[Joules23]

The rate law for the following reaction at some temperature is given below.


2 NOBr(g)  2 NO(g) + Br2(g)

(a) If the half-life for this reaction is 2.00 s when [NOBr]0 = 0.900 M, calculate the value of k for this reaction.
 .556 L/mol·s
(b) How much time is required for the concentration of NOBr to decrease to 0.137 M?
  s

I need help with part (b)...
would the change in [NOBr] be .900-.137=.763
and the Delta t = X-2 (x is what im solving for)
then set that equal to (.556)[.900]^2

[Yggdrasil]

To solve the problem you have to treat the rate law as a differential equation and integrate the differential equation to find an equation for the concentration of NOBr as a function of t.  It may be helpful to look up integrated rate equations in your chemistry text.

[Joules23]

ln[NOBr] = -kt + ln[NOBr]²_o

k=.556
[NObr]=.137
[NOBr]_o=1.8

yes?

[champ]

the order of this reaction is second so use the equation for 2nd order reaction.
1/[A]=1/[A]o+akt 
now u solve for "t" and you'll get your answer...

[LQ43]

1/[A]=1/[A]o+akt 
now u solve for "t" and you'll get your answer...

no a before the kt

[champ]

there is "a" before kt because in this reaction order its value is 2. i hope your answer is approximately 6 seconds...
is it? if not please correct me.. thanx