Question

3 attempts left Check my work Be sure to answer all parts. The reaction 2A + B is second order with a rate constant of 51.0/M min at 24°C. (a) Starting with (Alo = 8.90 X 10 M, how long will it take for (AJ, = 3.50 x 10-M? min (b) Calculate the half-life of the reaction. min

Answer 1

First we write the data which is given to us in the question.

   k= 51.0                     (rate constant of the reaction)

[A]ₒ= 8.90×10-3          (initial concentration of the reactant)

[A]t = 3.50 ×10-3        (concentration of the reactant at time t)

Now we need to find out

    t=?        (time taken for concentration of reactant becomes [A]t from [A]ₒ)

  t1⁄2=?      (half life of the reaction)

(a) We will use integrated equation of 2nd order reaction which is given below

      kt=(1⁄[A]t - 1⁄[A]ₒ)……equation 1

On putting the values of k, [A]ₒ, and [A]t from the given data in above equation we get

51.0×t=(1⁄3.50 ×10-3 - 1⁄8.90×10-3 )

On solving the above equation we get

t= 3.39 min

(b) For t1⁄2 we use the formula written below

t1⁄2= 1⁄[A]ₒk…….equation 2

We derived this formula from the integrated equation of 2nd law, as at t1⁄2 the concentration of the reactant becomes half of the initial concentration of the reactant.

So we can write

[A]t = [A]ₒ ⁄2

Put the above value in equation 1

We get the equation 2 which is

t1⁄2= 1⁄[A]ₒk

On putting the values of k and [A]ₒ from the given data

We get

t1⁄2=1⁄8.90×10-3 ×51.0

on solving the above equation we get

t1⁄2= 2.203 min

So the value of t is 3.39 min and value of t1⁄2 is 2.203 min.