A. the reactant concentration in a zero-order reaction was 8.00×10−2 m after 155s and 3.00×10−2 m after 355s . what is the rate constant for this reaction?
b. what was the initial reactant concentration for the reaction described in part a?
c. the reactant concentration in a first-order reaction was 7.60×10−2 {\it m} after 35.0 s and 5.50×10−3 {\it m} after 85.0 s . what is the rate constant for this reaction?
d. the reactant concentration in a second-order reaction was 0.510 {\it m} after 205 s and 5.10×10−2 {\it m} after 805 s. what is the rate constant for this reaction?
the integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line, y = mx + b.
.order integrated rate law graph slope
0 [a] = -kt+[a] [a]~vs.}~t -k
1 ln[a] = -kt+ln[a]_0 ln[a]~vs.}~t -k
2 [a] =kt+1/a_0} 1/ [a]}~vs k

A) The rate constant is 2.50 × 10⁻⁴ M/s.

B) The initial concentration of the reactant is 11.9 × 10⁻² M.

C) The rate constant is 0.0525 s⁻¹

D) The rate constant is 0.0294 M⁻¹ s⁻¹

Explanation:

Hi there!

A) The equation for a zero-order reaction is the following:

[A] = -kt + [A₀]

Where:

[A] = concentrationo f reactant A at time t.

[A₀] = initial concentration of reactant A.

t = time.

k = rate constant.

We know that at t = 155 s, [A] = 8.00 × 10⁻² M and at t = 355 s [A] = 3.00 × 10⁻² M. Then:

8.00 × 10⁻² M = -k (155 s) +  [A₀]

3.00 × 10⁻² M = -k (355 s) + [A₀]

We have a system of 2 equations with 2 unknowns, let´s solve it!

Let´s solve the first equation for [A₀]:

8.00 × 10⁻² M = -k (155 s) +  [A₀]

8.00 × 10⁻² M + 155 s · k = [A₀]

Replacing [A₀] in the second equation:

3.00 × 10⁻² M = -k (355 s) + [A₀]

3.00 × 10⁻² M = -k (355 s) + 8.00 × 10⁻² M + 155 s · k

3.00 × 10⁻² M - 8.00 × 10⁻² M = -355 s · k + 155 s · k

-5.00 × 10⁻² M = -200 s · k

-5.00 × 10⁻² M/ -200 s = k

k = 2.50 × 10⁻⁴ M/s

The rate constant is 2.50 × 10⁻⁴ M/s

B) The initial reactant conentration will be:

8.00 × 10⁻² M + 155 s · k = [A₀]

8.00 × 10⁻² M + 155 s · 2.50 × 10⁻⁴ M/s = [A₀]

[A₀] = 11.9 × 10⁻² M

The initial concentration of the reactant is 11.9 × 10⁻² M

C) In this case, the equation is the following:

ln[A] = -kt + ln([A₀])

Then:

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])

Let´s solve the first equation for ln([A₀]) and replace it in the second equation:

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])

ln(7.60 × 10⁻² M) + 35.0 s · k = ln([A₀]

Replacing ln([A₀]) in the second equation:

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln(7.60 × 10⁻² M) + 35.0 s · k

ln(5.50 × 10⁻³ M) -  ln(7.60 × 10⁻² M) = -85.0 s · k + 35.0 s · k

ln(5.50 × 10⁻³ M) -  ln(7.60 × 10⁻² M) = -50.0 s · k

(ln(5.50 × 10⁻³ M) -  ln(7.60 × 10⁻² M)) / -50.0 s = k

k = 0.0525 s⁻¹

The rate constant is 0.0525 s⁻¹

D) In a second order reaction, the equation is as follows:

1/[A] = 1/[A₀] + kt

Then, we have the following system of equations:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k

Let´s solve the first equation for 1/[A₀]:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/ 0.510 M - 205 s · k = 1/[A₀]

Now let´s replace 1/[A₀] in the second equation:

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k

1/5.10 × 10⁻² M = 1/ 0.510 M - 205 s · k + 805 s · k

1/5.10 × 10⁻² M - 1/ 0.510 M = - 205 s · k + 805 s · k

1/5.10 × 10⁻² M - 1/ 0.510 M = 600 s · k

(1/5.10 × 10⁻² M - 1/ 0.510 M)/ 600 s = k

k = 0.0294 M⁻¹ s⁻¹

The rate constant is 0.0294 M⁻¹ s⁻¹