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Equilibrium. Reactions are reversible. A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible As C and D build up, the reverse reaction speeds up while the forward reaction slows down.

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reactions are reversible
Reactions are reversible
  • A + B C + D ( forward)
  • C + D A + B (reverse)
  • Initially there is only A and B so only the forward reaction is possible
  • As C and D build up, the reverse reaction speeds up while the forward reaction slows down.
  • Eventually the rates are equal
slide3

Forward Reaction

Reaction Rate

Equilibrium

Reverse reaction

Time

what is equal at equilibrium
What is equal at Equilibrium?
  • Rates are equal
  • Concentrations are not.
  • Rates are determined by concentrations and activation energy.
  • The concentrations do not change at equilibrium.
  • Or if the reaction is verrrryslooooow.
law of mass action
Law of Mass Action
  • For any reaction
  • jA + kBlC + mD
  • K = [C]l[D]mPRODUCTSpower [A]j[B]kREACTANTSpower
  • K is called the equilibrium constant.
  • is how we indicate a reversible reaction
  • Only gases and aqueous solutions are included in the law of mass action
playing with k
Playing with K
  • If we write the reaction in reverse.
  • lC + mDjA + kB
  • Then the new equilibrium constant is
  • K’ = [A]j[B]k = 1/K[C]l[D]m
playing with k1
Playing with K
  • If we multiply the equation by a constant
  • njA + nkBnlC + nmD
  • Then the equilibrium constant is
  • K’ = [A]nj[B]nk= ([A] j[B]k)n=Kn[C]nl[D]nm ([C]l[D]m)n
manipulating k

From a practical point of view, here is how

K gets manipulated:

If a reaction gets doubled, the value of K is

squared. If it is tripled, K is raised to the

third power.

If the reaction gets cut in ½ then K is found by

taking the square root of K.

Manipulating K

the units for k
The units for K
  • This one is simple.
  • K has no units.
k is constant
K is CONSTANT
  • Temperature affects the equilibrium constant.
  • The equilibrium concentrations don’t have to be the same only K.
  • Equilibrium position is a set of concentrations at equilibrium.
  • There are an unlimited number of K values so the temperature needs to be stated.
equilibrium constant

Equilibrium Constant

One for each Temperature

calculate k
Calculate K
  • N2 + 3H2 3NH3

Initial At Equilibrium

  • [N2]0 =1.000 M [N2] = 0.921M
  • [H2]0 =1.000 M [H2] = 0.763M
  • [NH3]0 = 0 M [NH3] = 0.157M
calculate k1
Calculate K
  • N2 + 3 H2 2 NH3 (all gases)

Initial At Equilibrium

  • [N2]0 = 0 M [N2] = 0.399 M
  • [H2]0 = 0 M [H2] = 1.197 M
  • [NH3]0 = 1.000 M [NH3] = 0.157M
  • K is the same no matter what the amount of starting materials
equilibrium and pressure
Equilibrium and Pressure
  • Don’t forget your gas laws, they are often needed to get some information for the problem. Especially PV = nRT
  • Some reactions are gaseous
  • P = (n/V)RT
  • P = CRT
  • C is a concentration in moles/Liter
  • C = P/RT
equilibrium and pressure1
Equilibrium and Pressure
  • 2 SO2(g) + O2(g) 2 SO3(g)
  • Kp = (PSO3)2

(PSO2)2 (PO2)

