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Unformatted text preview: between Exp 1 and Exp 2, and Exp 2 and Exp 4, [A] and [B] triples, rate triplesthe order with respect to A and to B is 1 (both triple) k =0.010 1/Ms Example 163first order A, first order B, second order overallExp 2: [A] = 0.80 MExp 3: [B] = 0.20 MIntegrated Rate Equation1 st order ln [A] /[A] = a k td[A]/dt = a k t [A] = mol/L of A at time t=0 [A] = mol/L of A at time t k = specific T = 1/ak * ln([A] /[A])halflife (t 1/2 ) of the reactant is the time required for half of the reactant to be consumed, or the time at which [A] = [A]halflife is constant t = t 1/2 t 1/2 = 0.693/ ak ln [A] /[A] = a k tuse to find the time [A] = [A] eakt Example 164 t/2 = 0.075 s Example 165 0.030 g Example 166 k = 0.00101 hr1 [A] = 2.86 g left2nd order 1/[A] 1/[A] = a k t t 1/2 = 1/ ak[A]halflife depends on how much stuff was initially there Example 167...
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This note was uploaded on 10/04/2011 for the course CHEM 1212 taught by Professor Suggs during the Spring '08 term at University of Georgia Athens.
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