Chemical Equilibrium

[ ] [III Group]

A. TITLE : Chemical Equilibrium

B. DATE : March, 26th 2013

C. PURPOSE : To learn the ion equilibrium in the solution

D. BASIC THEORY :

Factors That Affect Chemical Equilibrium

Chemical equilibrium is a very delicate system that represents a perfect

balance between forward and reverse reaction. A small disturb in the equilibrium

may shift the equilibrium position either to right forming more products or to left

forming more reactants. This reaction by the system is of course temporary and

eventually the system will come back to equilibrium. This phenomenon can be

expressed in the form of Le Chatelier’s Principle.

Le Chatelier’s Principle

An important and very interesting qualitative principle governing the

equilibrium is the principle of Le Chatelier. This principle, which is named after

the French chemist Henry Louis Le Chatelier (1850-1936), may be stated as

follows: if an external stress is applied to a system at equilibrium, the system will

tend to react in such way as to relieve the applied stress and tries to reestablish the

equilibrium. In chemical reaction terminology, the “stress” means change in

concentration, pressure, volume or temperature. Le Chatelier’s principle can be

understood either qualitatively or quantitatively doing some problems. However,

we restrict ourselves only to qualitative explanation.

Change in Concentration

Consider the following equilibrium reaction

N2 + 3H2 2NH3

If we add either N2 or H2, we increase the collisions between N2 and H2 there

by forming more product NH3. This is how the equilibrium counteracts the

applied stress; we say the equilibrium shifts from left to right.

[ | INTERNATIONAL CHEMISTRY EDUCATION 2012 1

[ ] [III Group]

If we add more NH3, we increase the concentration of NH3. As a result, some

NH3 decomposes and forms more reactants. We say the equilibrium shifts from

right to left.

If we remove either N2 or H2, now there is less concentration of N2 and H2.

To offset the applied stress, the equilibrium shifts from right to left. If we remove

some NH3, the equilibrium shifts from left to right to counteract the applied

stress.

Changes in Pressure and Volume

Pressure haven’t any effect on concentrations of species that are present in

solid, liquid or solution form. On the other hand, the change in pressure affects

the concentrations of gases. According to ideal gas law, pressure and volume are

inversely proportional to each other; the greater the pressure, the smaller the

volume, and vice versa. Thus, it is just enough to understand the affect of change

in pressure on the equilibrium system.

Let us consider the following equilibrium reaction:

N2 + 3H2 2NH3

1 mol+ 3 mol 2 mol

1 vol + 3 vol 2 vol

In order to understand the affect of pressure, it is very important to understand

the total volume of reactants and products. According to Avogadro’s law, the

number of moles is directory proportional to the volume of the gas. Think that

one mole is one volume.

Therefore, there are 4 volumes (1 vol + 3 vol) on the reactant side and 2

volumes on the product side. The increase in pressure always affects the side that

has more volume. Hence, increase in pressure shifts the equilibrium from left to

right. The pressure has no effect if the total volume of reactants is equal to the

total volume of the products as in the following example.

H2 + I2 2HI


[ ] [III Group]

Changes in Temperature

A change in concentration, pressure or volume alters the position of the

equilibrium but not the magnitude (value) of the equilibrium constant. However,

the change in temperature changes the value of the equilibrium constant. To

understand the effect of temperature, we must know whether the reaction is

endothermic (absorption of heat) or exothermic( release of heat). Let us consider

the equilibrium reaction between dinitrogen tetroxide and nitrogen dioxide:

N2O4 2NO2

This reaction is endothermic in the forward direction and exothermic in the

reverse direction:

N2O4 2NO2 ΔH0 = 58.0 kJ/mol

2NO2 N2O4 ΔH0 = -58.0 kJ/mol

To understand the effect of temperature (heat), let us re-write the above equations

treating the heat as chemical reagent. Thus :

heat + N2O4 2NO2

2NO2 N2O4 + heat

Therefore, increase in temperature favors the endothermic reaction (forward

reaction, i.e. left to right) while decrease in temperature favors the exothermic

reaction (reverse reaction, i.e., right to left). What does it mean? It means that the

value of the equilibrium constant increases when the heat is added (increase in

temperature) and decreases when the heat is removed (cooling the system) that

can be explained by the following equilibrium constant expression:


[ ] [III Group]

E. DESIGN OF EXPERIMENT :a) Tools and Materials :

Tools :o Beaker glass 50 mL : 1 piece

o Test tube : 7 piece

o Rack : 1 piece

o Pipette

o Measurement glass 10 mL : 1 piece

Materials :o Fe(NO3)3 0,1 M

o K2Cr2O7

o NH4OH 0,5 M

o NaH2PO4

o KSCN

o NaOH

o NaNO3 0,1 M

o Pb(NO3)2 0,2 M

o NH4Cl 0,5 M

o H2SO4 concentrated/0,1 M

o MgCl2 0,2 M

o FeSO4

b) Experiment Step :

a. Equilibrium of Ferrum (III) Tyosinate

1. Added 5 mL KSCN 0,002 M into chemical glass, and then

added 2 drop Fe(NO3)3 0,1 M, shaked until flat.

