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# 10_all_ - 10.1 Calculate the heat of reaction H for the following reactions(a H2 F2 2HF(b CH4 F2 CH3F HF(c CH4 Cl2 CH3Cl HCl(d CH4 Br2 CH3Br HBr(e

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10.1 Calculate the heat of reaction, H ° , for the following reactions: (a) H 2 +F 2 2HF (b) CH 4 +F 2 CH 3 F+HF (c) CH 4 +Cl 2 CH 3 Cl+HCl (d) CH 4 +Br 2 CH 3 Br+HBr (e) CH 4 +I 2 CH 3 I+HI (f) CH 3 CH 3 +Cl 2 CH 3 CH 2 Cl+HCl (g) CH 3 CH 2 CH 3 +Cl 2 CH 3 CHClCH 3 +HCl (h) (CH 3 ) 3 CH+Cl 2 (CH 3 )CCl+HCl Answer: HH FF Cl Cl Br Br II H 3 CH 2 CC l (H 3 C) 2 HC Cl (H 3 C) 3 l H 3 CH H 3 CH 2 (H 3 C) 2 HC H DH=435KJmol -1 DH=159KJmol DH=243KJmol DH=192KJmol DH=151KJmol (H 3 C) 3 H 3 CF DH=435KJmol DH=410KJmol DH=395KJmol DH=381KJmol DH=452KJmol H 3 l H 3 CB r H 3 CI DH=349KJmol -1 DH=293KJmol DH=234KJmol -1 DH=341KJmol DH=339KJmol DH=328KJmol (a) H = 435+159-2*569 = - 544 KJ / mol (b) H = 435+159-452-569 = - 427 KJ / mol (c) H = 435+243-349-431 = - 102 KJ / mol (d) H = 435+192-293-366 = - 32 KJ / mol (e) H = 435+151-234-297 = 55 KJ / mol (f) H = 410+243-341-431= - 119 KJ / mol (g) H = 395+243-339-431 = - 132 KJ / mol (h) H = 381+243-328-431 = - 135 KJ / mol 10.2. List the following radicals in order of decreasing stability: CH 3 ,(CH 3 ) 2 CHCH 2 ,(CH 3 ) 2 CCH 2 CH 3 ,CH 3 CH 2 CHCH 3 CH 3 < (CH 3 ) 2 CHCH 2 < CH 3 CH 2 CHCH 3 < (CH 3 ) 2 CCH 2 CH 3 10.3 Suggest a method for separating the mixture of CH 4 , CH 3 Cl, CH 2 Cl 2 , CHCl 3 , and CCl 4 that is formed when methane is chlorinated. (You may want to consult a handbook.) What analytical method could be used to separate this mixture and give structural information about each component? How would the molecular ion peaks in their respective mass spectra differ on basis of the number of chlorines (remember that chlorine has two predominant isotopes, 35 Cl and 37 Cl)? Answer: The compounds all have different boiling points. They could, therefore, be separated by careful fractional distillation. Or, because the compounds have different vapor pressures, they could easily be separated by Gas Chromatography. GC / MS could be used to separate the compounds as well as provide structural information from their mass spectra. Their mass spectra would show contributions

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from naturally occurring 35 Cl and 37 Cl isotopes. The natural abundance of 35 Cl is approximately 75% and that of 37 Cl is approximately 25%. Thus, for CH 3 Cl, containing only one chlorine atom, there will be an M + . peak and M+2 peak in roughly a 3:1 ratio of intensities. For CH 2 Cl 2 , there will be M + ., M + .+2, M + .+4 peaks in roughly 9:6:1 ratio. For CHCl 3 , there will be M + ., M + .+2, M + .+4, M + .+6 peaks in roughly 27:27:9:1 ratio. For CCl 4 , there will be M + ., M + .+2, M + .+4, M + .+6, M + .+8 peaks in roughly 81:108:54:12:1 ratio, respectively. 10.4 When methane is chlorinated, among the products found are traces of chloroethane. How is it formed? Of what significance is its formation? Answer: A small amount of ethane is formed by the combination of two methyl radicals: CH 3 2 H 3 CC H 3 This ethane then reacts with chlorine in a substitution reaction (see Section 10.6) to form chloroethane. The significance if this observation is that it is the evidence for the proposal that the combination of methyl radicals is one of the chain-terminating steps in the chlorination of methane.
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## This note was uploaded on 02/23/2010 for the course CHEM 101 taught by Professor Smith during the Spring '10 term at Duke.

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10_all_ - 10.1 Calculate the heat of reaction H for the following reactions(a H2 F2 2HF(b CH4 F2 CH3F HF(c CH4 Cl2 CH3Cl HCl(d CH4 Br2 CH3Br HBr(e

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