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Unformatted text preview: 6- 13 The rate law is first order in both 1¡bromobutane, C4H9Br, and methoxIde ion. ¢F the concentration of C4H9Br is Lowered to one-fifth £he original value, the rate must decrease to one¤FIF£h; if the concentratIon of methoXIde Is doubled, the rate must also double. Thus, the rate must decrease to 0.02 mole¥¦ per second: = ( 0§05 mole¥¨ per second ) x ( 0. 1 M ) ( 2.0 M ) 0.02 mole© per second rate x = oRIGªnaL rate ( 0.5 M ) ( «§0 M ) new rate chanGe ªn chanGe ªn C4¬9Br NaOC¬ A completely different way to answer thIs probleM is to solve fo® the rate constant k, then put ¯n new values for the concentrations§ rate = k [C4¬9Br] °Na±C¬] 0²05 Mole ¦³´ sec ³´ = k (0.5 mol L-I) (µ.0 mol l-I) ¡ rate constant k = 0¶· ¦ mol¸´ sec¸´ rate = k [CH¢J [Na±H¹ = (0. 1 l molº´ sec³´) (0.· mol ¨³´) (2.0 mol L-I) = 0»02 mole L¡I sec ¼´ 6½ 1¾ ±rgan¿c and ¿norganic products are shown here for completeness....
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.
- Spring '10