2. Over that solution, apportionment into 4 test tube, that is :

Tube 1 : save as comparator

Tube 2 : adding 3 drop KSCN 1 M

Tube 3 : adding 3 drop Fe(NO3)3 0,1 M

Tube 4 : adding one small granule NaH2PO4

Observe and note all of change and write occur of reaction.


[ ] [III Group]


[ ] [III Group]

b. The Equilibrium of Sodium Dichromate

1. 2 test tube (I and II) each stuffed with 1 mL K2Cr2O7 0,1

M and save tube I for comparator.

2. Into test tube II added NaOH 0,5 M until there is occur the

change (account amount drops adding NAOH).

3. Into test tube II added HCL 0,5 M amount drops same

with NaOH (on procedure 1 b).

Is the color as previously?

4. Noted all observe and write occur of reaction

c. Experiment III

1. Into 1 mL solution MgCl2 0,2 M added 1 mL NH4OH 0,5

M, noted change occur

2. Into 1 mL solution of MgCl2 added 1 mL solution of

NaOH, then added 1 mL solution of NH4Cl. Noted change

and compare the result with result on step 2a!

d. Experiment IV

1. Into 1 mL solution NaNO3 added 5 drop of H2SO4 and 5

drop FeSO4 concentrate

2. Dropping in the wall of test tube 1 mL H2SO4 concentrate,

observe the change!

e. Experiment V

1. Added 1 mL Pb(NO3)3 0,5 M into test tube, then added

some drops H2SO4 1 M and some drops alcohol until there

is precipitate.

2. Heated precipitate until the precipitate dissolved.

Cooling down and observe what the precipitate dissolved

can form return.


[ ] [III Group]

c). Procedur :

1. Equilibrium of Ferrum (III) Tyosinate


5 ml of KSCN 0,002 M

Poured into beaker glass

Added by 2 drops of Fe(NO3)3 0,1 M

Mixed Distributed into 4 test

tube

A

Saved as comparator

B

Added by 3 drops of KSCN 1M

Noted the changing

C

Added by 3 drops of Fe(NO3)3

Noted the changing

D

Added by grams of NaH2PO4

Noted the changing

JINGGA ORANGE (++) ORANGE(+++)/BLACKISH RED

SMOOTH ORANGE

Compared

1 ml of K2Cr2O7 0,1 M

Some drops of NaOH 0,5 MCounted theamount of dropping NaOHObserved the color changingSome drops of HClAmount of dropping HCl as many as NaOH dropped

Orange solution1 ml of K2Cr2O7 0,1 M

1 ml of MgCl2 0,2 M 1 ml of MgCl2 0,2 M

Added by 1 ml of NH4OH 0,5 MNoted the changing

Added by 1 ml of NH4OH 0,5 MAdded by 1 ml of NH4Cl Noted the changing

Precipitate solution, like a gel.Colorless solution

COMPARED

[ ] [III Group]

2. The Equilibrium of Sodium Dichromate

3.


1 ml of NaNO3

Added by 5 drops of H2SO4Added by 5 drops of FeSO4 Added by 1 ml of H2SO4 concentrated (added in the wall of test tube drop by drop)Noted the changing

Colorless solution.Forming of brown ring.

1 ml of Pb(NO3)2 0,2 M

Poured into test tubeAdded some drops of H2SO4 0,1 MAdded some drops of C2H5OH until there is precipitateHeated until the precipitate dissolvedCooled downNoted the changing

White precipitate solution

[ ] [III Group]

4.

5.


[ ] [III Group]

F. EXPERIMENT RESULT :

NO

PROCEDUR RESULT REACTION CONCLUSIONBEFORE AFTER

1. Equilibrium of Ferrum (III) Tyosinate

The color of:

KSCN :Colorless

Add Fe(NO3)3 : Orange (+)

Add KSCN: Brown

Add Fe(NO3)3 : Orange (+++)

Add NaH2PO4

: Little more pure than orange/ smooth orange

If added

concentration in

reaction

equilibrium will

moving to

product


3KSCN (aq)❑ +Fe(NO¿¿3❑) →Fe (SCN ) +3KNO3 (aq)

❑3(aq)

❑3❑ ¿

[ ] [III Group]

2.

