Question

For the reaction

2 NOBr(g)----> 2 NO(g) + Br₂(g)

the rate law is second order in NOBr with a rate constant of 8.50 x 10^-2 M^-1·s^-1 at 35°C. If the initial concentration of NOBr is 1.00 M, how many seconds does it take for [NOBr] to decrease to 0.10 M?

A) 47 seconds

B) 72 seconds

C) 106 seconds

D) 163 seconds

E) 224 seconds

The answer is C (106 seconds) how do you get this? Please be detailed and show ever step.

Answer 1

The conversion of NaOBr into NO & Br2 is a second order reaction.

In this reaction rate of the reaction is dependends upon the square of the conc. of NaOBr

hence the euation applicable is,

k = 1   x 1. x x

t a (a--x )

t = 1   x 1. x x

k a (a--x )

t = 1   x    1_ x 0.9 M    (Consumed reactant conc.)

8.5x 10 ^{-2 (Sec.)} 1M (Initial Conc.) (0. 1M)   (Remaining Conc.)

By computing above , we get 105.8 = 106 seconds (Answer)