What is the half-life (in minutes) of this reaction?

Dr. A is correct. However I prefer to memorize just 1 equation and then work half-life problems using that equation.

ln(Ao/A) = kt (works for all 1st order reactions)

Ao = original concentration and A = concentration after time t

k is the rate constant

In this problem you are give the rate constant. At the half life, Ao/A will be 2 (100%/50%)

ln 2 = 0.00059 x t

[Note that you don't need the initial concentration in this case, because atfter 1 half life Ao/A will always be 2]

Supose you were given the half life was equal to 1175 sec and you were asked to find the amount remaining after 10.00 minutes. You need to calculate the rate constant-

ln 2 = k * 1175 s

k = 0.000590 s^-1

Now you can calculate the amount remaining after 10.00 minutes (600 seconds)-

ln(4.5x10^-2/A) = 0.000590 x 600

ln(4.5x10^-2/A) = 0.354

4.5x10^-2/A = 1.425

A = 3.16x10^-2

t 1/2 = ln 2 / k = 0.693 / 0.00059 = 1175 s