The rate constant for this first-order reaction is 0.0570 s–1 at 400 °C.

Question

A-->Products
After how many seconds will 14.0% of the reactant remain?

Answer 1

ln(No/N) = kt
If you use No = 100, then N = 14, substitute k and solve for t.

Answer 2

To expand on Dr.Bob's answer:
The usual formula is:
ln([A-t]/[A-0]) = -kt
but you're looking for t so you have to rearrange the formula:
ln(concentration over 100)/-(k) = t
[ln(14/100)]/-0.0570 = t
34.36834015 = t
Your answer should be 34.4 seconds