Question

@whpalmer4
starting concentration is 23 milliliters to start and 5 minutes later, concentration has dropped 4 milliliters per liter. what is the half life of this reaction?

Answer 1

\[C(t) = C_0 2^{-t/t_{1/2}}\]where \(C_0\) is initial amount and \(t_{1/2}\) is half life.

You know \[C_0 = 23\]\[C(5) = C_0 - 4 = 19\]

\[\frac{C(t)}{C_0} = 2^{-t/t_{1/2}}\]\[\ln \frac{C(t)}{C_0} = -t/t_{1/2} * \ln 2\]\[-\frac{1}{\ln 2} \ln\frac{C(t)}{C_0} = t/t_{1/2}\]\[t_{1/2} = -\frac{t\ln 2 }{\ln\frac{C(t)}{C_0}}\]Plug in \(t=5, C_0 = 23, C(5) = 19\) and evaluate for half life

Answer 2

What did you get for your result here?

Answer 3

I don't know how to i put the t half thing into my calc so I was trying to research it

Answer 4

oh, \(t_{1/2}\) is just the symbolic name for the half life. Was that the confusion, or are you not sure how to evaluate the other side of the equation?

Answer 5

It's part of the study guide but I don' t know how to input it into my calc I think it's too hard

Answer 6

\[t_{1/2} = -\frac{t \ln 2}{\ln \frac{C(t)}{C_0}} = -\frac{5*0.693147}{\ln \frac{19}{23}} = -\frac{3.46574}{-0.191055} \approx 18.14\]

I added a pair of grid lines showing the initial reading as well as the half-life reading

Answer 7

Just evaluate every individual part separately and write down the numbers, then do the fraction, which I'm sure you can evaluate without trouble. What brand/model calculator do you have?

Answer 8

Ti84 I'm just not so good at using it

Answer 9

I' m going to move on to a different problem for a bit