chemical equilibrium
DESCRIPTION
learn the ion equilibrium in the solution, chemical equilibrium, le catelier principal, basic chemistry, kimia dasar 2TRANSCRIPT
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[ ] [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.
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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
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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:
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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.
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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.
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c). Procedur :
1. Equilibrium of Ferrum (III) Tyosinate
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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
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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.
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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
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4.
5.
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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
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3KSCN (aq)❑ +Fe(NO¿¿3❑) →Fe (SCN ) +3KNO3 (aq)
❑3(aq)
❑3❑ ¿
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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
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The Equilibrium of Sodium Dichromate
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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.
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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
)
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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
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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.
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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
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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.
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Heating: reaction shift to the reactant (endothermic)
Cooling: productrection shift to the product (exothermic)
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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.
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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
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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).
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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.
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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
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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.
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[ ] [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.
[ | INTERNATIONAL CHEMISTRY EDUCATION 2012 23