Factors Affecting Reaction Rates Lab Report Dr. Z Edits (1)
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Factors Affecting Reaction Rates
Date: 11/19/2023
Data
Activity 1
Data Table 1: Calibration Trial
Na
2
S
2
O
3
(drops)
Reaction time (sec)
1
8 drops
29 seconds 2
6 drops 21 seconds 3
N/A
N/A
Question 1:
How many drops will be used in the remaining experiments? 6 drops of Na
2
S
2
O
3 will be used in the remaining experiments.
Activity 2
Data Table 2a. Effects of KI (I
-
) Concentration Trial
KI
(drops)
HCl
(drops)
Starch
(drops)
H
2
O
(drops)
Na
2
S
2
O
3
(drops)
H
2
O
2
(mL)
Time
1
(sec)
Time
2
(sec)
Average
time
(T
avg
)
(sec)
Average
Rate
(1/T
avg
)
1
8
2
4
0
6
0.4
24
(24 +
22)/2 = 23 1/23=
.043
2
8
2
4
0
6
0.4
22
3
6
2
4
2
6
0.4
33
(33
+36)/2
=34.5
1/34.5=
.029
Trial
KI
(drops)
HCl
(drops)
Starch
(drops)
H
2
O
(drops)
Na
2
S
2
O
3
(drops)
H
2
O
2
(mL)
Time
1
(sec)
Time
2
(sec)
Average
time
(T
avg
)
(sec)
Average
Rate
(1/T
avg
)
4
6
2
4
2
6
0.4
36
5
4
2
4
4
6
0.4
41
(41 +
39)/2 = 40
1/40=
.025
6
4
2
4
4
6
0.4
39
7
2
2
4
6
6
0.4
87
(87+83)/2
= 85
1/85=
.012
8
2
2
4
6
6
0.4
83
Data Table 2b. Effects of H
2
O
2
Concentration Trial
KI
(drops)
HCl
(drops)
Starch
(drops)
H
2
O
(mL)
Na
2
S
2
O
3
(drops)
H
2
O
2
(mL)
Time
1
(sec)
Time
2
(sec)
Average time
(T
avg
) (sec)
Average
Rate
(1/T
avg
)
1
8
2
4
0
6
0.4
24
(24 + 21)/2=
22.5
1/22.5=
.044
2
8
2
4
0
6
0.4
21
3
8
2
4
0.1
6
0.3
32
(32+30)/2=31
1/31=
.032
4
8
2
4
0.1
6
0.3
30
5
8
2
4
0.2
6
0.2
42
(42 + 48)/2 =
45
1/45=
.022
6
8
2
4
0.2
6
0.2
48
7
8
2
4
0.3
6
0.1
30
(30 + 27)/2
=28.5
1/28.5=
.035
8
8
2
4
0.3
6
0.1
27
©2016 2 Carolina Biological Supply Company
Data Table 2c. Effects of HCl (H
+
) Concentration
Trial
KI
(drops)
HCl
(%)
Starch
(drops)
H
2
O
(drops)
Na
2
S
2
O
3
(drops)
H
2
O
2
(mL)
Time
1
(sec)
Time
2
(sec)
Average
time
(T
avg
) (sec)
Average
Rate
(1/T
avg
)
1
8
100%
4
0
6
0.4
20
(20+25)/2=
22.5
1/22.5=
.044
2
8
100%
4
0
6
0.4
25
3
8
75%
4
0
6
0.4
24
(24+22)/2=
23
1/23=
.043
4
8
75%
4
0
6
0.4
22
5
8
50%
4
0
6
0.4
25
(25+28)/2=
26.5
1/26.5=
.038
6
8
50%
4
0
6
0.4
28
7
8
25%
4
0
6
0.4
22
(22 + 26)/2
=24
1/24 =
.042
8
8
25%
4
0
6
0.4
26
Activity 3
Data Table 3. Temperature Effects
Water bath trial
Temperature of the water
bath (°C)
Reaction time (sec)
Cold water
3
25 seconds Room-temperature water
20
18 seconds Hot water
56
6 seconds Question 2:
Explain how each of these treatments affected the reaction rate. Describe the effect at a molecular level.
a. Concentration b. Temperature ©2016 2 Carolina Biological Supply Company
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Activity 4
Data Table 4. Catalyst Trial
Water (drops)
Diluted CuSO
4
Solution (drops)
Reaction time (sec)
1
4
0
22 seconds 2
3
1
10 seconds 3
2
2
9 seconds 4
1
3
8 seconds 5
0
4
6 seconds Question 3:
How did the addition of copper (II) sulfate affect the reaction rate? Did the amount of catalyst affect the reaction rate?
