Question

< Question 26 of 38 > Consider the initial-rate data at a certain temperature in the table for the reaction described by 2 NO,() +0,(6)-,0,() +0,(8) [NO2). (M) 0.650 [03Jo (M) 0.800 0.800 Initial rate (M/s) 3.25 x 10 5.50 x 10 15.40 x 104 1.10 1.76 1.40 Determine the value and units of the rate constant, k. Units:

Answer 1

Given Reactio : No, + O2 + N2O5 t 02 XL Let the rate Law for this reaction can be written a - 8=K[NOT 1077 where se is the order with respect to . and Y is the order with respect to For experimento Rata, y = K [0.650] * [0.30018 or 3-25X104 = K [0*650)*(0.800] => 0 For experiment ☺ r= K[140]*[0 google 5.50x104 = K [1.10] * [0.80 970 For experiment (3 Y = [1.76]*[1.46 y 15.40X104 = K[1.763€ 140] : _ First we find the order wort No and Worst Oz - Use these equations to find the order wort No and Oz

To deteomine order woriz Nog 1.69 Dividing equ © by 0 69 5. Sex Loy LG Loga (osoogo в хру - к гостој* реста 1.69 = (1.69) Do (1169)' = (1.69) 2 (aa11 Thus order wirit No, is 1 To determine order woot Oz Dividing equo 3 by® 21 oro stot [1 % * (1907 550x1o4 =1 [1 toje [ongoolt 218 = (1,6)' (1.75)? since x = 1 :: 218 = (1.6) (175) op 20 (175) or 1.75 = 1.7594 of (1:75)' = (1.75) y=1) This order wooot Oz is,

The can be written on! Law ? rate, r= K[ No ] [ 0₂] r= k [nes] [] [r=K[NO2] [₃] or 3.25X10 = 6.25 XtO" üs To defeomine rate Constant, K From 1st experiment - K (0.650m) (0.800m) 3-25x10 = K 0-52) M? o K = 3.25x104 M 0.52 12,5 - date constant, K = 6.25 x 104 ñ š v From 2nd exp : 5150x10" - K Chom) C0.800m) 5.50x 104 = K (0.88) my? - Po K = 5.50X104 . 6125 x104 Ñ'ś 0788 M25 K = 6.25 x 104 m 5' Similadly from 3d exp, we will get the same answer of K. - - These fox / K = 6.25x10 v Lon't = m sv - MS= M.S