a first order reaction has a rate constant of 7.5 x 10^-3/s. The time required for th reaction to be 60% complete is

Use ln(No/N) = kt.

I wold call No = 100, then N = 40
Solve for t.

To find the time required for the reaction to be 60% complete, we can use the integrated rate law for first-order reactions:

ln[A]t = -kt + ln[A]0

where [A]t is the concentration of reactant A at time t, k is the rate constant, and [A]0 is the initial concentration.

Since the reaction is 60% complete, [A]t = 0.60[A]0.

ln(0.60[A]0) = -kt + ln[A]0

Simplifying the equation, we have:

ln(0.60) + ln([A]0) = -kt + ln([A]0)

ln(0.60) = -kt

t = -ln(0.60)/k

Plugging in the values, k = 7.5 x 10^-3/s, we can calculate the time required for the reaction to be 60% complete:

t = -ln(0.60)/(7.5 x 10^-3/s)

t ≈ 130.6 seconds

Therefore, the time required for the reaction to be 60% complete is approximately 130.6 seconds.

To determine the time required for the reaction to be 60% complete in a first-order reaction, we can use the integrated rate law for a first-order reaction:

ln(A/A₀) = -kt

Where:
A is the concentration of the reactant at time t
A₀ is the initial concentration of the reactant
k is the rate constant
t is the reaction time

In this case, we want to find the time (t) when the reaction is 60% complete. This means that the concentration (A) at time t is 60% of the initial concentration (A₀):

A/A₀ = 0.6

Substituting this into the equation:

ln(0.6) = -k * t

Now, we need to rearrange the equation to solve for t:

t = -ln(0.6) / k

Given that the rate constant (k) is 7.5 x 10^-3/s, we can substitute this value into the equation to find the time required:

t = -ln(0.6) / (7.5 x 10^-3/s)

Calculating this:

t ≈ 92.22 seconds

Therefore, the time required for the reaction to be 60% complete is approximately 92.22 seconds.

68 seconds