Dynamic Equilibrium Notes

1. What is Dynamic Equilibrium?
2. Why does equilibrium occur?
3. Is it an equilibrium or not?
4. Shifting an equilibrium - Le Chatelier's Principle
5. Keq
6. Keq meets Le Chatelier
7. Calculator stuff

Everything we talk about in dynamic equilibrium is directly connected to what you learned in the KINETICS unit.  It is particularly important that you know how to ready PE graphs like the one below, and identify whether the reaction is exothermic or endothermic.

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What is Dynamic Equilibrium?

• Dynamic equilibrium is a reaction which occurs both in the Forward direction AND in the Reverse direction
• the reaction is DYNAMIC because it continues to occur even though there are no observable macroscopic changes that can be observed
• the following properties will remain constant when equilibrium has been established:
• temperature
• pressure
• concentration of reactants and products
• colour
• volume
• the rate of the forward reaction = rate of reverse reaction at equilibrium
• equilibrium reactions do not go to completion because the reverse reaction uses up products

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Why does equilibrium occur?

You already know that equilibrium occurs because both the forward and the reverse reactions occur at the same time, and at the same rate.  BUT why do both reactions occur at all?

In the KINETICS unit, you learned:

• EXOTHERMIC reactions are SPONTANEOUS and go to completion
• ENDOTHERMIC reactions are NOT spontaneous and require a lot of energy in order to occur

So, why does the endothermic reaction in an equilibrium occur at all?

There are two forces that drive an equilibrium:

• ENTHALPY - the amount of energy released or consumed during a reaction
• ENTROPY - the amount of randomness in a system

Randomness is determined by the state the reactant or product is in.  Gases are more random than solutions.  Solutions are more random than liquids.  Liquids are more random than solids.

Randomness summarized from most to least random:

Gas >> Aqueous >> Liquid >> amorphous solids > crystaline solids

Reactions ALWAYS tend to MINIMUM enthalpy and MAXIMUM entropy.  If these two drives oppose each other then the reaction will be a dynamic equilibrium.  If these two forces push in the same direction, the reaction will either go to completion, or not occur.


MINIMUM ENTHALPY:   ΔH is NEGATIVE, or the heat term is on the PRODUCTS side of the equation.

MAXIMUM ENTROPY: ΔS is POSITIVE, or the products side is most random.

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Is it an equilibrium or not?

In order to decide if a reaction is a dynamic equilibrium, you will have to determine the direction that ENTHALPY pushes and the direction that ENTROPY pushes.  When these two driving forces oppose each other then there is a dynamic equilibrium.  When enthalpy and entropy both push in the forward direction, the reaction will go to completion.  When enthalpy and entropy both push in the reverse direction, the reaction will not occur.

In chemistry 12 you will be expected to determine whether a given reaction is likely to be an equilibrium or not.  You will not have to determine anything about the position of the equilibrium or anything tricky, just what direction each driving force pushes.

Some examples:

A_(l)  +  2B_(g)  →  AB_2(g)                               enthalpy pushes → and entropy pushes ← so reaction is an equilibrium

A_(g)  +  B_(g)  +  heat  →  AB_(g)                     enthalpy pushes ← and entropy pushes ← so there is NO reaction

AB_(s)  +  heat  →  A_(l)  +  B_(g)                      enthalpy pushes ← and entropy pushes → so reaction is and equilibrium

A₂B_(g)  →  2A_(g)  +  B_(g)  +  heat                 enthalpy pushes → and entropy pushes → so the reaction goes to completion

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Shifting an equilibrium - Le Chatelier's Principle

Le Chatelier's principle states:  An equilibrium system when subjected to a stress will shift to counteract the stress and a new equilibrium will be established.

So what does this really mean?  Well there are several stresses that a reaction can experience, the most common are changes in temperature, pressure, volume or concentration.  Pressure and volume changes only affect equilibria which include gas molecules.  Temperature changes will affect any equilibrium system.

Some analogies to help you understand what is going on:

• When you are too hot, your instinct is to do something to cool down, equilibrium reactions have the same instinct.  An equilibrium reaction will always be exothermic in one direction and endothermic in the opposite direction.  Since the endothermic direction "consumes heat" a reaction that has its temperature raised will start to shift so that the endothermic direction is favoured.  The reaction shifts AWAY from the heat term. 

• When you are cold, you want to get warmer, equilibrium reactions do the same thing.  An equilibrium reaction that is cooled down will always try to heat itself up.  Since the exothermic direction "produces heat" a reaction that is cooled down will start to shift so that the exothermic direction is favoured.  The reaction shifts TOWARDS the heat term.

• When you are under a lot of pressure... ok so are you starting to see a trend?  yes?  no?  Let's see if you get this...  Will you try to relieve pressure or add more?  Equilbrium reactions will always try to relieve any build up of pressure by shifting to the side of the reaction with fewer gas molecules.  Fewer gas molecules take up less space and so the pressure will decrease.