  • K = [SO3]2 [SO2]2 [O2]
slide16
Equilibrium and Pressure: For those that are a glutton for punishment and want to see the full treatment.
  • K =(PSO3/RT)2 (PSO2/RT)2(PO2/RT)
  • K =(PSO3)2 (1/RT)2 (PSO2)2(PO2) (1/RT)3
  • K = Kp(1/RT)2= Kp RT (1/RT)3
general equation
General Equation
  • jA + kBlC + mD
  • Kp= (PC)l (PD)m= (CC x RT)l (CD x RT)m (PA)j (PB)k (CA x RT)j(CB x RT)k
  • Kp=(CC)l (CD)m x (RT)l+m(CA)j(CB)k x (RT)j+k
  • Kp=K (RT)(l+m)-(j+k) = K (RT)Dn
  • Dn=(l+m)-(j+k) = Change in moles of gas
homogeneous equilibria
Homogeneous Equilibria
  • So far every example dealt with reactants and products where all were in the same phase.
  • We can use K in terms of either concentration or pressure.
  • Units depend on reaction.
heterogeneous equilibria
Heterogeneous Equilibria
  • If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change.
  • As long as they are not used up we can leave them out of the equilibrium expression.
  • For an example, check out the next slide.
for example
For Example
  • H2(g) + I2(s) 2 HI(g)
  • K = [HI]2 [H2][I2]
  • But the concentration of I2 does not change, so it factors out of the equilibrium expression and looks like this.
  • K = [HI]2 [H2]
the reaction quotient
The Reaction Quotient
  • Tells you the direction the reaction will go to reach equilibrium
  • Calculated the same as the equilibrium constant, but for a system not at equilibrium
  • Q = [Products]coefficient [Reactants] coefficient
  • Compare value to equilibrium constant
what q tells us
What Q tells us
  • If Q < K
    • Not enough products
    • Shift to right
  • If Q > K
    • Too many products
    • Shift to left
  • If Q = K system is at equilibrium. We like these ones – there is nothing to do !!
example
Example
  • For the reaction:
  • 2 NOCl(g) 2 NO(g) + Cl2(g)
  • K = 1.55 x 10-5 M at 35ºC
  • In an experiment 0.10 molNOCl, 0.0010 mol NO(g) and 0.00010 mol Cl2 are mixed in 2.0 L flask.
  • Which direction will the reaction proceed to reach equilibrium?
solving equilibrium problems1
Solving Equilibrium Problems
  • Given the starting concentrations and one equilibrium concentration.
  • Use stoichiometry to figure out other concentrations and K.
  • Learn to create a table of initial and final conditions.
  • Check out the next slide for an actual problem.
slide26
Consider the following reaction at 600ºC
  • 2 SO2(g) + O2(g) 2 SO3(g)
  • In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO3 were found to be present. Calculate the equilibrium concentrations of O2 and SO2, K and KP
the ice box
The ICE BOX

The table we create is called the ICE Box.

2 SO2(g) + O2(g) 2 SO3(g)

Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated.

Initial 2 1.5 3

Change

Equilibrium 3.5

the ice box1
The ICE BOX

The table we create is called the ICE Box.

2 SO2(g) + O2(g) 2 SO3(g)

Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated.

Initial 2 1.5 3

Change - 2x - x + 2x

Equilibrium 3.5

where are we at
Where are we at ?

The values in the CHANGE row are a combination of stoichiometry and equilibrium.

Notice that the numbers are the same as the coefficients. We can’t forget stoichiometry.

The [SO3] increases so it has to be a + 2x.

This tells us it shifts to the right and that the value of the reactants has to decrease.

We combine the + or – sign along with the stoichiometry to set up the equation.

the ice box2
The ICE BOX

The table we create is called the ICE Box.

2 SO2(g) + O2(g) 2 SO3(g)

Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated.

Initial 2 1.5 3

Change - 2x - x + 2x

Equilibrium 2 – 2x 1.5 – x 3.5

where are we at part 2
Where are we at ? Part 2

The values of the equilibrium are known algebraically. Here is our Law of Mass Action. We can solve that x = .25

K = [ SO3] 2 = (3.5) 2 [ O2] [ SO2]2 (1.25) (1.5)2

K = 4.36

slide32

A New Problem

  • Consider the same reaction at 600ºC

2 SO2(g) + O2(g) 2 SO3(g)

  • In a different experiment .500 mol SO2 and .350 mol SO3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O2 are present.
  • Calculate the final concentrations of SO2 and SO3 and K.
let s set up the ice box
Let’s set up the ICE Box

2 SO2(g) + O2(g) 2 SO3(g)

Initial .5 0 .35

Change + 2x + x - 2x

Notice the signs. Since there is no oxygen present at the start, the reaction must shift to the left and that means products consumed and reactants created.

Equilibrium .5 +2x.045 .35 – 2x

This tells us that x = .045 and therefore the concentrations must be as follows.

[SO2] = .5 + 2x or .5 + .09 = .59

[SO3] = .35 – 2x or .35 - .09 = .26

K = (.26)2 / (.59)2 (.045) = 4.32

what if you re not given equilibrium concentration
What if you’re not given equilibrium concentration?
  • The size of K will determine what approach to take.
  • First let’s look at the case of a LARGE value of K ( >100).
  • We can make simplifying assumptions.
5 rule
5 % rule

Now is as good a time as any to introduce the 5 % rule.

Equilibrium concentrations are often a bit of an approximation. Therefore it’s allowable to simplify the math to make it more workable.

The 5 % rule states that if x is small in relation to the stated amount, and the K, then it can be ignored in the calculation.

Generally works if the Keq is less than 10-3 or greater than 103

example1
Example
  • H2 (g) + I2 (g) 2 HI (g) K = 7.1 x 102 at 25ºC
  • Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.9 g of H2 and 294 g I2 .
  • [ H2 ]0 = (15.7g/2.02)/5.00 L = 1.56 M
  • [ I2 ]0 = (294g/253.8)/5.00L = 0.232 M
  • [ HI ]0 = 0
slide37

Let’s set up the Ice Box !!

