Question

Using the given data, determine the rate constant of this reaction and the correct units.

A + 2B --> C + D

              [A](M)   [B](M) Rate (M/s)

Trial 1     .360     .370    .0211

Trial 2     .360     .740     .0211

Trial 3     .720     .370     .0844

Answer 1

Solution :-

Using the data given in the table first we need to calculate the order of the each reactant

Calculating the order with respect to the reactant A by using the trial 1 and 3

Rate 3/Rate 1 = ([A]3/[A]1)^m

0.0844/0.0211 = (0.720/0.360)^m

4= 2^m

Log 4 = m* log 2

m= log 4 / log 2

m = 2

so its it second order with A

Now lets calculate order with B using the data from trial 1 and 2

Rate /Rate 1 = ([B]2/[B]1)^n

0.0211 / 0.0211 = (0.74/0.37)^n

1 = 2^n

Log 1 = 0

Therofre it is zero order with B

So the rate law for the reaction is

Rate = K [A]^2

Now lets calculate the rate constant K

K = rate / [A]^2

    = 0.0211 M/s / [0.360]^2

    = 0.163 M-1.s-1

So the rate constant of ther eaction is 0.163 M-1 .s-1