Verified answer

The concentration of the reactant in a first order reaction as a function of time is given by:

C(t) = C(0)*exp(-k*t)

where C(t) is the concentration at time t, and C(0) is the initial concentration.

The definition of the halflife (t_1/2) is the time at which the concentration has fallen to half it's initial value, so:

C(t_1/2) = C(0)/2 = C(0)*exp(-k*t_1/2)

1/2 = exp(-k*t_1/2)

ln(1/2) = -k*t_1/2

ln(2) = k*t_1/2 ......(remember that ln(1/x) = -ln(x))

ln(2)/t_1/2 = k

Plugging in the value you are given and dividing that into the value for ln(2) gives:

ln(2)/146*10^2 sec = 4.748*10^-5 sec^-1