  • Q > K so more product will be formed.
  • Since K is large, rxn. will go to completion.
  • Another great clue is that when there is a component with a concentration of zero, the reaction must shift in that direction.
  • Stoichiometry tells us I2 is LR, it will be smallest at equilibrium let it be x
  • Set up table of initial, final and change in concentrations.
slide38

This equation needs manipulating

H2(g) + I2(g)  2 HI(g) initial 1.56 M 0.232 M 0 M change - x - x + 2xequilibrium 1.56 - x .232 - x 2x

  • Choose x so it is small.
  • Since we know the I2 is going to lose most of its concentration, x is actually going to be a very high percentage of the amount of I2
  • There is a way around this dilemma.
here we go
Here we go.

In the work below, we are going to assume that the reaction goes to completion.

Let’s combine the following facts and ideas and solve the problem.

H2(g) + I2(g)  2 HI(g) initial 1.56 M .232 M 0 M change - .232 - .232 + .464 equilibrium 1.328 0 .464

piecing it together
Piecing it together

1. We didn’t change the conditions of the rxn. (ie. Temp. ) so the Keq remains unchanged.

2. The Keq is large so we know there will be more product than reactant at equilibrium.

3. Since we assumed the reaction went to completion we now have no Iodine and the reaction must move to the left.

4. X must be small because there needs to be more product to satisfy Keq. X will be negligible compared to the amounts given.

5. However, x can’t be ignored for the I2 because it is at zero and x is 100% of the total amount of I2

slide41

H2(g) + I2(g)  2 HI(g)

initial 1.328 0 .464

change + x + x -2x

equilibrium 1.328 + x x.464 – 2x

Here is our new Law of Mass Action after we remove the ‘x’ according to the 5% rule.

Remember we can’t remove the x in the I2 because that is all that there is of it.

710 = (.464)2 / (1.328) ( x ) = 2.28 x 10-4

checking the assumption
Checking the assumption
  • The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid.
  • If not we would have had to use the quadratic equation
  • More on this later.
  • Our assumption was valid.
practice
Practice
  • For the reaction Cl2 + O2 2ClO(g) K = 156
  • In an experiment 0.100 mol ClO, 1.00 mol O2 and 0.0100 mol Cl2 are mixed in a 4.00 L flask.
  • If the reaction is not at equilibrium, which way will it shift? Q = ?
  • Calculate the equilibrium concentrations of all components.
process is the same
Process is the same
  • Set up table of initial, change, and final concentrations. Your ICE Box
  • Choose X to be small.
    • Sounds familiar so far, I hope !!
  • For this case it will be a product.
  • For a small K the product concentration is small.
slide46

Let’s start the set up.

  • For the reaction 2 NOCl 2 NO + Cl2
  • K= 1.6 x 10-5 Make note the small K favors more reactant
  • If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl2 are mixed in a 1 L container
  • What are the equilibrium concentrations ?
  • Q = [NO]2[Cl2] = (0.45)2(0.87) = 0.15 [NOCl]2 (1.20)2
  • Q > K so it shifts to the left.
slide47

Here is the ICE Box set up

Once again we can manipulate the components to make the math easier and we do this by assuming the reaction goes to completion according to stoichiometry.

  • We can now use the Equilibrium concentrations to rework the problem. Remember that the K does not change. K favors the reactants. Having [ NO ] = 0 means that the reaction will shift to the right. Combining the two means that the change will be small so x will be small and we can use the 5 % simplification rule. This is a Happy Day !

2 NOCl  2 NO + Cl2

Initial 1.20 0.45 0.87

Change +.45 -.45 -.225

Equilibrium 1.65 0 .645

where are we at1
Where are we at ?

We can now use the Equilibrium concentrations to rework the problem.

Remember that the K does not change.

K favors the reactants.

Having [ NO ] = 0 means that the reaction will shift to the right. Combining the two means that the change will be small so x will be small and we can use the 5 % simplification rule.

This is truly a Happy Day !

slide49

Here is the new ICE Box

2 NOCl  2 NO + Cl2

Initial 1.65 0 .645

Change - 2x +2x +x

Equilibrium 1.65 – 2x 2x .645 + x

Equilibrium (5%) 1.65 2x .645

Solving: 1.6 x 10-5 = (.645) (2x)2

(1.65)2

The algebra is trickier but doable x = .0042

slide50

Always verify the 5 % rule

  • If x = .0042, we compare that to the concentration. .0042 / .87 = .48 %
  • We are good !!
practice problem
Practice Problem
  • For the reaction 2 ClO(g) Cl2(g) + O2(g)
  • K = 6.4 x 10-3
  • In an experiment 0.100 mol ClO(g), 1.00 mol O2 and 1.00 x 10-2 mol Cl2 are mixed in a 4.00 L container.
  • What are the equilibrium concentrations.
no simplification
No Simplification
  • Choose X to be small.
  • Can’t simplify so we will have to solve the quadratic formula
  • H2 (g) + I2(g) 2 HI (g) K = 38.6
  • What is the equilibrium concentrations if 1.800 mol H2, 1.600 mol I2 and 2.600 mol HI are mixed in a 2.000 L container?
here is the set up
Here is the set - up

Q = 2.34 so Q < K & it shifts to the right.

