NO2(g) --> NO(g) + 1/2 O2(g) The reaction is 2nd order in NO2 with a rate consta

Question

NO2(g) --> NO(g) + 1/2 O2(g) The reaction is 2nd order in NO2 with a rate constant of 0.543(Ms)^-1 at 300 degrees C. If the initial [NO2] is 0.260M, itwill take ____ s for the concentration to drop to 0.100M. (a) 0.299 (b) 11.3 (c) 8.8 x 10^-2 (d) 3.34 (e) -0.611 NO2(g) --> NO(g) + 1/2 O2(g) The reaction is 2nd order in NO2 with a rate constant of 0.543(Ms)^-1 at 300 degrees C. If the initial [NO2] is 0.260M, itwill take ____ s for the concentration to drop to 0.100M. (a) 0.299 (b) 11.3 (c) 8.8 x 10^-2 (d) 3.34 (e) -0.611

Explanation / Answer

For a second order reaction , rate law in integral form is K 2= ( 1/ t ) [ x / ( a * ( a-x ) ] Given initial concentration a = 0.26 M                                    ( a- x ) = 0.1 M            So, x = a - ( a-x )                     = 0.26 - 0.1                      0.16 M t = time taken K 2 = rate constant =  0.543 (Ms)^-1 substituting values we get , t = 11.33 s

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