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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

Answer :

mauricioalessandrell

Answer:

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

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

Zero order reaction:

It is a chemical reaction wherein the rate does not vary with the increase or decrease in the concentration of the reactants

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

[A] = -kt + [A₀]

Given:

t = 155 s

[A] = 8.00 × 10⁻² M

At t = 355 s

[A] = 3.00 × 10⁻² M.

Substituting the values in the above formula:

8.00 × 10⁻² M = -k (155 s) +  [A₀].........(1)

3.00 × 10⁻² M = -k (355 s) + [A₀].........(2)

Solving for equation 1:

8.00 × 10⁻² M = -k (155 s) +  [A₀]

8.00 × 10⁻² M + 155 s · k = [A₀]............(3)

Using equation 3 in 2:

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 concentration 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])

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])............(4)

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])............(5)

Solving for equation 4:

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

ln(7.60 × 10⁻² M) + 35.0 s · k = ln([A₀])............(6)

Using equation 6 in 5:

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) For second order the reaction is as follows:

1/[A] = 1/[A₀] + kt

1/ 0.510 M = 1/[A₀] + 205 s · k............(7)

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k..........(8)

Solving for equation 7:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/ 0.510 M - 205 s · k = 1/[A₀]...........(9(

Using equation 9 in 8:

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.

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