Question

The following reaction is second order in [A] and the rate constant is 0.025 M-1s-1: The concentration of A was 0.65 M at 33 s. The initial concentration of A was ________ M.

Select one:

a. 0.24

b. 1.2 � 10-2

c. 0.27

d. 2.4

e. 1.4

Answer 1

Answer:

e. 1.4.

Explanation:

Hello,

In this case, for a second-order reaction, the integrated rate law is:

$\frac{1}{[A]} =kt+\frac{1}{[A]_0}$

In such a way, for the given data, we compute the initial concentration as shown below in molar units (M):

$\frac{1}{[A]_0} =\frac{1}{[A]}-kt=\frac{1}{0.65M} -\frac{0.025}{M*s}*33s \\\\\frac{1}{[A]_0} =\frac{0.713}{M}$

$[A]_0=\frac{1M}{0.713} =1.4M$

Therefore, answer is e. 1.4 M.

Regards.

Answer 2

The initial concentration of the reaction is 1.4 M.

Using the formula;

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

Where;

[A] = concentration at time t =  0.65 M

t = time = 33 s

k = rate constant = 0.025 M-1s-1

[A]o = initial concentration = 0.025 M-1s-1

Hence;

1/ 0.65 = (0.025 × 33) +  1/[A]o

1/[A]o = (0.65)^-1 -  (0.025 × 33)

1/[A]o = 1.54 - 0.83

[A]o = 1.4 M

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