Determine the rate constant for a first order reaction that has a half-life of 26.7 min.?

For a first order reaction :
half life = ln(2) / k

Then,
Rate constant, k = ln(2) / (half life) = ln(2) / (26.7 min) = 0.0260 min⁻¹

k = ln(2)/26.7 (min^-1)= 0.026 min^-1

t(1/2) = 0.693/k

Its that simple. If you know the half-life you know the rate constant and vice versa

The other folks here are giving you the equation k = ln(2) / half life... that's correct. So instead of just plopping down that equation for you, I'll show you where it came from.

from my answer here
.. https://answers.yahoo.com/question/index?qid=20160119213750AA2vPBo

you'll find this table

.. .. .order.. .. .. . non-integrated.. .. .. . ..integrated
.. .. . .. 0.. . .. . ...rate = k x [A]°... .. .. . . ..[At] = -kt + [Ao]
.. . .. .. 1.. . .. . ...rate = k x [A]¹... .. .. . . ln[At] = -kt + ln[Ao]
.. . .. . .2... . .. .. .rate = k x [A]²... .. .. . . 1/[At] = +kt + 1/[Ao]

memorize it. notice the difference between non-integrated and integrated rate equations
.. non-integrated... ... relates RATE with CONCENTRATION
... .. ..integrated.. ... .relates TIME with CONCENTRATION

**********
back to your problem...
.. first order
.. half life is TIME
so we use the equation
.. ln([At]) = -kt + ln([Ao])

rearranging
.. kt = ln( [Ao] / [At] ).... .using the rule of logs that ln(a) - ln(b) = ln(a/b)

from the definition of half life... which is t½ = time when [At] = ½[Ao]
.. k * t½ = ln( [Ao] / ½[Ao] )
.. k * t½ = ln(2)
.. k = ln(2) / t½

**************
and the rest is the same as the other folks
.. k = ln(2) / (26.7 min)