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.

Eventually the rates are equal

Rea

ctio

n R

ate

Time

Forward Reaction

Reverse reaction

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 very slow.

LAW OF MASS ACTION

For any reaction jA + kB lC + mD K = [C]l[D]m PRODUCTSpower

[A]j[B]k REACTANTSpower

K is called the equilibrium constant.

is how we indicate a reversible reaction

PLAYING WITH K

If we write the reaction in reverse. lC + mD jA + kB Then the new equilibrium constant is

K’ = [A]j[B]k = 1/K

[C]l[D]m

PLAYING WITH K

If we multiply the equation by a constant njA + nkB nlC + 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

THE UNITS FOR K

Are determined by the various powers and units of concentrations.

They depend on the reaction. Are usually dropped, unlike rate constant

units which are specifically asked for.

K IS CONSTANT

At any temperature. Temperature affects rate. 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.

EQUILIBRIUM CONSTANT

UNIQUE FOR EACH TEMPERATURE

CALCULATE K

N2 + 3H2 2NH3

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 K

N2 + 3H2 2NH3

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

Some reactions are gaseous PV = nRT P = (n/V)RT P = CRT C is a concentration in moles/Liter C = P/RT

EQUILIBRIUM AND PRESSURE

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

Kp = (PSO3)2

(PSO2)2 (PO2)

K = [SO3]2

[SO2]2 [O2]

EQUILIBRIUM AND PRESSURE

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 jA + kB lC + mD

Kp= (PC)l (PD)m= (CCxRT)l (CDxRT)m (PA)j

(PB)k (CAxRT)j(CBxRT)k

Kp= (CC)l (CD)mx(RT)l+m

(CA)j(CB)kx(RT)j+k

Kp = K (RT)(l+m)-(j+k) = Kc (RT)n

n=(l+m)-(j+k)=change in moles of gas

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

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 EXAMPLE

H2(g) + I2(s) 2HI(g)

K = [HI]2 [H2][I2]

But the concentration of I2 does

not change. K[I2]= [HI]2 = K’

[H2]

SOLVING EQUILIBRIUM PROBLEMS

Type 1

THE REACTION QUOTIENT Tells you the directing 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 Q value to the equilibrium constant

WHAT Q TELLS USIf Q<K K > Q

Not enough productsShifts right

If Q>K K < QToo many productsShifts left

If Q=K system is at equilibrium

EXAMPLE

For the reaction

2NOCl(g) 2NO(g) + Cl2(g)

K = 1.55 x 10-5 M at 35ºC

In an experiment 10.0 mol NOCl, 1.0 mol NO(g)

and .10 mol Cl2 are mixed in 2.0 L flask.

Which direction will the reaction proceed to reach

equilibrium?

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.

Consider the following reaction at 600ºC 2SO2(g) + O2(g) 2SO3(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 SO3 were present. Calculate the equilibrium concentrations of O2 and SO2, K and KP

Initial

Change

Equilibrium

PROBLEMS WITH SMALL K

K< .01

PROCESS IS LIKE LAST YEAR’S ACID/BASE PROBLEMS

Set up table of initial, change, and final concentrations.

For a small K the product concentration is small.

FOR EXAMPLEnitrogen + oxygen yields nitrogen monoxide

K= 4.10 x 10-4

If .50 moles of nitrogen and .86 moles of oxygen are mixed in a 2.0 L container, what are the equilibrium concentrations of all 3

substance?

Q =

K =

Because K is small, assume x is negligible compared to .25 and .43

N2 O2 2NO

Initial

Change

Equilibrium

Simplified K expression:

K =

Always check the assumption.X must be less than 5% of the original

concentration.

LARGE K

K >100

WHAT IF YOU’RE NOT GIVEN EQUILIBRIUM CONCENTRATIONS?

The size of K will determine what approach to take.

First let’s look at the case of a LARGE value of K ( >100).

Allows us to make simplifying assumptions.

EXAMPLE

H2(g) + I2(g) 2HI(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 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

Q= 0<K so more product will be formed.