The addition of copper (II) sulfate increased the reaction rate significantly. The greater the amount of catalyst added (paired with a decrease in the amount of water added each trial), the lower the reaction time (in seconds) was each trial. Activity 5: Data Table 5. Orders of Reactants in the Rate Law
Determine the actual Rate Law: R = k[I
–
]
m
[H
2
O
2
]
n
[H
+
]
p
Calculated Reaction order (X) (e.g. (
concentration
1
concentration
2
)
X
=
rate
1
rate
2
)
You must show your full calculation work for credit. ©2016 2 Carolina Biological Supply Company
Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
KI (I
)
8 drops
.043
(1.369
+ .366 +
1.059)/2 =
.931
m = 1
1.369
KI (I
)
6 drops
.029
.366
KI (I
)
4 drops
.025
1.059
KI (I
)
2 drops
.012
You must show your full calculation work for credit. Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
H
2
O
2
0.4 mL
.044
(.831
+ .924
-.544)/3
=
.404
n = 0
.831
H
2
O
2
0.3 mL
.032
©2016 2 Carolina Biological Supply Company
Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
.924
H
2
O
2
0.2 mL
.022
-.544
H
2
O
2
0.1 mL
.035
You must show your full calculation work for credit. Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
HCl (H
+
)
100%
.044
(.08 + .305
- .144)/3 =
.08
p = 0
.08 HCl (H
+
)
75%
.043
.305
HCl (H
+
)
50%
.038
-.144
HCl (H
+
)
25%
.042
©2016 2 Carolina Biological Supply Company
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Rate Law
R = k
[I–]^1 [H
2
O
2
]^0 [H ]^0
Overall Order of the Reaction
1 + 0 + 0 = 1 Question 4:
Use the rate-law expression you
determined for Data Table 5 to answer the following questions. Answer using doubled, tripled, quadrupled, halved, quartered, no change, etc. A)
If the concentration of I
–
was doubled, how would that affect the reaction rate?
Double.
B)
If the concentration of H
2
O
2
was halved, how would that affect the reaction rate?
No change. C)
If the concentrations of I
–
and H
2
O
2
were both doubled, how would that affect the reaction rate?
It would double due to the concentration of I-, but the reaction rate would not be impacted by the concentration of H2O2.
D)
If the concentration of H
+
(HCl) was doubled, how would that affect the reaction rate?
No change. ©2016 2 Carolina Biological Supply Company
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Review | Constants | Periodic Table
The half-life of a reaction, t1/2. is the time it takes
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2. For the reaction A + B + C → products, the following data were collected:
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[A]. [B]. [C]
Experiment
2.06
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4.00
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3.05
4.00
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0.50 0.50
1.00
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1
2
3
4
Initial rate (mmol • L-1.s-1)
3.7
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The initial rate given in the table above is for the rate of loss of A.
(a) Write the rate law for the reaction.
(b) Determine the overall order of the reaction.
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- y courses / S22 CHEM 161_01_04 / 27 March - 2 April/ Quiz 9 Time left 0:53:13 Calculate the mass of oxygen that will be prepared through the decomposition of 50 kg of potassium chlorate ( KCIO3 , Molar mass= 122.55 g/mol) 2 KCIO3 →2 KCI + 3 O2 Select one: O a. 612 kg O b. 12.25 kg O c. 9.8 kg O d. 19.6 kg Next pagearrow_forward#2arrow_forwardUsing the following chemical equation and data: 3A + B + 2C --> D + 2E Experiment Initial [A] Initial [B] Initial [C] Initial Rate 1 1.0 x 10-2 4.0 x 10-3 2.0 x 10-2 5 mM/s 2 1.0 x 10-2 8.0 x 10-3 2.0 x 10-2 10 mM/s 3 2.0 x 10-2 8.0 x 10-3 2.0 x 10-2 40 mM/s 4 2.0 x 10-2 8.0 x 10-3 1.0 x 10-2 40 mM/s Determine the initial rate of reaction if the initial concentrations of A, B, C are 3 x 10-2 M, 5 x 10-3 M, 5 x 10-5 M, respectively, for the reaction.arrow_forward
