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13.1 Equilibrium ConditionsPowerPoint Presentation

13.1 Equilibrium Conditions

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13.1 Equilibrium Conditions

- When a system is at equilibrium it may appear that everything has stopped; however, this is NOT the case.
- Think of chemical equilibrium like the cars on the Golden Gate Bridge and that the traffic flow in both directions is the same.
No net change

Chemical Equilibrium

- The state where the concentrations of all reactants and products remain constant with time.
- On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

What is equilibrium?

- Equal rates
- Concentrations are not
- Rates are determined by concentrations and activation energy
- Concentrations do not change at equilibrium or if the reaction is verrrrrry slooooooow.

13.2 The Equilibrium Constant

- For any reaction
- Where K is the equilibrium constant
- A,B,C & D are the concentrations of the “chemical species”a@ equilibrium
- j,k,l & m are the coefficients in the balanced equation

Try this:

- Write the equilibrium expression for
4NH3 + 7O24NO2 + 6H2O

And if I knew…

- …the equilibrium concentrations for each of the reaction components, I could calculate K.
Try this reaction:

N2 + 3H2 ↔ 2NH3

given:

[NH3]=3.1 x 10-2 mol/L

[N2]=8.5 x 10-1 mol/L

[H2]=3.1 x 10-3 mol/L

What would K be for the reverse reaction?

2NH3 ↔ N2 + 3H2

K’ = 2.6x10-5

- How about if only 1mol of NH3 was produced?
N2 + H2 ↔ 1NH3

K’’=1.9 x 102

Big conclusions!!

- By comparing K for the forward reaction to K for the reverse reaction, notice that
- And if I multiply (or divide) my reaction by a factor then where n is the factor

K is constant…???

- The equilibrium constant K always has the same value at a given temperature.
- A change in the temperature will change the rate and thus a new K.
- A set of concentrations at equilibrium is called an equilibrium position.
- There are an unlimited (infinite) number of equilibrium positions.

13.3 Equilibrium Expressions Involving Pressures

- So far we’ve been talking about equilibrium in terms of concentration…
- But gases can also be described by pressures as well.
PV = nRT

Or

Where C is molar concentration of the gas

For: N2 + 3H2 ↔ 2NH3

The relationship

- The relationship between K and Kp comes from the fact that for an ideal gas C = P/RT.
- But for the general reaction
the relationship between K and Kp is

Where ∆n is the sum of the coefficients of the gaseous products MINUS the sum of the coefficients of the gaseous reactants.

∆n = change in moles of gas

13.4 Heterogeneous Equilibria

- Remember that homogeneous equilibria is when all the products and reactants are gases
2SO2(g) + O2(g) ↔ 2SO3(g)

0.14

0.012 x 0.054

[COCl2]

[CO][Cl2]

Let’s look at a problem with homogeneous equilibria- The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

= 220

Kc=

CO (g) + Cl2 (g) ↔ COCl2 (g)

Kp = Kc(RT)Dn

Kp= 220 x (0.0821 x 347)-1 = 7.7

Heterogeneous equilibria

- Heterogeneous equilibria is when you have reaction components that may not be in the gaseous state.
CaCO3(s) ↔CaO(s) + CO2(g)

CaCO3 (s) CaO (s) + CO2 (g)

Kc =

Kc =

[CO2]

[CaO][CO2]

[CaCO3]

Kp= PCO

2

More importantly remember…- …that the concentrations of pure solids and liquid can not change therefore they are not included in the expression for the equilibrium constant

[CaCO3] = constant

[CaO] = constant

K reactionp = P

NH3

P

H2S

Try this one.- Consider the following equilibrium at 295 K:
- The partial pressure of each gas is 0.265 atm. Calculate Kp and Kcfor the reaction?
Hint: this time start with Kpand then find Kc

NH4HS (s)↔ NH3(g) + H2S (g)

= 0.265 x 0.265 = 0.0702

Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4

13.5 Applications of the Equilibrium Constant reaction

- Knowing the equilibrium constant for a reaction allows us to predict several important features of the reaction:
- Tendency of the reaction to occur
- If concentrations given represent equilibrium
- Equilibrium position from the initial concentrations

Important feature 1:Tendency of a reaction to occur reaction

- The tendency of a reaction to occur can be indicated by the magnitude of K

K >> 1

Lie to the right

Favor products

K << 1

Lie to the left

Favor reactants

Reaction goes to completion

Reaction does not occur to any significant extent

Remember reaction

- The size of K and the time it takes for the reaction to reach equilibrium are NOT directly related.
- The time it takes to reach equilibrium depends on the reaction rate and therefore ultimately the activation energy

Reaction Quotient - Q reaction

- Tells us the direction the reaction needs to go in order to reach equilibrium
- It is calculated the same as the equilibrium concentration; however, it uses initial concentrations rather than equilibrium concentrations.

