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for the reaction, 2NOBr -> 2 NO+ Br 2 , the rate law israte= k[NOBr] 2 and the r

ID: 686136 • Letter: F

Question

for the reaction, 2NOBr -> 2 NO+ Br2, the rate law israte= k[NOBr]2 and the rate constant is 0.80 l/Ms. Whatis the concentration of NOBr after 22s if the startingconcentration is 0.086 M?
I have the answer I just need a detailed explanation on how todo it
I have the answer I just need a detailed explanation on how todo it

Explanation / Answer

for the reaction, 2NOBr -> 2 NO+ Br2, the rate law israte= k[NOBr]2 and the rate constant is 0.80 l/Ms we know that rate of reaction is defined as the rate of change inconcentration of reactant with respect to time. so now rate= k[NOBr]2 ( rate constant is 0.80 l/Ms i.ek) this is a second order reaction.                                                          then if 2A           ->            product                                 initialconc                         a                                     0                                 finalconc                           a-x                                   x then rate equation is: k =( 1/t)*(x/(a*(a-x))) putting the vallues in this equation where a is initial conc and tis time we get 0.801 = (1/22)(x/(0.086(0.086-x))) so x = 0.0518 so the concentration after 22 sec = a-x                                                   = 0.086 - 0.051 conc after 22sec                     = 0.035 M (ans)