chemistry! - first order rate?

Question

in a first order reaction, how many times greater is the time required for 99.9% of the reactant to be consumed than the time required to consume 50% of the same reactant?

Answer 1

For a first order reaction, if you satrt with Co amount of a reactant then the amount remaining C at time t is

C=Co*e^(-kt) =>
C/Co=e^(-kt)

When 99.9% is consumed you have 0.1% remaining thus
C1/Co= 0.001. so 0.001= e^-(kt1) (1)

When 50% is consumed, 50% remains thus C2/C0=0.5 and
0.5=e^-(kt2) (2)

Divide (2) by (1) and you get
500= e^-(kt2)/ e^-(kt1) = e^-(t2/t1) =e^(t1/t2) =>
t1/t2= ln(500) = 6.21