Question

A second order reaction, A>3B+C, has a rate constant of 1.37*10^-2M^-1s^-1. What will be the concentration of B after 50.0s, if [A]initial=0.250M, [B]initial=0.164 and [C]initial=0.51.

Answer 1

step 1: find concentration of A after 50.0 s
we have:
[A]o = 0.25 M
t = 50.0 s
k = 1.37*10^-2 M-1.s-1
Given:
[A]o = 0.25 M

use integrated rate law for 2nd order reaction
1/[A] = 1/[A]o + k*t
1/[A] = 1/(0.25) + 1.37*10^-2*50
1/[A] = 4 + 1.37*10^-2*50
1/[A] = 4.685
[A] = 0.2134 M

step 2:
decrease in A = 0.250 - 0.2134 = 0.0366 M
increase in B = 3*decrease in A
= 3*0.0366 M
= 0.1098 M

step 3:
[B] = initial + increase
= 0.164 + 0.1098
= 0.274 M

Answer: 0.274 M