Equil Notes Honors

8/10/2019 Equil Notes Honors

1/31

Chemical Equilibrium..

Until now, weve treated reactions as though they can only go

in one direction.with all of the reactants turning into

products

Most reactions that we have witnessed appear this way

Are all reactions like this?

Is it possible to reverse a reaction after it is completed?

Can we turn CO2and H2O back into gasoline and O2after it

burns?

2 C8H18+ 25 O2 16 CO2+ 18 H2O

16 CO2+ 18 H2O 2 C8H18+ 25 O2?
http://localhost/var/www/apps/conversion/tmp/scratch_10/gasoline.flvhttp://localhost/var/www/apps/conversion/tmp/scratch_10/gasoline.flv


2/31

Chemical equilibrium

Most chemical reactions are reversible!

Reactions will continue to occur until they finally

maintain a ratio of products versus reactants thatwill not change

This ratio is called the Equilibrium ConstantExpression, or Keq, or Kc, and is equal to the final

product concentration over the final reactantconcentration

K = Products

Reactants


3/31

Lets take the following chemical reaction:

aA + bB cC + dDWhen a chemical reaction reached equilibria, the

Equilibrium Constant Expression, or Keq, is:

Keq= [C]c[D]d

[A]a[B]b


4/31

Chemical Equilibrium

(continued)

A, B, C, and D represent the concentration of eachchemical at the end of the reaction

This concentration is measured in Molarity, or M

Once the reaction reaches equilibrium, the concentrationsof these molecules WILL NOT CHANGE

Keq= [C]c[D]d

[A]a[B]b


5/31

Lets examine the following

reaction

This reaction is reversible

At equilibrium, the constant, or Keq, is equal to:

Keq= [NO2]2

[N2O4]1

BUT WHAT DOES THIS TELL US?

N2O4(g) 2 NO2(g)at 250C


6/31

It tells us whether we have more products or

reactants at equilibrium!!!!Looking at the same reaction:

N2O4(g) 2 NO2(g) at 250C

If the concentration of N2O4(g)at equilibrium was .0445 M

and the concentration of NO2(g)was .0161 M, then:

Keq= [NO2]2 = [.0161]2 = .00582 = 582

[N2O4] [.0445] 1 100,000

This is the same as: 5.82

1000

Is this a exothermic

or endothermic

reaction.and why?


7/31

Suppose we have this reaction:

CO(g)+ H2O(g) H2(g)+ CO2(g)

The Keq= 5.10 at 5270C.

1

This means that products are favored over reactants almost 5to 1!!

Lets say we are at some point in time, NOT ATEQUILIBRIUM and we measure the concentrations of eachof the molecules in the reaction


8/31

Is it possible to predict if a reaction

is at equilibrium?CO(g)+ H2O(g) H2(g)+ CO2(g)

The Keq= 5.10 at 5270C.

The concentrations are [CO] = .15 M,

[H2O] = .25 M, [H2] = .42 M, and

[CO2] = .37 M.

Can you set up a ratio of concentrations?

DO IT NOW!!!!!


9/31

Does your ratio look like this?

Q = [.42]1[.37]1

[.15]1

[.25]1

This is called the REACTION

QUOTIENT, OR Q

It is set up the same way as Keq, but itis not Keq

Why is it not the same as Keq?


10/31

What does Q tell you???

The reaction quotient, or Q, tells you where you

are in the reaction. You compare it to your Keq!

Q = [.42]1[.37]1 = 4.14

[.15]1[.25]1 1

Q = 4.14, which is less than Keq= 5.10


11/31

Q = 4.14, which is less than Keq= 5.10

1 1

If Q < Keq, then the reaction will make more

products to reach equilibrium

If Q > Keq, then the reaction will turn productsback into reactants to reach equilibrium

It Q = Keq, then the reaction is..

AT EQUILIBRIUM!!!!!

That means it is possible for a reaction to reverse

itself and turn product back into reactant!


12/31

AT EQUILIBRIUM

- The forward reaction, where molecules A

and B are forming molecules C and D

A + B --------> C + D

- Occurs at the same rate as the reverse

reaction, the reaction where molecules C

and D are forming molecules A and B

C + D --------> A + B

That is how the ratio at equilibrium never changes!


13/31

Chemical Equilibrium does not

mean: That there are equal

concentrations of

reactant and productwhen the reaction

reaches equilibrium!

It simply means that

there is eventually aratio reached that will

not change!


14/31


Lets look at problem #12 in the homework.

The reaction of elemental hydrogen and fluorine to form

hydrofluoric acid has an equilibrium constant of 1.15 x 102at a

certain temperature. In one experiment, 3.00 mol of all 3

components, including the product, was added to a 1.50-liter

flask. Calculate the equilibrium concentrations of all species.

Write the reaction:

H2 (g)+ F2(g)--------> 2 HF (g)

This is an odd problem, because there is actually some of the

product present before the reaction actually begins! This is not

commonproduct isnt usually present until the reaction

happens!


15/31


We want to set up an ICE:

1 H2 (g)+ 1 F2(g)--------> 2 HF (g)

I

C

E

Remember, I mean initial, C means change, and E means end or

equilibrium!

And we use concentrationnot moles, or amount!

What are the initial concentrations of each chemical?


