Question

AB(g) → A(g) + B(g) rate = k[AB]2 and k = 0.20 L/mol·s. If the initial concentration of AB is 1.50 M, what is [AB] after 14.6 s?

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.

Step 2: Extract the data from question

Given

initial concentration = [A]0 = 1.50 M

K = 0.20 M-1 s-1

time(t)= 14.6 s and concentration at time t = [A]t = we need to calculate

Step 3: Calculation of  the concentration at time 14.6 s

So by using the second order kinetics equation

[ Note : By using the unit of rate constant we can identify that which kinetic order equation we need to use

like:

Zeroth order -----> M s-1

1st order ----> s-1

2nd order ---> M-1 s-1 = L/mol.s

So by using the second order kinetics equation

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

on substituting the value

=> (1 / [A]t) = 0.20 M-1 s-1 × 14.6 s + (1/ 1.50 M )

=>  (1 / [A]t) = 2.92 M-1 + ( 0.666666667 M-1 )

=>  (1 / [A]t) = 3.5866666667 M-1

=> [A]t = 1 / 3.586666667 M-1

=>  [A]t = 0.279 M

Hence, the concentration after 14.6 s = 0.279 M