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C14 Problems worked in lecture

# C14 Problems worked in lecture - with k = 0.810 M-1 s-1 at...

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Using Method of Initial Rates to Determine Rate Law A + 2B AB 2 Rate = k[A] x [B] y Exp [A],M [B],M Rate M, s -1 (Initial formation rate AB 2 ) 1 0.01 0.01 1.5x10 -4 2 0.01 0.02 1.5x10 -4 3 0.02 0.03 6.0x10 -4 2A + B 2 + C A 2 B + BC Rate = k[A] x [B 2 ] y [C] z [A],M [B],M [C],M Rate M,min -1 1 0.2 0.2 0.2 2.4x10 -6 2 0.4 0.3 0.2 9.6x10 -6 3 0.2 0.3 0.2 2.4x10 -6 4 0.2 0.4 0.6 7.2x10 -6 Integrated Rate Laws The reaction below obeys the rate law k[N 2 O 5 ], in which the specific rate constant is 0.00840 s -1 at a certain temperature. If 2.50 moles of N 2 O 5 were placed in a 5.00 liter container at that temperature, how may moles of N 2 O 5 would remain after 1.00 minute? 2 N 2 O 5 (g) 2 N 2 O 4 (g) + O 2 (g) Compound A decomposes to form B and C in a reaction that is first order with respect to A and first order overall. At 25 o C, the specific rate constant for the reaction is 0.0450 s -1 . What is the half-life of A at 25 o C? A B + C The gas phase decomposition of NOBr is second order in [NOBr],

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Unformatted text preview: with k = 0.810 M-1 s-1 at 10 o C. We start with 4.00 X 10-3 M NOBr in a flask at 10 o C. How many seconds does it take to use up 1.50 X 10-3 M of NOBr? 2 NOBr(g) → N 2 (g) + O 2 (g) + Br 2 (g) Compounds A and B react to form C and D in a reaction that was found to be second order in A. The rate constant at 30 o C is 0.622 liters per mole per minute. What is the t 1/2 of A when 4.10 X 10-2 M A is mixed with excess B? A + B → C + D Temperature and Rates The rate constant is tripled when T increases from 298K to 308K. Find E a . Mechanisms NO 2 + CO → NO + CO 2 by experiment: Rate = k[NO 2 ] 2 2 NO + Br 2 → 2 NOBr Rate = k[NO] 2 [Br 2 ] 3 rd order (1) NO + Br 2 ⇔ NOBr 2 (fast) k 1 /k-1 (2) NOBr 2 + NO → 2 NOBr (slow) k 2 2 NO + Br 2 → 2 NOBr...
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C14 Problems worked in lecture - with k = 0.810 M-1 s-1 at...

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