Compare Q to reactionKc

- Qc < Kcsystem proceeds from left to right to reach equilibrium
Not enough products

Forward reaction occurs

- Qc = Kcthe system is at equilibrium
No net change

- Qc > Kcsystem proceeds from right to left to reach equilibrium
Too many products

Reverse reaction occurs

Try this… reactionN2 (g)+ 3H2 (g)↔ 2NH3 (g)

- At the start of a reaction, there are 0.249 mol N2, 3.21x10-2 mol H2, and 6.42x10-4 mol NH3 in a 3.50 L reaction vessel at 375oC. If the equilibrium constant (Kc) for the reaction is 1.2 at this temperature, decide whether the system is at equilibrium. IF it is not, predict which way the net reaction will proceed.

How did you do? reaction

- First: find the molarity of each gas.

- Then find Q
- Not at equilibrium
- Shift left to right

Important feature 2: given the equilibrium constant and initial concentrations, find equilibrium concentrations/pressures of products & reactants

- Let’s do a few problems
- Gaseous N2O4 was placed in a flask and allowed to reach equilibrium @ a temperature where Kp=0.133. At equilibrium, the pressure of N2O4 was found to be 2.71atm. Calculate the equilibrium pressure of NO2.
N2O4 (g)↔NO2(g)

Here’s how initial concentrations, find equilibrium concentrations/pressures of products & reactants

- First: Given Kpand PN2O4 = 2.71atm

Here’s another type of problem initial concentrations, find equilibrium concentrations/pressures of products & reactants

- At a certain temperature a 1.00L flask initially contained 0.298M PCl3 and 8.7x10-3M PCl5. After the system had reached equilibrium, 2.00x10-3M Cl2 was found in the flask. Gaseous PCl5 decomposes according to the following reaction:
PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

Calculate the equilibrium concentrations for all species and Kc

The steps to calculating equilibrium concentrations – ICE diagrams

- Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
- Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
- Having solved for x, calculate the equilibrium concentrations of all species.

Let’s set one up diagrams

Since the concentration of Cl2 we know has changed by 2x10-3, the balanced reaction tells us that since it is a 1-1-1 reaction that the reactants decompose by 2x10-3 and the products concentration increases by 2x10-3

Once we know the equilibrium concentrations… diagrams

- …we can find the equilibrium constant.

Let’s try another using diagramsan ICE diagram

- There are some instances when you may not be given ANY equilibrium concentrations.
Here’s how to do those.

Let’s get started!!

H diagrams2 (g) + I2 (g) ↔ 2HI (g)

- A mixture of 0.00623M H2, 0.00414M I2 and 0.0224M HI was placed in a 1.00L stainless-steel flask at 430oC. The equilibrium constant for the reaction is 54.3 at that temperature. Calculate the concentrations of each species at equilibrium.

The first step diagrams

- Calculate Q

Since 19.5<<<54.3, the reaction will proceed from left to right. The hydrogen and iodine will be depleted and there will be a gain in HI.

H2 (g) + I2 (g) ↔ 2HI (g)

Set up an ICE diagram diagrams

Now for the Algebra!!

Next: the equilibrium diagramsconstant is…

After multiplying all of this out…

Now do the quadratic formula…

And the x values are… diagrams

x = 0.0114M and x = 0.00156M

The first value of x can NOT be correct since it is larger than the original concentration of hydrogen and iodine.

Therefore x = 0.00156

This is the change in concentration.

Now back to the ICE diagram!!

- You can check your work by plugging the values in to find K diagramsc.
Remember Kc = 54.3

13.6: Solving equilibrium problems diagramsA more complete set of steps to solve equilibrium problems.

- Write the balanced equation for the reaction.
- Write the equilibrium expression using the law of mass action
- List the initial concentrations
- Calculate Q, and determine the direction of the shift to equilibrium
- Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations.
- Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown
- Check your calculated equilibrium concentrations by making sure the give the correct value of K.

Let’s do a few more ICE problems diagrams

Consider the reaction represented by the equation

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

- Trial #1
6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M.

What is the value for the equilibrium constant for this reaction? You’ll need this for the next few problems!!

Fe diagrams3+(aq) + SCN-(aq) ↔ FeSCN2+(aq)

- Initial 6.00 10.00 0.00
- Change -4.00 -4.00 +4.00
- Equilibrium 2.00 6.00 4.00

Consider the reaction represented by the equation diagrams

Fe3+(aq) + SCN-(aq) ↔ FeSCN2+(aq)

- Trial #2:
Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq)

Equilibrium: ? M FeSCN2+(aq)

Consider the reaction represented by the equation diagrams

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

- Trial #3:
Initial: 6.00 M Fe3+(aq) and 6.00 M SCN−(aq)

Equilibrium: ? M FeSCN2+(aq)

Consider the reaction represented by the equation diagrams

Fe3+(aq) + SCN-(aq) ↔ FeSCN2+(aq)

Fe3+ SCN- FeSCN2+

Trial #1 9.00 M5.00 M 1.00 M

Trial #2 3.00 M2.00 M 5.00 M

Trial #3 2.00 M9.00 M6.00 M

Find the equilibrium concentrations for all species.

