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?

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To solve these questions, we will use the first-order kinetic equation:

ln([N2O5]/[N2O5]0) = -kt

Where:
[N2O5] is the concentration of N2O5 at time t
[N2O5]0 is the initial concentration of N2O5
k is the rate constant
t is the time

(a) To find the half-life, we need to find the time it takes for [N2O5] to decrease to half of its initial concentration [N2O5]0. In this case, [N2O5]0 = [N2O5]/2.

Substituting these values into the equation, we have:

ln([N2O5]/([N2O5]/2)) = -kt

Simplifying the equation, we get:

ln(2) = -kt

Now, we can solve for t:

t = -ln(2)/k

Given that k = 0.15 s-1, we can calculate the half-life.

t = -ln(2)/0.15

(b) To find the concentration of N2O5 after 4.0 seconds, we will use the same equation and substitute the values into the equation with t = 4.0 s:

ln([N2O5]/[N2O5]0) = -kt

ln([N2O5]/0.0579) = -0.15 * 4.0

Solving for [N2O5]:

[N2O5] = 0.0579 * e^(-0.15 * 4.0)

(c) To find the time it takes for the concentration of N2O5 to decrease from 0.0579 mol·L-1 to 0.0143 mol·L-1, we need to find the time t where [N2O5] = 0.0143 mol·L-1.

Using the same equation:

ln([N2O5]/0.0579) = -0.15 * t

ln(0.0143/0.0579) = -0.15 * t

Solving for t:

t = -ln(0.0143/0.0579)/0.15

Finally, to convert the time from seconds to minutes, divide t by 60.