Question

The decomposition of XY is second order in XY and has a rate constant of 6.94×10−3 M−1⋅s−1 at a certain temperature.

If the initial concentration of XY is 0.160 M , how long will it take for the concentration to decrease to 6.00×10−2 M ? Express your answer using two significant figures.  

If the initial concentration of XYXY is 0.050 MM, what is the concentration of XYXY after 500 ss ?

Express your answer using two significant figures.

Answer 1

Answer:

Step 1: Explanation

We can Use the following Second order kinetic equation for second-order reactions:

(1 / [A]t) = kt + (1/[A]0)

where

k= rate constant, t= time, [A]t= concentration after the elapsed time and [A]0 =initial concentration.

(a)

Step 2: Calculation of  the time (t)

Given

initial concentration = [A]0 = 0.160 M

Rate constant(K) = 6.94 × 10-3 M-1.s-1

time(t)= we need to find

concentration at time t = [A]t =  6.00 × 10-2 M

So by using the second order kinetics equation

=> (1 / [A]t) = kt + (1/[A]0)

on substituting the value

=> (1 / 6.00 × 10-2 M ) = 6.94 × 10-3  M-1 s-1 × t + (1/ 0.160 M )

=>  (16.6666667 M-1) =   6.94 × 10-3  M-1 s-1 × t   + ( 6.25 M-1 )

=>   (16.6666667 M-1) - ( 6.25 M-1 )  =   6.94 × 10-3  M-1 s-1 × t  

=> (16.6666667 M-1) - ( 6.25 M-1 )  =   6.94 × 10-3  M-1 s-1 × t  

=> 10.41666667 M-1 = 6.94 × 10-3  M-1 s-1 × t  

=> t =  10.41666667 M-1 /  6.94 × 10-3  M-1.s-1

=> t = 1.5 × 103

hence, the time taken = 1.5 × 103

(b)

Step 3: Calculation of  the concentration at time 500 s

Given

initial concentration = [A]0 = 0.050 M

K = 6.94 × 10-3 M-1.s-1

time(t)= 500 s and concentration at time t = [A]t = we need to calculate

So by using the second order kinetics equation

(1 / [A]t) = kt + (1/[A]0)

on substituting the value

=> (1 / [A]t) = 6.94 × 10-3 M-1.s-1 × 500 s + (1/ 0.050 M )

=>  (1 / [A]t) = 3.47 M-1 + ( 20 M-1 )

=>  (1 / [A]t) = 23.47 M-1

=> [A]t = 1 / 23.47 M-1

=>  [A]t = 4.3 × 10-2 M

Hence, the concentration after 500 s = 4.3 × 10-2 M