Dinitrogen pentoxide, N2O5, decomposes by first-order kinetics with a rate constant of 0.15 s-1 at 353 K.
(a) What is the half-life (in seconds) for the decomposition of N2O5 at 353 K?
(b) If [N2O5]0 = 0.0579 mol·L-1, what will be the concentration of N2O5 after 4.0 s?
(c) How much time (in minutes) will elapse before the N2O5 concentration decreases from 0.0579 mol·L-1 to 0.0143 mol·L-1?
To solve this problem, we can use the integrated rate law for a first-order reaction:
ln([A]/[A]0) = -kt
where [A] is the concentration at a given time, [A]0 is the initial concentration, k is the rate constant, and t is the time.
(a) To find the half-life, we can rearrange the integrated rate law equation as follows:
ln(1/2) = -kt1/2
The half-life (t1/2) is the time it takes for the concentration to decrease by half.
Substituting the given values into the equation:
ln(1/2) = -0.15 s-1 * t1/2
ln(1/2) = -0.15 t1/2
Now, we can solve for t1/2 using natural logarithms:
t1/2 = ln(1/2) / -0.15
t1/2 ≈ 4.61 seconds
Therefore, the half-life for the decomposition of N2O5 at 353 K is approximately 4.61 seconds.
(b) To find the concentration of N2O5 after 4.0 seconds, we can use the integrated rate law again:
ln([N2O5]/[N2O5]0) = -kt
Rearranging the equation:
[N2O5] = [N2O5]0 * e^(-kt)
Substituting the given values:
[N2O5] = 0.0579 mol·L-1 * e^(-0.15 s-1 * 4.0 s)
[N2O5] ≈ 0.0415 mol·L-1
Therefore, the concentration of N2O5 after 4.0 seconds is approximately 0.0415 mol·L-1.
(c) To find the time it takes for the N2O5 concentration to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1, we can rearrange the equation and solve for t:
ln([N2O5]/[N2O5]0) = -kt
t = ln([N2O5]/[N2O5]0) / -k
Substituting the given values:
t = ln(0.0143 mol·L-1 / 0.0579 mol·L-1) / -0.15 s-1
t ≈ 8.86 seconds
Since there are 60 seconds in a minute, we can convert the time from seconds to minutes:
t = 8.86 s * (1 min / 60 s)
t ≈ 0.148 minutes
Therefore, it takes approximately 0.148 minutes for the N2O5 concentration to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1.
To solve these problems, we can use the first-order kinetics equation:
ln([N2O5]t/[N2O5]0) = -kt,
where [N2O5]t is the concentration of N2O5 at time t, [N2O5]0 is the initial concentration, k is the rate constant, and t is the time.
(a) The half-life (t1/2) is the time taken for the concentration to decrease to half of the initial concentration ([N2O5]0/2). We can use the formula for half-life of a first-order reaction:
t1/2 = ln(2)/k.
Substituting the given value of k = 0.15 s-1, we can calculate:
t1/2 = ln(2)/0.15 ≈ 4.6229 s.
Therefore, the half-life of the decomposition of N2O5 at 353 K is approximately 4.6229 seconds.
(b) To calculate the concentration of N2O5 after 4.0 seconds, we can rearrange the first-order kinetics equation:
[N2O5]t = [N2O5]0 * e^(-kt).
Substituting the given values [N2O5]0 = 0.0579 mol·L-1, k = 0.15 s-1, and t = 4.0 s:
[N2O5]t = 0.0579 * e^(-0.15*4) ≈ 0.0452 mol·L-1.
Therefore, the concentration of N2O5 after 4.0 seconds is approximately 0.0452 mol·L-1.
(c) To find the time it takes for the concentration to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1, we can rearrange the first-order kinetics equation:
t = -(ln([N2O5]t/[N2O5]0))/k.
Substituting the given values [N2O5]0 = 0.0579 mol·L-1, [N2O5]t = 0.0143 mol·L-1, and k = 0.15 s-1:
t = -(ln(0.0143/0.0579))/0.15 ≈ 14.9 s.
Since we want the time in minutes, we divide by 60:
t = 14.9/60 ≈ 0.2483 min.
Therefore, it will take approximately 0.2483 minutes for the N2O5 concentration to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1.
a. k = 0.693/t1/2
b. ln(No/N) = kt.
No = given
N =?
k = from part a.
t = 4 s.
c.
same equation as b but
No = given
N = given
k = from a
t = ?
Note that part a is given in seconds, part c wants the answer in min.