Question

The rate constant for this first-order reaction is 0.610 s–1 at 400 °C.

A--->products

How long (in seconds) would it take for the concentration of A to decrease from 0.690 M to 0.260 M?

The rate constant for this second-order reaction is 0.590 M–1·s–1 at 300 °C.

How long (in seconds) would it take for the concentration of A to decrease from 0.950 M to 0.330 M?

The rate constant for this zero-order reaction is 0.0230 M·s–1 at 300 °C

How long (in seconds) would it take for the concentration of A to decrease from 0.890 M to 0.240 M?

Answer 1

First order reaction a= 0.690     a-x= 0.260        k₁ =0.610 s^-1      t=?

    FORMULA    TIME t=[2.303/k] log[0.690/0.260]

                                 = 2.303/0.610   log [69/26]

                             =   3.77541 x[log69-log26]

                       =   3.77541 [1.8388-1.41497]

                     = 3.77541x0.42383

                   = 1.600 seconds

FOR second oder reaction

K₂=1/t[(1/a-x)-1/a]      k₂=       a-x= 0.330        a= 0.950

t=1/0.590[(1/0.330)-(1/0.950)]

           t=1/0.590[3.030-1.052]

       t=1/0.590[1.978]

    t=3.3525 seconds

For Zero ORDER REACTION

a= 0.890     a-x= 0.240        k₀=0.0230    

   t=1/k₀[a-(a-x)]

   t=[1/0.0230] x[0.890-0.240]

   t= [1/0.0230][0.650]

    t= 28.260secs