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CHEMISTRY LABORATORY
REPORT
EXPERIMENT 5: FACTORS AFFECTING
REACTION RATE
Group: 1
Section: 5 Date: 19/10/2019
Group members:
Seq.
Full name
Student ID
%
contribution
(total =
100%)
1
Trần Minh Duy
ITITIU18230
25%
2
Nguyễn Thị Hà
ITITIU18059
25%
Signature
Duyên
3
Lê Duy Anh
ITDSIU19002
25%
4
Trần Quỳnh Anh
ITDSIU18046
25%
Total score: ________/15__
Present well: + 1 point
Clear picture: + 1 point
Total = 15 points
The report must be typed and handed in together with the signed data sheet by the deadline
Score
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Introduction (1 pt)
- This experiment tells us the way to identify and observe the change of the
reversilbe reaction at quilibrium when applying stresses and explain those
changes by the Le Chatelier's Principle. When the rate of the forward/
reaction becomes equal to the rate of the backward reaction, the
reversilbe is at equilibrium. If a system at equilibrium is subjected to a
change of pressure, temperature of the concentration of the components,
there will a be a tendency for the net reaction in the direction that against
the effect of this change(Le Chatelier's Principle).
I.
II.
Experimental (1 pt)
(Describe what experiments you did in this report, write down each step of your
experiments)
A: Instruments:
Thirty test tubes
One 50mL graduated cylinder
One test tube rack
One stop watch
Three 150mL beakers
One stop watch
Buret, clamp and ring
One medicine dropper
Buret , clamp and ring
Stirring rod
B : Experimental Procedure:
Part 1: Effect pf concentration on reaction time
The solutions to be used are as follows:
Prepare solution A: 0.2M potassium iodide
Prepare solution B: 0.005M sodium thiosulfate. This solution also contains starch that
will act as an indicator to detect the presence of iodine.
Prepare solution C: 0.1M ammonium peroxydisulfate.
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In this reaction, solution B will be the limiting reagent. The system consists of those
reaction:
Reaction 1: 2I- + S2O82-  I2 + 2SO4Iodide ions + peroxydisulfate  iodine + sulfate ion
Reaction 2: I2 + 2S2O3-  2I- + S4O62Iodine + thiosulfate ion  iodide ion + tetrathionate
Reaction 1 is relatively slow. As the iodine is formed it’s quickly used in reaction 2,
which is relatively fast. The limiting reaction ( solution B ) is a source of the thiosulfate
ions. When solution B is used up, the excess iodine formed will react with starch to
form a deep blue solution.
In this experiment, we have vary the concentration of solution A and C. The
temperature will remain constant at room temperature.
Combine the solution in 11 different combinations. The procedure for each of the
reaction is the same.
-
-
Step 1: Label 11 test tubes #1-11
Step2 : place 5mL of solution B in each test tube.
Step 3: add solution A and C into the test tube with the volume as shown in the table
below. Begin to time via using a stop watch. Stir the solution with a clean stirring rod.
At the first sign of color, stop timing. Record the results.
Step 4: Calculations
For example: iodide ion : (10mL x 1.2mol?l) / 25mL = 0.08 mol/L
Peroxydisulfate: (10mL x 0.1 mol/L ) / 25mL = 0.04 mol/L
-
Step 5: make graph
Part 2: Effect of temperature on the reaction rate
The reaction rate of oxidation- reduction between potassium permanganate, KMnO4 and
oxalic acide – H2C2O4 – can be measured by observing the time elapsed for the purple color
of the permanganate ion – MnO4 – to disappear.
5H2C2O4 (aq) + 2KMNO4 (aq) + 3H2SO4  2MnSO4 –(aq) + K2SO4(aq) +
10CO2(g) + 8H2O
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Prepare the reaction system:
-
Pipet 1mL of 0.01M KMnO4 and 5mL of 3M H2SO4 into a clean test tube
Pipet 5mL of 0.33M H2C2O4 into a second clean test tube .
Observe the reaction at room temperature:
Pour the H2C2O4 solution into the KMnO4 solution, observe and record the time for the
purple color of the permanganate ion to disappear.
Observe the reaction at high temperature:
-
-
Place a second KMnO4- H2C2O4 pair of test tubes in warm water ( 500C ) bath until
thermal equilibrium is established. Pour the H2C2O4 solution into the KMnO4 solution,
mix well and return the reaction system to be warm water bath. Record the time for
the purple color to disappear.
Repeat the same procedure, but increase the temperature of the water bath to about
900C. Record the change.
Part 3: Effect of a catalyst on the reaction rate
Hydrogen peroxide is relatively, but readily decomposes in the presence of a catalyst. In this
part, we observe which reagent(s) act as a catalyst for the decomposition of hydrogen
peroxide.
2H2O2  2H2O +O2
-
III.
Label 8 test tube #1-8
Place 5mL of the 3% H2O2 solution into each of the 8 test tubes.
Add a “pinch” of each of the following reagents to separate test tubes.
