisions with more energy)• bad for equilibrium, reaction is exothermic

– optimum pressure is 2 atm• not good for kinetics (less collisions)• low pressure favors reactants

– however, this is a very efficient yield (99%) even at low pressure» therefore, avoids the need for expensive and

dangerous equipment