Question

After 62.0 min, 33.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

Answer 1

We set up the first order reaction rate:
\[\LARGE [A]_{t}=[A]_{0}e ^{-k _{r}~t}\]
As we do not know the concentration A to the time t nor the concentration to the t=0 we can use the percentage to cancel the variables. Mathematically we say:
\[\LARGE [A]_{t}=[A]~0.33\]
Substutute:
\[\LARGE [A]_{0}*0.33=[A]_{0}e ^{-k _{r}~t}\]
The [A]0 cancel and we have a expression of know variables, solve for kr as the half life can be expressed as:
\[\LARGE t _{1/2}=\frac{ \ln(2) }{ k _{r} }\]

Answer 2

Correction:
\[\LARGE [A]_{t}=[A]_{0}~0.33\]

Answer 3

wait sorry most be 0.77.

Answer 4

because we know that there most be 77% to the time t.

Answer 5

using this find K then find half time