Let’s see what happen here… diagrams

H2(g) + I2(g) ↔ 2HI(g)

K=7.1 x 102

Calculate the equilibrium concentrations if a 5.00L container initially contains 15.8g of H2 and 294g of I2

[HI]0 = zero

Since Q = 0 and Q < K, more product will be formed. diagrams

We know that:

K is large so the rxn will almost go to completion

I2 is the limiting reagent. It will be the smallest at equilibrium…let it’s equilibrium concentration = x

Fill in ICE based on diagramsstoichiometry

Now use K = 7.1 x 102 and algebra to find the value of x.

BUT…we know x will be very small…sooo..

- Because x is so small, it is negligible when added to another number and will essentially have little effect on the outcome. Giving us the following for our equation…
- Making the algebra MUCH, MUCH easier.
- Now , find the other concentrations.

The end result! another number and will essentially have little effect on the outcome. Giving us the following for our equation…

- Whew!!

Checking the assumption: another number and will essentially have little effect on the outcome. Giving us the following for our equation…(x is negligible)

- The rule of thumb is that if the value of x is less than 5% of all the smallest concentrations, our assumption is valid.
- If not, we would have used the quadratic equation to solve the problem.

What if K is really small? another number and will essentially have little effect on the outcome. Giving us the following for our equation…

- Begin the problem exactly the same way.
- For small K values, the product concentration is small
- Choosing a product to be x.

Here we go… another number and will essentially have little effect on the outcome. Giving us the following for our equation…

- Gaseous NOCl decomposes to form the gases NO and Cl2. At 35oC the equilibrium constant is 1.6x10-5. If 1.20 mol NOCl, 0.45 mol of NO, and 0.87 mol Cl2 are mixed into a 1.0L flask, what are the equilibrium concentrations?
2NOCl(g) ↔ 2NO(g) + Cl2(g)

First find Q

- Choose a product to be x at equilibrium another number and will essentially have little effect on the outcome. Giving us the following for our equation…
- NO will be the limiting reagent. It will be the smallest at equilibrium…let it’s equilibrium concentration = x

Fill in ICE based on another number and will essentially have little effect on the outcome. Giving us the following for our equation…stoichiometry

- And add to get the equilibrium “values”
Use these values and Kc = 1.6 x 10-5 to find the value of x.

- Let’s again assume that x is negligible and ignore it. another number and will essentially have little effect on the outcome. Giving us the following for our equation…
Now use this to find final concentrations

13.7 Le another number and will essentially have little effect on the outcome. Giving us the following for our equation…Chatlier’s Principle

- If there is a change imposed (a stress) on a system at equilibrium, the position of equilibrium will shift in a direction that tends to reduce that change (or stress)

Concentration another number and will essentially have little effect on the outcome. Giving us the following for our equation…

- Change the amounts of products or reactants…

Remove another number and will essentially have little effect on the outcome. Giving us the following for our equation…

Remove

Add

Add

aA + bBcC+ dD

Le Châtelier’s PrincipleConcentrationChange

Shifts the Equilibrium

Increase concentration of product(s)

Makes Q > K

left

Remember the effect on Q will tell you the direction of the shift

Decrease concentration of product(s)

Makes Q < K

right

Increase concentration of reactant(s)

Makes Q < K

right

left

Decrease concentration of reactant(s)

Makes Q > K

The effect of pressure another number and will essentially have little effect on the outcome. Giving us the following for our equation…

- There are three ways to change pressure
- Add or remove a gaseous product or reactant
- Add an inert gas (one not involved in the reaction
- Change the volume of the container

- We already looked at #1, and #2 won’t have any effect on equilibrium, so let’s look at effects of pressure by changing volume.

A another number and will essentially have little effect on the outcome. Giving us the following for our equation…(g) + B (g) C (g)

Changes in volume and pressureChange

Shifts the Equilibrium

Increase pressure

Side with fewest moles of gas

Decrease pressure

Side with most moles of gas

Increase volume

Side with most moles of gas

Decrease volume

Side with fewest moles of gas

Changes in temperature another number and will essentially have little effect on the outcome. Giving us the following for our equation…

- The big key to all of the previous changes are that they effect the equilibrium position but NOT the equilibrium constant.
- However, K, the equilibrium constant is changed with temperature. And it affects both the forward and the reverse reactions.
- The direction of the shift depends on whether or not it is exo- or endothermic

Changes in temperature another number and will essentially have little effect on the outcome. Giving us the following for our equation…

Change

Exothermic Rxn

Heat is product

Endothermic Rxn

Heat is reactant

- equilibrium video

Increase temperature

Kdecreases

Shift to the left

Kincreases

Shift to the right

Decrease temperature

Kincreases

Shift to the right

Kdecreases

Shift to the left

Catalyst & Le another number and will essentially have little effect on the outcome. Giving us the following for our equation…Châtlier

- Adding a catalyst
- Does not change K
- Does NOT shift the position of an equilibrium system
- System will reach equilibrium sooner!

- Catalyst lowers Ea for both forward and reverse reactions
- Catalyst does NOT change equilibrium constant or shift equilibrium.

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