The rate constant for this first order reaction is 0.0470 s-1 at 400 C.
A TO Products
After how many seconds will 11.0% of the reactant remain?
This is the long way around but this is what I would do.
ln(No/N) = kt
I would call No = 100
Then N would be 11.
Substitute k and solve for t in seconds.
To solve this problem, 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
We are given:
k = 0.0470 s^(-1)
We need to find t when 11.0% of the reactant remains, which means [A]t = 0.110[A]0
Substituting these values into the first order reaction equation, we get:
ln(0.110[A]0/[A]0) = -0.0470 s^(-1) * t
Simplifying the equation, we have:
ln(0.110) = -0.0470 s^(-1) * t
Now we can rearrange the equation to solve for t:
t = ln(0.110) / -0.0470 s^(-1)
Using the natural logarithm function in a calculator, we can find the value of ln(0.110) and divide it by -0.0470 s^(-1) to get the value of t. This will give us the time in seconds after which 11.0% of the reactant remains.