H2 (g) + I2(g) 2 HI (g)

Initial .9 .8 1.3

Change -x -x + 2x

Equilibrium .9 – x .8 – x 1.3 + 2x

K = 38.6 = (1.3 + 2x)2 / (.9 – x) (.8 – x) = ??

Fortunately, we don’t see them this type often.

You are encouraged to discover the many uses of your TI – 84 or equivalent.

problems involving pressure
Problems Involving Pressure
  • Solved exactly the same, with same rules for choosing X but use pressures due to Kp
  • For the reaction N2O4 (g)  2NO2 (g)

Kp = .131 atm. What are the equilibrium pressures if a flask initially contains 1.000 atm N2O4?

N2O4 (g)  2 NO2 (g)

Initial 1 0

Change - x + 2x

Equilibrium 1-x 1+ 2x

le chatelier s principle
Le Chatelier’s Principle
  • If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress, and establish a new equilibrium.
  • 3 Types of stress
    • Temperature changes
    • Pressure changes
    • Concentration changes
the basic steps
The Basic Steps
  • Here is the best way to always get LeChatelier’s problems correct.
  • Sample Problem A + B  C + D

How is the system going to change if more B is added ?

  • 1. Identify the stress: The addition of B
  • 2. What is the opposite: The removal of B
  • 3. Which way must the reaction shift to achieve step 2: It must shift to the right
change amounts of reactants and or products
Change amounts of reactants and/or products
  • Adding product makes Q > K
  • Removing reactant makes Q > K
  • Adding reactant makes Q < K
  • Removing product makes Q < K
  • Determine the effect on Q, will tell you the direction of shift: remember that it is important to know what Q is equal to.
change in pressure 1
Change in Pressure # 1
  • This is usually done by changing volume.

Decreasing the volume

increases the pressure

  • System shifts in the direction that has the less moles of gas: less gas = less pressure
  • Because partial pressures (and concentrations) change a new equilibrium must be reached and established.
  • System tries to minimize the moles of gas.
change in pressure 2
Change in Pressure # 2

This is usually done by changing volume.

Increasing the volume

decreases the pressure

System shifts in the direction that has more moles of gas: more gas = more pressure

Because partial pressures (and concentrations) change a new equilibrium must be reached and established.

System tries to maximize the moles of gas.

change in pressure 3
Change in Pressure # 3
  • By the addition of an inert gas.
  • Partial pressures of reactants and product are not changed .
  • No effect on equilibrium position.
  • This is a common question on the AP exam so be aware of it. It should be easy points.
  • It might phrase it as an inert gas or it might use one of the Noble Gases:
    • He Ne Ar Kr XeRn
change in temperature
Change in Temperature
  • Affects the rates of both the forward and reverse reactions.
  • Doesn’t just change the equilibrium position, changes the equilibrium constant.
  • The direction & magnitude of the shift depends on whether the reaction is exothermic or endothermic.
  • On many questions, it states a temperature that K is measured in. This is more to make the question technically correct. Look back on our solved problems. Temperature was not calculated in.
change in temperature exothermic
Change in Temperature :Exothermic
  • DH < 0 It is a negative number.
  • The reaction releases heat.
  • Think of heat as a product since you know an equation can be written with a heat term:

A + B  C + D + Heat

  • Raising temperature (adding heat) effectively increases the amount of heat (product)
  • Shifts to left.
  • Remember the steps to solving LeChatelier’s problems.
change in temperature endothermic
Change in Temperature :Endothermic
  • DH > 0 It is a positive number.
  • The reaction absorbs heat.
  • Think of heat as a reactant since you know an equation can be written with a heat term.

A + B + Heat  C + D

Lowering temperature push toward reactants.

  • Shifts to left.
  • Remember the steps to solving LeChatelier’s problems.
changes in temperature summary
Changes in Temperature: Summary
  • Place the heat term in the reaction.
  • Determine if heat is being added or taken away from the reaction.
    • If heat is being added to an exothermic reaction, the reaction shifts to the left.
    • If heat is being added to an endothermic reaction, the reaction shifts to the right.
    • If heat is being taken away from an exothermic reaction, the reaction shifts to the right
    • If heat is being taken away from an endothermic reaction, the reaction shifts to the left.