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