- Using the following chemical equation and data: 3A + B + 2C --> D + 2E Experiment Initial [A] Initial [B] Initial [C] Initial Rate 1 1.0 x 10-2 4.0 x 10-3 2.0 x 10-2 5 mM/s 2 1.0 x 10-2 8.0 x 10-3 2.0 x 10-2 10 mM/s 3 2.0 x 10-2 8.0 x 10-3 2.0 x 10-2 40 mM/s 4 2.0 x 10-2 8.0 x 10-3 1.0 x 10-2 40 mM/s Determine the rate equation.arrow_forwardUsing the following chemical equation and data: 3A + B + 2C --> D + 2E Experiment Initial [A] Initial [B] Initial [C] Initial Rate 1 1.0 x 10-2 4.0 x 10-3 2.0 x 10-2 5 mM/s 2 1.0 x 10-2 8.0 x 10-3 2.0 x 10-2 10 mM/s 3 2.0 x 10-2 8.0 x 10-3 2.0 x 10-2 40 mM/s 4 2.0 x 10-2 8.0 x 10-3 1.0 x 10-2 40 mM/s Calculate the specific rate constant (k).arrow_forwardCalculation Tables Table 4- Logarithms [l]o (M) -log[I]o -log(rate) [H2O2]o (M) 0.0100 -log[H2O2]o 2.00 Experiment # Rate (M/s) 1 2.63 x10-6 5.58 0.0200 1.70 5.18 x 106 0.0100 0.0400 0.0600 8.01 x 10-6 1.62 x 105 0.0100 4 0.0200 0.0600 3.21 x 105 0.0400 0.0600 Table 5- Rate Constants Experiment [H2OzJo(M) [H2O2]om Initial rate k # (M/s) 2.63 x10-6 1 0.0100 0.0200 2 0.0100 0.0400 5.18 x 10-6 0.0100 0.0600 8.01 x 106 4 0.0200 0.0600 1.62 x 105 5 0.0400 0.0600 3.21 x 105 Value of m (2 significant figures): Value of n (2 significant figures): verage value of k (3 significant figures):arrow_forward
- Analyze the following reaction mechanism: 1. H2O2 → H2O + O 2. O + CF2CI2 → CIO + CF2CI 3. CIO + O3 → CI + 202 4. Cl + CF2CI → CF2CI2 From the mechanism provided, write the overall balanced equation for this chemical reaction. Ignore phases. 4- 2. 2+ 3+ 4+ 1 2 3 4 6 8 04 O6 Do 1 3 8. (s) (1) (g)|(aq) + F CI H C Reset • x H2O || Delete 3.arrow_forwardPre-lab question #10-1: A reaction between substances A and B has been found to give the following data: 3 A + 2 B → 2 C + D [A], M [В], М Rate of Appearance of С, М/hr 1.0 x 10-2 1.0 0.300 x 10-6 1.0 x 10-2 3.0 8.10 x 10-6 2.0 x 10-2 3.0 3.24 x 10-5 2.0 x 10-2 1.0 1.20 х 10-6 3.0 x 10-2 3.0 7.30 x 10-5 Using the data above, determine the following: Pre-lab question #10-1: a) Order of the reaction with respect to A and B 1st order with respect to A, and 2nd order with respect to B 2nd order with respect to A, and 2nd order with respect to B 2nd order with respect to A, and 3rd order with respect to B 3rd order with respect to A, and 2nd order with respect to Barrow_forwardPls help ASAP. Pls show all work.arrow_forward
- What is m and narrow_forward# 3 E D C Energy diagrams for two reactions are shown. Macmillan Learning Energy (kJ/mol) e 150 100 50- What is the heat of reaction for Reaction A? AHrxnA = 4 FER 20 What is the activation energy for Reaction A? R F V Reaction progress Reaction A 5 T G Search or type URL G B tv 6 MacBook Pro Y Ⓒ H 201 7 N U S J kJ mol 00 8 M I Energy (kJ/mol) 150 T 100- NIMZAO ✪ 50 What is the heat of reaction for Reaction B? AHxnB = What is the activation energy for Reaction B? K ( - O 9 < O I * ) -C O Reaction progress Reaction B L command ») P . V = : P W ; 242) { option [ SEC e = ? O kJ mol 1 delete retuarrow_forwardLab 6: Chemical Reactions Post-Lab Questions Station V: Heating Cu with Atmospheric O₂ Chemical equation: Complete ionic equation: Net ionic equation: General reaction type: Station VI: Reacting CuSO4 Solution with Steel Wool (Fe)arrow_forward
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