The rate constant for this first order reaction is 0.0470s-1 at 400 C.

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After how many secs will 11.0% of the reactant remain?

Josh, I worked this for you a couple of days ago.

To find out how many seconds it will take for 11.0% of the reactant to remain, we need to use the first-order reaction equation:

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

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

In this case, we want to find out the time required for 11.0% of the reactant to remain, so [A]t will be 11.0% of [A]0.

Let's plug in the values we have:

0.110[A]0/[A]0 = e^(-0.0470s^-1 * t)

Simplifying further:

0.11 = e^(-0.0470s^-1 * t)

To solve for t, we can take the natural logarithm (ln) of both sides to eliminate the exponential term:

ln(0.11) = ln(e^(-0.0470s^-1 * t))

ln(0.11) = -0.0470s^-1 * t

Now we can rearrange the equation to solve for t:

t = ln(0.11) / (-0.0470s^-1)

Using a scientific calculator, calculate ln(0.11) and divide it by -0.0470 to find the value of t. The resulting value will give you the number of seconds required for 11.0% of the reactant to remain.