Assumption since K is large reaction will go to completion.

Stoichiometry tells us I2 is LR, it will be smallest

at equilibrium let it be x

Set up table of initial, change in and equilibrium concentrations.

Choose X so it is small. For I2 the change in X must be X-.232 M

Final must = initial + change

H2(g) I2(g) 2HI(g)

initial 1.56 M 0.232 M 0 M changeeq X

Initial + change = final, Change = final – initial

Using to stoichiometry we can find:

Change in H2 = = x -.232 MChange in HI = (2(.232-x) = (.464-2x)

H2(g) I2(g) 2HI(g) initial 1.56 M 0.232 M 0 M change X-0.232

eq X

Now we can determine the final concentrations by adding.

Now plug these values into the equilibrium expression

K = (0.464-2X)2 = 7.1 x 102 (1.328+X)(X) Now you can see why we made sure X was small.

H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 X-0.232 0.464-2X

eq 1.328+x X .464 – 2x

WHY WE CHOSE X K = (0.464-2X)2 = 7.1 x 102

(1.328+X)(X)

Since X is going to be small, we can ignore it in relation to 0.464 and 1.328

Avoid algebra nightmare and quadratic equation!

So we can rewrite the equation 7.1 x 102 = (0.464)2 (1.328)(X)X = 2.8 x 10-4 back to ICE table,

substitute into equilibrium line

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………..anyone have it

programmed already??????

Our assumption was valid.

PRACTICE

For the reaction Cl2 + O2 2ClO(g)

K = 156In 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?

Calculate the equilibrium concentrations of all species.

Cl2 O2 2 ClO

Initial

Change

equilibrium

MID-RANGE K’S

.01<K<100

QUADRATIC EQUATION TI-82 PROGRAM

Hit PRGM key

Using right arrow, highlight NEW and hit

ENTER

Give the program a name (i.e. QUAD) and hit

ENTER

Hit PRGM key

Using right arrow, highlight I/O

Highlight Input and hit ENTER

Hit ALPHA key and then hit the letter A, then

hit ENTER

Repeat for B and C

Hit PRGM key

Using right arrow, highlight I/O

Using down arrow, highlight Disp and hit

ENTER

Type in the following formula:

(-B+√(B^2-4*A*C))/(2*A) and hit ENTER

Repeat for the following: (-B-√(B^2-4*A*C))/(2*A) and hit ENTER

To execute the program: Hit 2nd and Quit to exit the editing of the program Hit PRGM key Highlight EXEC and use down arrow to

highlight QUAD and hit ENTER

PRGMQUAD should show up at the top of the screen –hit ENTER

ENTER first coefficient after ? (i.e. 1)ENTER second coefficient after ? (i.e. -5) ENTER constant term after ? (i.e. 6)

The answers of 3 and 2 DONE should show up in the lower right corner of the

screen.

NO SIMPLIFICATION Choose X to be small. Can’t simplify so we will have to solve the

quadratic

H2(g) + I2 (g) HI(g) K=38.6

What are 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?

H2 I2 2 HI

Initial

Change

equilibrium

PROBLEMS INVOLVING PRESSURE

Solved exactly the same, with same rules for choosing X depending on 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 2 NO2

Initial

Change

equilibrium

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.

3 Types of stress1. Changing amounts of reactants or products2. Changing pressure3. Changing temperature

1. 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

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

2. CHANGE PRESSURE

By changing volumeSystem will move in the direction

that has the least moles of gas.Because partial pressures (and

concentrations) change a new equilibrium must be reached.

System tries to minimize the moles of gas.

CHANGE IN PRESSURE

By adding an inert gasPartial pressures of reactants

and product are not changedNo effect on equilibrium

position

3. 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 of the shift depends on whether it is exothermic or endothermicThink about ∆G and its relation to K and

∆H.

EXOTHERMIC

H<0Releases heatThink of heat as a productRaising temperature pushes

toward reactants.Shifts to left.

ENDOTHERMIC

H>0requires heatThink of heat as a reactantRaising temperature pushes

toward products.Shifts to right.