Question

1.)For a second-order reaction, the initial reactant concentration is 0.62 M. After 36.3 min, the concentration is 0.22 M. What is k in M-1min-1?

2.) For a second order reaction, the initial reactant concentration, [A]o, is 0.84 M. After 12.6 s, the concentration is 0.62 M. What is [A] after 81 s? Hint given in feedback

3.)For the zero-order rate of reaction A → B + C, what is the concentration of A (in M) after 31.0 s if [A]o = 0.80 M and k = 0.010 Ms-1?

Answer 1

Answer – 1) We are given, [A]o = 0.62 M , time, t = 36.3 min, [At] = 0.22 M

Rate constant, k = ?

We know the second order integrated rate law

1/[A] = kt + 1/[A]o

So, 1/0.22 M = k * 36.3 min + 1/0.62 M

4.55 = k * 36.3 min + 1.61 M^-1

4.55 M^-1 - 1.61 M^-1 = k *36.3 min

So, k = 2.93 M^-1 / 36.3 min

         = 0.0808 M^-1. Min^-1

2) We are given, [A]o = 0.84 M , time, t = 12.6 s, [A] = 0.62 M

At time, t = 81 s , [A] = |?

First we need to calculate the Rate constant, k = ? ,

We know the second order integrated rate law

1/[A] = kt + 1/[A]o

So, 1/0.62 M = k * 12.6 s + 1/0.84 M

1.61 = k * 12.6 s + 1.19 M^-1

1.61 M^-1 - 1.19 M^-1 = k *12.6 s

So, k = 0.422 M^-1 / 12.6 s

         = 0.0335 M^-1. s^-1

Now the [A] after time, t = 81 s

1/[A] = kt + 1/[A]o

So, 1/[A] = 0.0335 M^-1.s^-1 * 81 s + 1/0.84 M

1/[A] = 3.91 M^-1

[A] = 0.26 M

3) We are given, time, t = 31.0 s , [A]o = 0.80 M , [A] = ? k = 0.010 M.s^-1

We know the zero-order integrated law –

[A] = -kt + [A]o

[A] = -0.010 M.s^-1 * 31.0 s + 0.80 M

      = 0.49 M