3.

K2Cr2O7 : Orange (+)

Mg Cl2 :Colorless

Add NaOH :Yellow Bright

Add HCl :Back to Orange

Add NH4OH:Colorless

Add NH4Cl :Colorless

The equilibrium

move to the

reactant because

the addition of

volume.

K2Cr2O7 + NaOH :

in base

condition

And + HCl : in

acid condition

The equilibrium

moving to the

reactant

because the

addition of

concentration

NH4Cl in product


The Equilibrium of Sodium Dichromate

1 ml of NaNO3

Added by 5 drops of H2SO4Added by 5 drops of FeSO4 Added by 1 ml of H2SO4 concentrated (added in the wall of test tube drop by drop)Noted the changing

Colorless solution.Forming brown ring.

White precipitate solution

1 ml of Pb(NO3)2 0,2 M

Poured into test tubeAdded some drops of H2SO4 0,1 MAdded some drops of C2H5OH until there is precipitateHeated until the precipitate dissolvedCooled downNoted the changing

[ ] [III Group]

NO PROCEDUR RESULT HYPOTESIS CONCLUSIONBEFORE AFTER

4.

5.

The color of :

NaNO3 :Colorless

Pb(NO3)2 :Colorless

Add H2SO4 :Colorless

Add FeSO4 :Colorless

Add H2SO4 :Colorless

Add H2SO4 : turbid and there is precipitate (+)

Add C2H5OH turbid and there is precipitate (++)

There is a

brown ring a

in the ground

of the solution,

but it should

be in the upper

of the solution.

Function of

C2H5OH for

decreasing

solubility of

Pb(NO3)2.

Increase

temperature

to the

endothermic

reaction.


NaN

O3(

aq) +

FeS

O4(

aq) +

H2S

O4(

aq) ➝

Fe(

SO

) 4(a

q) +

H2O

(l) +

N

O2(

aq) +

Na 2

SO

4(aq

)

Fe2+

+ N

O➝

[(F

e(N

O))

]3 br

own

ring

[(F

e(N

O))

]3+ a

re f

orm

ed

Pb(

NO

3)2(

aq)+

H2S

O4(

aq) ➝

2HN

O3(

aq) +

Pb(

SO

4)(s

)

[ ] [III Group]

G. DATA ANALYSIS :

For the first experiment, we react 5 ml of KSCN 0,002 M and 2 drops of

Fe(NO3)3. After the solution reacted, we distribute it to 4 test tube. In the first test

tube, saved as comparator solution. The color of comparator solution is orange(+).

The reaction is :

KSCN (aq) + Fe(NO3)3 (aq) KNO3 (aq) + Fe(SCN)3 (aq)

And the second test tube, we add 2 drops of KSCN 0,1 M, in this second test tube,

there is concentration adding of KSCN and it affect the equilibrium shift to the

reactant. The equilibrium shift to reactant because when we add KSCN in product,

the equilibrium is already reached the product equilibrium shifted. And there is a

color changing from orange to brown.

In the third test tube, we add 3 drops of Fe(NO3)3. In this test tube, there is volume

adding of Fe(NO3)3 that caused the equilibrium shift to the reactant. And the color

changes from orange (+) to orange (++).

In the last test tube, we add grain of NaH2PO4. The function of NaH2PO4 is

damage the equilibrium of Fe(SCN)3. The result, the solution become little more

pure from orange (+).

For the second experiment, we react 1 ml of K2Cr2O7 with some drops of

NaOH till the color of K2Cr2O7 is changing. We need 10 drops of NaOH till the

color changes from orange to yellow bright. After this, we add drops of HCl that

has same amount with NaOH, it is 10 drops. After that the color of solution

become orange almost the same with the first solution.

K2Cr2O7 (aq) + 2NaOH (aq) 2KCrO4(aq) + Na2CrO4(aq) + H2O(aq)

The third experiment, MgCl2 that it color is colorless reacted with NH4OH

colorless in first test tube and the result is keep colorless

MgCl2 (aq) + 2NH4OH (aq) Mg(OH)2 (aq) + 2NH4Cl(aq)

And the second test tube is reacted MgCl2 with NH4OH and NH4Cl and the result

is colorless.

