CHEM:ATOM FOC 2E CL (TEXT)
2nd Edition
ISBN: 9780393284218
Author: Stacey Lowery Bretz, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss
Publisher: WW Norton & Co
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Don't used hand raiting don't used Ai solution
Don't used hand raiting and don't used Ai solution
13.84. Chlorine atoms react with methane, forming HCI
and CH3. The rate constant for the reaction is
6.0 × 107 M¹ s¹ at 298 K. When the experiment
was run at three other temperatures, the following data
were collected:
T (K)
k (M-1 s-1)
303
6.5 × 107
308
7.0 × 107
313
7.5 x 107
a. Calculate the values of the activation energy and the
frequency factor for the reaction.
b. What is the value of the rate constant in the lower
stratosphere, where T = 218 K?
Chapter 2 Solutions
CHEM:ATOM FOC 2E CL (TEXT)
Ch. 2 - Prob. 2.1VPCh. 2 - Prob. 2.2VPCh. 2 - Prob. 2.3VPCh. 2 - Prob. 2.4VPCh. 2 - Prob. 2.5VPCh. 2 - Prob. 2.6VPCh. 2 - Prob. 2.7VPCh. 2 - Prob. 2.8VPCh. 2 - Prob. 2.9VPCh. 2 - Prob. 2.10VP
Ch. 2 - Prob. 2.11VPCh. 2 - Prob. 2.12VPCh. 2 - Prob. 2.13QACh. 2 - Prob. 2.14QACh. 2 - Prob. 2.15QACh. 2 - Prob. 2.16QACh. 2 - Prob. 2.17QACh. 2 - Prob. 2.18QACh. 2 - Prob. 2.19QACh. 2 - Prob. 2.20QACh. 2 - Prob. 2.21QACh. 2 - Prob. 2.22QACh. 2 - Prob. 2.23QACh. 2 - Prob. 2.24QACh. 2 - Prob. 2.25QACh. 2 - Prob. 2.26QACh. 2 - Prob. 2.27QACh. 2 - Prob. 2.28QACh. 2 - Prob. 2.29QACh. 2 - Prob. 2.30QACh. 2 - Prob. 2.31QACh. 2 - Prob. 2.32QACh. 2 - Prob. 2.33QACh. 2 - Prob. 2.34QACh. 2 - Prob. 2.35QACh. 2 - Prob. 2.36QACh. 2 - Prob. 2.37QACh. 2 - Prob. 2.38QACh. 2 - Prob. 2.39QACh. 2 - Prob. 2.40QACh. 2 - Prob. 2.41QACh. 2 - Prob. 2.42QACh. 2 - Prob. 2.43QACh. 2 - Prob. 2.44QACh. 2 - Prob. 2.45QACh. 2 - Prob. 2.46QACh. 2 - Prob. 2.47QACh. 2 - Prob. 2.48QACh. 2 - Prob. 2.49QACh. 2 - Prob. 2.50QACh. 2 - Prob. 2.51QACh. 2 - Prob. 2.52QACh. 2 - Prob. 2.53QACh. 2 - Prob. 2.54QACh. 2 - Prob. 2.55QACh. 2 - Prob. 2.56QACh. 2 - Prob. 2.57QACh. 2 - Prob. 2.58QACh. 2 - Prob. 2.59QACh. 2 - Prob. 2.60QACh. 2 - Prob. 2.61QACh. 2 - Prob. 2.62QACh. 2 - Prob. 2.63QACh. 2 - Prob. 2.64QACh. 2 - Prob. 2.65QACh. 2 - Prob. 2.66QACh. 2 - Prob. 2.67QACh. 2 - Prob. 2.68QACh. 2 - Prob. 2.69QACh. 2 - Prob. 2.70QACh. 2 - Prob. 2.71QACh. 2 - Prob. 2.72QACh. 2 - Prob. 2.73QACh. 2 - Prob. 2.74QACh. 2 - Prob. 2.75QACh. 2 - Prob. 2.76QACh. 2 - Prob. 2.77QACh. 2 - Prob. 2.78QACh. 2 - Prob. 2.79QACh. 2 - Prob. 2.80QACh. 2 - Prob. 2.81QACh. 2 - Prob. 2.82QACh. 2 - Prob. 2.83QACh. 2 - Prob. 2.84QACh. 2 - Prob. 2.85QACh. 2 - Prob. 2.86QACh. 2 - Prob. 2.87QACh. 2 - Prob. 2.88QACh. 2 - Prob. 2.89QACh. 2 - Prob. 2.90QACh. 2 - Prob. 2.91QACh. 2 - Prob. 2.92QACh. 2 - Prob. 2.93QACh. 2 - Prob. 2.94QACh. 2 - Prob. 2.95QACh. 2 - Prob. 2.96QACh. 2 - Prob. 2.97QACh. 2 - Prob. 2.98QACh. 2 - Prob. 2.99QACh. 2 - Prob. 2.100QACh. 2 - Prob. 2.101QACh. 2 - Prob. 2.102QACh. 2 - Prob. 2.103QACh. 2 - Prob. 2.104QACh. 2 - Prob. 2.105QACh. 2 - Prob. 2.106QACh. 2 - Prob. 2.107QACh. 2 - Prob. 2.108QACh. 2 - Prob. 2.109QACh. 2 - Prob. 2.110QACh. 2 - Prob. 2.111QACh. 2 - Prob. 2.112QACh. 2 - Prob. 2.113QACh. 2 - Prob. 2.114QACh. 2 - Prob. 2.115QACh. 2 - Prob. 2.116QA
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- My Organic Chemistry textbook says about the formation of cyclic hemiacetals, "Such intramolecular reactions to form five- and six-membered rings are faster than the corresponding intermolecular reactions. The two reacting functional groups, in this case OH and C=O, are held in close proximity, increasing the probability of reaction."According to the book, the formation of cyclic hemiacetals occurs in acidic conditions. So my question is whether the carbonyl group in this reaction reacts first with the end alcohol on the same molecule or with the ethylene glycol. And, given the explanation in the book, if it reacts first with ethylene glycol before its own end alcohol, why would it? I don't need to know the final answer. I need to know WHY it would not undergo an intermolecular reaction prior to reacting with the ethylene glycol if that is the case. Please do not use an AI answer.arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forwardHighlight in red each acidic location on the organic molecule at left. Highlight in blue each basic location on the organic molecule at right. Note for advanced students: we mean acidic or basic in the Brønsted-Lowry sense only. Cl N شیخ x Garrow_forward
- Q4: Draw the mirror image of the following molecules. Are the molecules chiral? C/ F LL CI CH3 CI CH3 0 CI CH3 CI CH3 CH3arrow_forwardComplete combustion of a 0.6250 g sample of the unknown crystal with excess O2 produced 1.8546 g of CO2 and 0.5243 g of H2O. A separate analysis of a 0.8500 g sample of the blue crystal was found to produce 0.0465 g NH3. The molar mass of the substance was found to be about 310 g/mol. What is the molecular formula of the unknown crystal?arrow_forward4. C6H100 5 I peak 3 2 PPM Integration values: 1.79ppm (2), 4.43ppm (1.33) Ipeakarrow_forward
- Nonearrow_forward3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forward
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