16/31


What does our ICE look like?

1 H2 (g)+ 1 F2(g)--------> 2 HF (g)

I 2M 2M 2M

C -x -x +2x

E 2 - x 2 - x 2 + 2x

Kc 115 = [HF]2 = [2 + 2x]2

1 [H2

]1[F2

]1 [2-x]1[2-x]1

Remember, we insert the concentrations into our K at

equilibrium!

We want to solve for the amounts at equilibriumthis is going

to require some algebra!


17/31


Kc 115 = [HF]2 = [2 + 2x]2

1 [H2]1[F2]

1 [2-x][2-x]

We have to expand this..

115 = 4 + 8x + 4x2

1 4 - 4x + x2

460460x + 115x2= 4 + 8x + 4x2

Rearranging,

0 = 456468x + 111x2 or 111x2468x + 456 = 0

How do we solve this?


18/31


111x2468x + 456 = 0

Use the quadratic formula!

You can use the program on your calculator to solve this

You can go to http://www.wolframalpha.com/and enter in the

equationit will solve it for you!

What are the solutions for x?

X = 1.5284 and 2.6878

One of these solutions cant work.

Which one? Why?
http://www.wolframalpha.com/http://www.wolframalpha.com/


19/31


H2 (g)+ F2(g)--------> 2 HF (g)

I 2M 2M 2M

C -x -x +2x

E 2 - x 2 - x 2 + 2x

X = 1.5284 and 2.6878

It is impossible to lose 2.6878 when you only start with 2!

This is the non-real solution

So x = 1.5284

What are the amounts at equilibrium then?


20/31


H2 (g)+ F2(g)--------> 2 HF (g)

I 2M 2M 2M

C -x -x +2x

E .4716 .4716 5.0568

Lets plug these values back into K to see if they equal 115/1!

Kc 115 = [HF]2 = [5.0568]2 114.975/1

1 [H2]1[F2]

1 [.4716]1[.4716]1


21/31

We can set up a ratio of pressures instead of

concentrations too! THIS IS MORE CONVENIENT

FOR GASES, BECAUSE PRESSURE IS EASIERTO MEASURE THAN CONCENTRATION!

This is called Kp

Lets look at number 24

Hydrogen gas and iodine vapor react to form hydrogen

iodide gas. The equilibrium constant, or Kc, is 1.00 x

102at 250C. Suppose hydrogen at 5.0 x 10-1atm,

iodine at 1.0 x 10-1 atm, and hydrogen iodide at 5.0 x

10-1atm are added to a 5.0 L flask. Calculate the

equilibrium pressures of all species.

What is the reaction?


22/31

Hydrogen gas and iodine vapor react to form hydrogen

iodide gas. The equilibrium constant, or Kc, is 1.00 x 102

at 250

C. Suppose hydrogen at 5.0 x 10-1

atm, iodine at 1.0x 10-1 atm, and hydrogen iodide at 5.0 x 10-1atm are added

to a 5.0 L flask. Calculate the equilibrium pressures of all

species.

H2 (g)+ I2 (g) --------> 2 HI (g) What would Kplook like?

Kp= [P HI]2

[P H2]1[P I2]

1

Where P is the pressure, instead of concentration, of the gas!But the problem gives us the Kc, and not the Kpare they the

same? No!


23/31

Kp= Kc(RT)Dn

What is delta n?

It is the moles of gas produced minus the moles of gas youstart with

In this reaction, you make 2 moles of HI, and start with one

mole of H2, and one mole of I2

So, Dn = 22 = 0

Kp= 100 [(.0821)(298)]0

Kp= 100

In this case, the Kp

actually did equal the Kc

!

So..


24/31

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

I .5 atm .1 atm .5 atm

C -x -x +2x

E .5x .1x .5 + 2x

Same exact setup as beforebut we are using pressures instead

of concentrations, so we use a Kpand not a Kc!

Kp= [.5+2x]2 = 100

[.5-x]1[.1-x]1

Can you do the algebra?


25/31

Remember, if concentrations are given, use

a Kcif pressures are given, use a K

p

To convert between them:

Kp= Kc(RT)Dn


26/31


27/31

Le Chateliers Principle:

If a system at equilibrium is disturbed by a

change in temperature, pressure, or the

concentration of one of the molecules, thesystem will shift its equilibrium position to

counteract the disturbance

WHAT DOES THAT MEAN.?


28/31

Fritz Haber was a German Scientist who utilized

LaChateliers Principle in making ammonia for

explosives in World War I:N2(g)+ 3H2(g) ------> 2 NH3(g)

Keq= 9.60

1

at 3000C and 200 atm

At equilibrium, products are

favored, but how do we get the

reaction to continueto make

NH3, or ammonia?


29/31

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

By adding more N2(g) or H2(g), the reaction

would..?

Shift to the right and make more ammonia - but

this requires more reactant and more money!

By adding more NH3(g), the reaction would.?

Shift to the left and form more N2(g)and H2(g),which is NOT what we want!


30/31


31/31

Kc, or Keq, only applies to chemicals reacting in the

solid, liquid, or gaseous state. Kponly applies to

gases. They do not apply to aqueous solutions

(reactions that occur in water).

Kspdeals with aqueous reactions, or reactions

that occur in water!

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Equil Notes Honors