Mg(OH)2(aq) + 2NH4Cl(aq) MgCl2(aq) + 2NH4OH(aq)

For the first tube, there is no adding of NH4Cl. NH4Cl just as the product of

reaction. This is cause Mg(OH)2 can not be dissolved at all. But in the second test


[ ] [III Group]

tube, there is adding of NH4Cl after reaction. Mg(OH)2 can be dissolved if there is

ammonium salt. So, the greater the amount of ammonium salt, the higher amount

of Mg(OH)2 that can be dissolved. This experiment is proved because the second

test tube is colorless solution which indicate that Mg(OH)2 has been dissolved.

The forth experiment we react 1 ml NaNO3 with 5 drops H2SO4 there is not

changing color, so the color is colorless and than we add it with 5 drops FeSO4.

After all the substances mixed, we added 10 drops of H2SO4 concentrated. To

drop it to test tube, the pipette sticks to the wall of test tube, and the result is in it

colorless solution there is something like a clump, where clump’s color is brown

2NO3- + 4H2SO4 + 6Fe2+ 6Fe3+ + 2NO + 4SO4

2- + H2O

2NaNO3(aq) + 6FeSO4(aq) + 4H2SO4(aq)3Fe2(SO)4(aq) + 4H2O(l) + 2NO2(aq) + Na2SO4(aq)

The fifth experiment, we react 1 ml of Pb(NO3)2 colorless with some drops

of H2SO4 and the result is the solution become turbid and there is precipitate (+)

Pb(NO3)2(aq) + H2SO4(aq) 2HNO3(aq) + Pb(SO4)(s)

And after that we drop some drops of C2H5OH. There result , there is precipitate

and precipitate is more. The next step, we heated the solution until the precipitate

disappear. And when we cooled it down, the precipitate is apparently appear.


[ ] [III Group]

H. DISCUSSION :

In our first experiment, Base on the theory Fe3+ + 3SCN- ↔ [Fe

(CNS)3]2+ with the color blackish red solution. But the result of our

experiment the color was bright orange (+). It can to be like this because the

concentration of the KSCN is so little (0.002 M) so the color was dull, so

the color of result solution did not match with the hypothesis caused by the

concentration of KSCN is too small that is 0,002 M. To make the color as

same as with the hypothesis we need KSCN 0.1 M

The second experiment K2Cr2O7 first color is orange reacted with NaOH

the color changed to be yellow bright solution and after we added again with HCl

with the same volume the color changed to be orange solution again, same with

the initial color of K2Cr2O7. The reason is

Cr2O2-7 + 2OH- 2 CrO2-

4 + H2O

Dichromate changed to be chromate so the color was changed to be yellow bright

solution

And when we added HCl to the result, we will get

2CrO2-4 + 2H+ Cr2O2-

7 + H2O or

2CrO2-4 + 2H+ 2HCrO-

4 Cr2O2-7 + H2O

Chromate changed to be dichromate, and the color of the solution turn back the

initial color, and because the reaction shift from the product to reactant.

The third experiment, in tube first base on the theory the MgCl2 added 1

ml NH4OH 0,5 M the result is white precipitate of Mg(OH)2 ), the result of our

experiment was keep colorless and there is not precipitate so the theory is not

prove, maybe because less careful my group in doing the experiment,

And for second tube MgCl2 added by 1 ml of NH4OH and than added by

NH4Cl base on the hypothesis the result is colorless solution. And in our

experiment the result was colorless solution, so the hypothesis is prove. It can to


[ ] [III Group]

be like this because when added NH4Cl the reaction shift to the reactant so it

make the solution change same the initial or reactant condition.

The fourth experiment Base on the theory after NaNO3 added by H2SO4 and

FeSO4 and after added H2SO4 concentrated with drop it to test tube, the pipette

sticks to the wall of test tube will form a brown ring

2NO3- + 4H2SO4 + 6Fe2+ 6Fe3+ + 2NO + 4SO4

2- + 4H2O

Fe2+ (aq) + NO(aq) [Fe(NO)]2+ (aq)

A brown ring will form at the zone of contact of the two liquids. The brown ring

is due to the formation of the [Fe(NO)]2+. In our experiment the result was

something like a clump, where clump’s color is brown that not perfect that was in

the corner not in center of the test tube. Maybe the mistook in this experiment was

when we added concentrated sulphuric acid to the solution in which the

concentration of sulphuric acid not reacted in the center but in the corner and

because we who drop H2SO4 concentrated is not correct.

The fifth experiment base on the theory

Pb(NO3)2 + H2SO4 PbSO + HNO3 ∆H = -919.94 kJ/mole.

