Chapter 2 Discussion Section Problems Key

Chapter 2 Discussion Section Problems Key - 4. a. First...

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Chapter 2 Discussion Section Problems Key 1. 2. a. S o is positive; one molecule became two smaller molecules that have greater freedom in motion b. S o is negative; two small molecules produced a single molecule that has less freedom of motion c. S o cannot be predicted; the number of molecules in products and reactants is the same 3. . a. When the [CH3Cl] is doubled, the rate doubles, which is 2 to the first power. The reaction is thus first order with respect to chloromethane b. When [-CN] is tripled, the reaction rate triples, which is 3 to the first power. Thus, the reaction is first order with respect to cyanide ion c. 1 st order + 1 st order = 2 nd order overall d. Rate = Kr[CH3Cl][-CN]
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Unformatted text preview: 4. a. First order: the exponent of [(CH3)3CCl] in the rate law is = 1 b. Zero order: [CH3OH] does not appear in the rate law (i.e. its exponent is zero) c. First order: the sum of the exponents in the rate law is: 0 + 1 = 1 Acids/Bases 1. 2. 3. 4. 5. Functional groups Alkane Properties 1. . a. n-Octane has a higher boiling point because linear molecules boil higher than branched molecules of the same molecular weight b. 2-methylnonane has a higher boiling point than n-heptane because it has a significantly higher MW c. n-nonane has a higher boiling point. Same reasoning as in (a) 2. Nomenclature Conformations Challenge Problems 1....
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This note was uploaded on 04/05/2012 for the course CHEM 140A taught by Professor Whiteshell during the Winter '04 term at UCSD.

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Chapter 2 Discussion Section Problems Key - 4. a. First...

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