Mix well and observe the change with the production of gas bubbles. Record each
reation rate as fast, slow, very slow, or none in data table
Results and discussion
1. EFFECT OF CONCENTRATION ON REACTION TIME :(5 pts)
Reaction 1:
2I-+ S2O82-
Reaction 2:
I2+ 2S2O32-→ I- +s4O62-
→ I2 +2SO42-
Calculate the initial concentrations of I- and S2O82- ions:
Mixture # 5:
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(1 pt)
[I-] = (𝑽𝑰− ×𝑪𝑴 𝑰− )/Vtotal =(4ml×0.2mol)/25ml=0.032mol/L
[S2O82-] = (𝑽𝑺𝟐𝑶𝟖 𝟐− ×𝑪𝑴 𝑺𝟐 𝑶𝟖𝟐− )/Vtotal = (10ml×0.1mol)/25ml=0.04mol/L
Mixture
Iodide ion
Peroxydisulfate
Time in seconds
1
0.08
0.04
57
2
0.068
0.04
110
3
0.056
0.04
184
4
0.044
0.04
245
5
0.032
0.04
297
6
0.02
0.04
501
Mixture
Iodide ion
Peroxydisulfate
7
0.08
0.034
154
8
0.08
0.028
203
9
0.08
0.022
216
10
0.08
0.016
257
11
0.08
0.01
469
Time in seconds
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- Mixtures # 1-6: Graph (1 pt)
Plotting the concentration of iodide ion versus time: [Note: X – axis: time; Y – axis:
concentrations].
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The order of reaction with respect to iodide ion? (0.5 pts)
-
-
The order of reaction with respect to iodide ion followed: Mixture 1>2>3>4>5>6
Comments: (0.5 pts)
If a greater concentration of reactant atoms and molecules( concentration of Iodine
ion increases and concentration of peroxydislfate is constant) is present, the
greater chance for collisions will occur among them. More collisions mean a higher
reactions rate. Thus decreasing the concentration of Iodine in this reactions results
in lower reaction rate as well as the time for the reaction is longer.
- Mixtures # 1, 7, 8, 9, 10, and 11: Graph (1 pt)
The order of reaction with respect to peroxydisulfate ion? (0.5 pts)
-
The order of reaction with respect to peroxydisulfate ion followed:Mixture
1>7>8>9>10>11
Comments: (0.5 pts)
-
If a greater concentration of reactant atoms and molecules( concentration of
peroxydisulfate increases and concentration of Iodine ion is constant) is present,
the greater chance for collisions will occur among them. More collisions mean
higher reactions rate. Thus decreasing the concentration of peroxydisulfate in this
reactions results in lower reaction rate as well as the time for the reaction is longer.
2. EFFECT OF TEMPERATURE ON THE REACTION RATE :(2 pts)
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Reaction System:5H2C2O4(aq) + 2KMnO4(aq) + 3H2SO4 -> 2MnSO4(aq) + K2SO4(aq) +
10CO2(g) + 8H2O
Description of
conditions
Room
Predicted
outcome
Observation
Reaction
time
Explanation
The color
changes
14s
-At room temperature, the reaction
occurs with the normal conditions
The color
changes
7.4s
-When increasing temperature, the
atoms and molecules of reactants are
provided more energy so they move
faster and increase the collision
among them. Hence, the reaction rate
is higher and the less time is
consumed
The color
changes
0.05s
-And the opposite trend was true
when we decreasing the temperature.
temperature
500C
900C
Comment: (1pt)
The order of reaction: mixture 3>2>1
All reaction velocities increase with an increase in temperature. Higher temperature implies higher
average kinetic energy of molecules and more collisions per unit time, so the faster the reactants
particles move, the greater the chance they will collide and reaction with each other. Thus increasing
the temperature in this reactions results in higher reaction rate as well as the time for the reaction is
shorter.
3. EFFECT OF A CATALYST ON THE REACTION RATE: (3 pts)
Reaction System: 2H2O2(aq) -> 2 H2O + O2(g)
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Trial
Description of
conditions
Predicted
outcome
1
+ MnCl2
Slow
2
+ MnO2
Very fast
3
+ NaCl
4
Observation
(Reaction rate)
Explanation
Fast (1.5s)
-MnCl2 is not a catalyst for this
reaction
Very fast (1s)
MnO2 is a good catalyst for this
reaction
Slow
Slower than
MnCl2 (2s)
Nacl is not a catalyst for this
reaction
+ CaCl2
Slow
Faster than
MnCl2 (1.25s)
CaCl2 is not a catalyst for this
reaction
5
+ Zn
Slow
Slow
Zn is not a catalyst for this reaction
6
+ KNO3
Slow
Very slow
KNO3 is not a catalyst for this
reaction
7
+ Fe(NO3)3
Fast
Very fast (0.05s)
Fe(NO3)3 is a good catalyst for this
reaction
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The order of catalyst activity: (0.5 pts)
The order of catelyst activity: MnO2 > Fe(NO3)3 > CaCl2 >MnCl2 > Zn >KNO3> NaCl
Comments: (0.75 pts)
-
Catalyst plays role to initiates or accelerates a reaction without itself being affected
by reducing the amount of energy required to break and form a new chemical
bonds. If we use the right catalyst for the reaction, it will happens faster and less
energy is used.
IV.
Conclusions: (1 pt)
- After accessing with 3 sub-experiments above, we can know the method to examine
the effect of concentration, temperature, and catalyst on the reaction rates. Thus we
will able to maintain the reaction, speed up and slow down the reaction rate as well by
adding more concentration for the reactants or substances possibly react with the
components of the reactions, increasing or decreasing the temperature, and choosing
the suitable catalyst for the reactions.
The report must be typed and handed in together with the signed data sheet by the deadline