Pb(NO3)2 added by H2SO4 produce precipitate and after added by C2H5OH it

will be more precipitate, when heated it will be decrease the precipitate because

the reaction shift to the reactant (endothermic) and when it cooled it will increase

the precipitate because the reaction shift to the product (exothermic) . Base from

our experiment when we added Pb(NO3)2 + H2SO4 there was precipitate, when

we added by C2H5OH it be more precipitate. when heated it will be decrease the

precipitate and when cooled it increase the precipitate, so the theory is prove.


Heating: reaction shift to the reactant (endothermic)

Cooling: productrection shift to the product (exothermic)

[ ] [III Group]

I. CONCLUSSION :

Base on the experiment, we can conclude that :

Chemical Equilibrium is achieved when the rates of the forward and reverse

reactions are equal and the concentrations of the reactants and products

remain constant.

Base from our experiment addition concentration of the product should shift

the equilibrium towards formation of reactant, forward direction so as to

reduce the amount of product, similarly addition of the reactants should shift

the equilibrium towards formation of products.

Increase of temperature should shift the reaction in the direction of absorbing

the heat the backward direction. Similarly, decrease in temperature will shift

the equilibrium in the forward direction. It can analys base from the changed

condition.


[ ] [III Group]

J. ANSWER of QUESTION :

QUESTION

1. Assuming equilibrium for the reaction system:

H2 + I2 2HI

if 23 grams I2 and 0.5 grams of H2 heated at 450℃ until equilibrium is reached, determine if the heavy weight of I2 early - first calculate the concentration of 8.95 grams of HI and H2 in the mixture if the system volume of 1 liter!

2. Where the equilibrium system will shift direction if :a. Volume incrasedb. System temperature is raised

ANSWER

1. Known : mass of I2 = 23 grams Mr of I2 = 53 x 2 = 106 mass of H2 = 0,5 grams Mr of H2 = 1 x 2 = 2

Mole H2 : n = massmr

= 0,52

= 0,25

Mole I2 : n = massmr

= 23

106 = 0,22

H2 + I2 2HI M : 0,25 0,22 - B : 0,22 0,22 0,44

S : 0,03 - 0,44

[HI ] = 0,44

1 = 0,44 M

[H 2 ] = 0,03

1 = 0,03 M


[ ] [III Group]

2. a). If the pressure is reduced = the volume is increase, the equilibrium will

shift toward the large reaction coefficient.

b). When the system equilibrium temperature is raised, then the

equilibrium will shift towards the need of heat (endothermic reaction

direction).


[ ] [III Group]

BIBLIOGRAPY :

Tim Kimia Dasar.2013.Petunjuk Praktikum Kima Lanjut.Unesa:Unipress.

Brady, James E. 1990. General Chemistry:Principle and Structure 5 thed. United

State of America.

Chang, Raymond.General Chemistry: The Essential Concept/Raymond Chang-

3rded.America:Von Hoffman Press, Inc.

Hein, Morris and Susan Arena. 2007. Foundation of Collage Chemistry. Twelfth

Edition. USA: John Wiley&Sons, Inc.

Svehla,G. 1979. Vogel:Longman Group Limited. London.


[ ] [III Group]

ATTACHEMENT

Picture Explanation

5 mL KSCN after added by 2 drops of Fe(NO3)3 and the color become jingga.

KSCN+ Fe(NO3)3 divided into same drops and move to reaction tube.

The result of first experiment after adding by 3 drops of KSCN, Fe(NO3)3, NH4OH.

The Second experiment the color of K2Cr2O7

After added by NaOH

The Second experiment the color of K2Cr2O7 after added by NaOH + HCl the color is back to orange


[ ] [III Group]

The Second experiment the comparison color between K2Cr2O7 and K2Cr2O7 after added by NaOH + HCl.

The result of third experiment the solution still colorless.

The Fourth experiment the result after added NaNO3 and H2SO4.

The Fourth experiment the result after added NaNO3 +H2SO4 + FeSO4.

The Fourth experiment the result after added NaNO3 +H2SO4 + FeSO4 and concentrate H2SO4 there is a brown ring in the ground of the solution.


[ ] [III Group]

The fifth experiment after added Pb(NO3)2 + H2SO4 there is a lot of precipitate on the solution and make the solution become turbid.

After heated the solution the precipitate became less than before we heated and the precipitate only in the groud od the solution n the upper solution become clear, not turbid anymore.


Chemical Equilibrium

Documents

equilibrium system

ion equilibrium

equilibrium position

affect of pressure

volume pressure havent

applied stress

concentration of n2

chateliers principle