HW1_30 - [1.00] y----------- = ----------------------...

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Homework due 1/30 For the reaction 2 NO + Cl 2 -----> 2 NOCl the following kinetic data was obtained. Trial [NO] [Cl 2 ] Initial Rate 1 0.50 0.50 1.14 2 1.00 0.50 4.56 3 1.00 1.00 9.12 Calculate the rate law, the order with respect to each reactant and the overall order, and the rate constant. The rate law has the form : rate = k [NO] x [Cl 2 ] y . In order to get the value of x use the ratio of two trials in which the concentration of NO varies which and the concentration of Cl 2 is kept constant (trials 1 & 2 in this problem). To get the value of y use two trials in which the concentration of Cl 2 is varied and the concentration of NO is kept Rate2 k[NO] 2 x [Cl 2 ] 2 y 4.56 k[1.00] x [0.50] y 4.56 [1.00] x ----------- = ---------------------- -------- = ------------------------ ------- = ------------ 4 = 2 x x = 2 Rate1 k[NO]1x[Cl 2 ] 1 y 1.14 k[0.50] x [0.50] y 1.14 [0.50] x Rate3 k[NO] 3 x [Cl 2 ] 3 y 9.12 k[1.00] x [1.00] y 9.12
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Unformatted text preview: [1.00] y----------- = ---------------------- -------- = ------------------------------- = ------------ 2 = 2 y y = 1 Rate2 k[NO] 2 x [Cl 2 ] 2 y 4.56 k[1.00] x [0.50] y 4.56 [0.50] y Rate law: Rate = k[NO] 2 [Cl 2 ] 1 The order in each reactant is the power to which that concentration is raised. Thus, this reaction is second order in NO and first order in Cl 2 . The over all order is the sum of the orders of the different reactants, thus 2 + 1 or third order over all. The value of k is found by solving the rate equation for k and then substituting in the numbers for a given trial. If the rate law is correct, all of the trials will give the same value of k Rate k = ------------------------- For trial 1 k = 1.14/(0.50) 2 (0.50) = 9.12 [NO] 2 [Cl 2 ] Substituting in the data for trials 2 and 3 will also give the same value for the rate constant....
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This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.

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