Consider the following reaction: CH 3 X + Y → CH 3 Y + X At 25°C, the following two experiments were run, yielding the following data: Experiment 1: [Y] 0 = 3.0 M [CH 3 X] (mol/L) Time(h) 7.08 × 10 −3 1.0 4.52 × 10 −3 1.5 2.23 × 10 −3 2.3 4.76 × 10 −4 4.0 8.44 × l0 −5 5.7 2.75 × l0 −5 7.0 Experiment 2: [Y] 0 = 4.5 M [CH 3 X] (mol/L) Time(h) 4.50 × 10 −3 0 1.70 × 10 −3 1.0 4.19 × 10 −4 2.5 1.11 × 10 −4 4.0 2.81 × l0 −5 5.5 Experiments also were run at 85°C. The value of the rate constant at 85°C was found to be 7.88 × 10 8 (with the time in units of hours), where [CH 3 X] 0 = 1.0 × 10 −2 M and [Y] 0 = 3.0 M. a. Determine the rate law and the value of k for this reaction at 25°C. b. Determine the half-life at 85°C. c. Determine E a for the reaction. d. Given that the C8X bond energy is known to be about 325 kJ/mol, suggest a mechanism that explains the results in parts a and c.
Consider the following reaction: CH 3 X + Y → CH 3 Y + X At 25°C, the following two experiments were run, yielding the following data: Experiment 1: [Y] 0 = 3.0 M [CH 3 X] (mol/L) Time(h) 7.08 × 10 −3 1.0 4.52 × 10 −3 1.5 2.23 × 10 −3 2.3 4.76 × 10 −4 4.0 8.44 × l0 −5 5.7 2.75 × l0 −5 7.0 Experiment 2: [Y] 0 = 4.5 M [CH 3 X] (mol/L) Time(h) 4.50 × 10 −3 0 1.70 × 10 −3 1.0 4.19 × 10 −4 2.5 1.11 × 10 −4 4.0 2.81 × l0 −5 5.5 Experiments also were run at 85°C. The value of the rate constant at 85°C was found to be 7.88 × 10 8 (with the time in units of hours), where [CH 3 X] 0 = 1.0 × 10 −2 M and [Y] 0 = 3.0 M. a. Determine the rate law and the value of k for this reaction at 25°C. b. Determine the half-life at 85°C. c. Determine E a for the reaction. d. Given that the C8X bond energy is known to be about 325 kJ/mol, suggest a mechanism that explains the results in parts a and c.
Solution Summary: The author explains that the rate law for the given reaction is to be determined by the use of the concentration verses time data.
At 25°C, the following two experiments were run, yielding the following data:
Experiment 1: [Y]0 = 3.0 M
[CH3X] (mol/L)
Time(h)
7.08 × 10−3
1.0
4.52 × 10−3
1.5
2.23 × 10−3
2.3
4.76 × 10−4
4.0
8.44 × l0−5
5.7
2.75 × l0−5
7.0
Experiment 2: [Y]0 = 4.5 M
[CH3X] (mol/L)
Time(h)
4.50 × 10−3
0
1.70 × 10−3
1.0
4.19 × 10−4
2.5
1.11 × 10−4
4.0
2.81 × l0−5
5.5
Experiments also were run at 85°C. The value of the rate constant at 85°C was found to be 7.88 × 108 (with the time in units of hours), where [CH3X]0 = 1.0 × 10−2M and [Y]0 = 3.0 M.
a. Determine the rate law and the value of k for this reaction at 25°C.
b. Determine the half-life at 85°C.
c. Determine Ea for the reaction.
d. Given that the C8X bond energy is known to be about 325 kJ/mol, suggest a mechanism that explains the results in parts a and c.
8. (16 pts) Provide the stepwise mechanism for the synthesis of the following compound via an enamine
Draw the titration curve of (i) weak acid vs. strong base; (ii) weak acid vs. weakbase; (iii) diprotic acid with strong base (iii) triprotic acid with strong base.
Complete the reaction in the drawing area below by adding the major products to the right-hand side.
If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead.
Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule
to represent each pair of enantiomers, using line bonds at the chiral center.
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No reaction.
my
ㄖˋ
+
1. Na O Me
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drawing a structure.
2. H
+
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