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Unformatted text preview: In mechaNism 2, Initiaion sep 2) Is exotheic BY aBout 0 Kj/mole (24 kcalmole); mecansm 2 is prefeRed 451 C l _ l c: l H + :O H O l : I U I i H H H a carBene 452 This itical eqUation Is the keY to this proBlem 1 G = 1 h T 1 S At 1400 the equiLiBIu constant is 1; therefore c f G = 0 > Assuming f H Is aBout he same at 1400 as it is at caloimeter temperaure -137 k/mole = 140 = 37000 1400 98 /mole (-23 calmole) jmole . . :L : This is a large decrease in entropY consIstent wit two molecules comBining into one 75...
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- Spring '10