I need a fast response to these questions. Will reward with best answer and alot
ID: 824581 • Letter: I
Question
I need a fast response to these questions. Will reward with best answer and alot of points.
Provide a copy of all of your data tables involved in the planning and experimental section of your experiment (examples are given below) (6 points)
Table 1.1: Volume of chemicals
Trial
Acetone (mL)
HCl (mL)
I2 (mL)
H2O (mL)
Total Volume (mL)
1
10
10
10
20
50
2
20
10
10
10
50
3
10
10
20
10
50
Table 1.2: Molarities
Trial
Acetone (mol/L)
I2 (mol/L)
1
0.8
2.36 x 10-4
2
1.6
2.36 x 10-4
3
0.8
4.72 x 10-4
Table 1.3: Determination of the Rate of Iodine
Trial
1st trial time (sec)
2nd trial time (sec)
Avg. Time (sec)
?[I2] (M)
Rate = ?[I2]/time (M/s)
1
41.95
57.72
49.85
-2.36 x 10-4
4.73 x 10-6
2
20.85
23.44
22.15
-2.36 x 10-4
1.07 x 10-5
3
97.91
88.69
93.3
-4.72 x 10-4
5.06 x 10-6
Analyze your data by completing the table below and answering the subsequent questions. Show all calculations. (5 points)
Table 1.4: Rate Relationship
Trial
Acetone (M)
Iodine (M)
Rate = ?[I2]/time (M/s)
1
0.8
2.36 x 10-4
4.73 x 10-6
2
1.6
2.36 x 10-4
1.07 x 10-5
3
0.8
4.72 x 10-4
5.06 x 10-6
a. What is the order for I2? ______________________
b. What is the order for acetone? ___________________
c. What is the rate constant (including units)? ____________________________
2) Use the rate law to predict the theoretical rate of reaction if the initial molarity of Iodine is 5.6 x 10-3 and the initial molarity of acetone is 0.95 M.
3) Why is it important to keep the total volume 50 ml? If more water had been introduced to one of the solutions (given a total of 60), would you expect the reaction rate to increase or decrease?
Trial
Acetone (mL)
HCl (mL)
I2 (mL)
H2O (mL)
Total Volume (mL)
1
10
10
10
20
50
2
20
10
10
10
50
3
10
10
20
10
50
Explanation / Answer
a) keep the acetone conc constant
Rate = k [ I2[ ^x [ acetone ] ^y
4.73 x 10-6 / 5.06 x 10-6 = ( 2.36 x 10-4 / 4.72 x 10-4 ) ^x
x= 0
so order with respect to iodine is 0
2) now keep the iodine conc constant
4.73 x 10-6 / 1.07 x 10-5 = ( 0.8 / 1.6 ) ^y
y = 1
so the order with respect to acetone is 1
c) so the rate equation is given by
rate = k [ acetone ]
4.73 x 10-6 = k x 0.8
k = 5.9125 x 10-6
rate constant is 5.9125 x 10-6
2) rate of reaction = 5.9125 x 10-6 x 0.95
rate of reaction = 5.61 x 10-6
3) the rate will decrease as the conc of acetone decreases if you add more water .
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