What you need to do with a Le Chatelier's principle question:

1. Identify the stress
2. Decide how the stress will affect the equilibrium
3. Determine what direction the equilibrium will have to shift in order to reduce the stress on the system
4. Determine whether rates for the forward and reverse reactions will increase, decrease, or stay the same

Some examples:

1)           N₂O₄   +    heat    ↔   2NO₂

• As temperature rises the equilibrium shifts AWAY from the heat term => more NO₂ is produced.  The NO₂ looks BROWN whereas the N₂O₄ is colourless.
• As temperature goes down the equilibrium shifts TOWARDS the heat term => more N₂O₄ is produced => colour fades
• As pressure increases the equilibrium shifts to the side with FEWER gas molecules because this will relieve the pressure => this equilibrium will shift to the reactants side => colour fades
• As pressure decreases the equilibrium shifts to the side with MORE gas molecules because this will increase the pressure => this equilibrium will shift to the products side => colour gets darker
• If [N₂O₄] is increased there will be too much reactant so the equilibrium shifts AWAY from the reactant side and towards the products side
• If [NO₂] is decreased there will be too little product so the equilibrium shifts TOWARDS the products side.

2) Haber Process (see page 56 of Hebden)

3) Making CaO from limestone (see page 56 of Hebden)

The GRAPHS...

You need to be able to understand the "concentration vs time graphs" that show how concentration changes as a result of a stress.  This information is on pages 50 to 53 of Hebden, and in your class notes.  Make sure you can do question 27 on page 55 of Hebden.

Example

For the equilibrium:                 A_(g)  +  B_(g) ↔  AB_(g)  +  heat 

                   Temperature was increased      [A] was increased                    Pressure was increased             

Summary of the graphs:

• all changes are gradual = TEMPERATURE change
• one concentration is a sudden increase or decrease = CONCENTRATION change
• all sudden increase or sudden decrease = PRESSURE change
  

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The Equilibrium Constant - K_eq

The equilibrium constant K_eq is the ratio of the .  A K_eq that is LARGER than 1 means that the equilibrium favours products over reactants.  A K_eq that is SMALLER than 1 means that the equilibrium favours reactants over products.

Writing a K_eq expression for an equation - only use species that are either gases or aqueous, all other species do NOT appear in K_eq expression.

Examples:

H_2(g)  +  I_2(g)  ↔  2HI_(g)

CaCO_3(s)  ↔  CaO_(s)  +  CO₂

N_2(g)  +  3H_2(g)  ↔  2NH_3(g)

REMINDER: 

• The value of K_eq is ONLYaffected by TEMPERATURE! 
  
• We assume that K_eq is constant when the concentration or gas pressure changes!

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K_eq Meets Le Chatelier

K_eq is ratio of so if you have conditions that result from an equilibrium being shifted, then you can predict from the K_trial what direction the reaction needs to shift in order to reach equilibrium.  You need to either be given the K_eq value or be given the conditions at equilibrium before the shift occurs.

• If K_trial is LARGER than K_eq then there is too much product and not enough reactant ⇒ shifts to reactants
• If K_trial is SMALLER than K_eq then there is not enough product and too much reactant ⇒ shifts to products
• If K_trial is equal to K_eq then you are already at equilibrium ⇒ does NOT shift

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Calculator Stuff

Examples:

1) Calculating K_eq

A 3.0L bulb contains 4.0mol gas A, 6.0mol gas B, and 2.3mol gas A₂B
What is K_eq for the equilibrium:  2A_(g)  +  B  ↔  A₂B_(g)

2) Calculating K_eq from initial and some equilibrium values

5.0mol of gas A₂B₃ is introduced into a 3.0L bulb. The following equilibrium is established:   A₂B_3(g)  ↔  2A_(g)  +  3B_(g)

At equilibrium there are 1.5mol gas A.  Determine K_eq.

NOTE: Be careful with significant figures sometimes the + − rule will affect the significant figures when using ICE chart.

3) Calculating initial concentrations using K_eq

a) Gas AB₂ was introduced into a 1.5L bulb. The following equilibrium was attained:   AB_2(g)  ↔  A_(g)  +  2B_(g)

At equilibrium [A] = 0.15M.  If K_eq = 0.89, how many moles of AB₂ was originally present?

b) Consider the equilibrium:   2A₂B_(g)  ↔  2A_2(g)  +  B_2(g)
A₂B is placed in a 3.0L flask. At equilibrium [B₂] = 0.87M
How many moles of A₂B were present initially if K_eq = 1.5

4) Determine whether a system is at equilibrium using K_trial and direction of shift to establish equilibrium

K_eq = 0.78 for the equilibrium   A_2(g)  +  B_{2(g)  ↔  2AB(g)
If 1.5mol A2, 1.5mol B2 and 7.0mol AB are put in a 3.0L flask which direction will the reaction shift to reach equilibrium?}

_{5) Calculating Equilibrium concentration of species}

_{Consider the equilibrium:   AB(g)  +  CD(g)  ↔  AD(g)  +  CB(g)}

K_eq = 2.8

If 3.5mol AB and 3.5mol CD are placed in a 2.0L bulb and the system is allowed to come to equilibrium, what is the equilibrium concentration for all species?

6) Finding New Equilibrium [ ] when an existing equilibrium is shifted

a) Consider the equilibrium:   A_2(g)  +  B_2(g)  ↔  2AB_(g)

At equilibrium there are 3.0moles A₂, 4.5moles B₂ and 5.2moles AB in a 2.0L flask. 1.5moles A₂ is added and the system is allowed to re-establish equilibrium. Calculate the concentration of all species when the new equilibrium is established.

b) Consider the equilibrium:   2A₂B_(aq)  ↔  2A_2(g)  +  B_2(aq)
At equilibrium there are 3.00moles A₂B, 2.00moles A₂ and 1.00moles B₂ in a 1.0L flask. How many moles of A₂ need to be removed to increase [B₂] to 1.5M

_{For more examples and explanations please refer to HEBDEN pages 63 - 72.}

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