The rate constant for this first-order reaction is 0.0990 s–1 at 400 °C.

A --> products

After how many seconds will 10.0% of the reactant remain?

ln(No/N) = kt

No = 100
N = 10
k from the problem
t = ? solve for this.

To find the time it takes for a certain percentage of the reactant to remain in a first-order reaction, we can use the formula for the first-order reaction kinetics:

ln([A]t/[A]0) = -kt

Where:
[A]t is the concentration of the reactant at time t
[A]0 is the initial concentration of the reactant
k is the rate constant for the reaction
t is the time

In this case, we have the rate constant (k) as 0.0990 s^(-1) at 400 °C, and we want to find the time (t) at which 10.0% of the reactant remains. Let's assume the initial concentration [A]0 is 100%, and [A]t is 10.0%.

We can rearrange the equation to solve for t:

ln(10.0%/100%) = -0.0990 s^(-1) * t

Taking the natural logarithm of both sides:

ln(0.1) = -0.0990 s^(-1) * t

Using the fact that ln(0.1) ≈ -2.303, we can substitute this value into the equation:

-2.303 = -0.0990 s^(-1) * t

Now, isolate t by dividing both sides of the equation by -0.0990 s^(-1):

t = -2.303 / -0.0990 s^(-1)

t ≈ 23.3 s

Therefore, it will take approximately 23.3 seconds for 10.0% of the reactant to remain.