Question

What is the half life of compound if 75% of the given sample decomposes in 60min, assume first order kinetics.

This may sound silly but hear me out
if you assume 1 gram 75% would be .75 grams can you times that by 60 and the half life would be 45min?

Answer 1

@Photon336 can you help me with this?

Answer 2

@paki Can you help?

Answer 3

\(\sf A_t=A_o*e^{-kt}\)

At is the amount of the compound at time "t"
Ao is the initial amount of compound
k is the decay constant
t is time

Solve for k, then use it in: \(\sf t_{1/2}=\dfrac{ln(2)}{k}\)

Answer 4

if 75% has decayed, then that means 25% must \(remain\). How many times must you cut a sample \(in \space half\) so that you have 25% remaining?

Answer 5

@JFraser You would have to cut it in half twice, correct?

Answer 6

correct, and since the \(total\) time is 60 minutes, how long is each half-life if \(two\) half-lives takes 60 minutes?

Answer 7

@jfraser would it be 30 minutes

Answer 8

it would be, GJ

Answer 9

Awesome , Thanks @JFraser

Answer 10

@aaronq I know I closed this but I came back because I want to know how to do this with the equation as well and see if i get the same answer.

\[A _{t}=A _{o}*e ^{-kt}\] I used 100 g so 75% is 75g
\[75=100 * e ^{-k(3600s)}\]

now this is where I'm confused...you would divide 75/100 so you get
\[.75=e ^{-k(3600)}\]
Now do you take the ln of both sides ? or what do you do?

@Directrix @Photon336 maybe you could help to explain this as well?

Answer 11

@Savannah_Lynn13 your \(A_t\) isn't 75, it should be 25, because there's only 25% left of the compound

Answer 12

@JFraser Ok, so the .75 would be .25 but still how do you do the last part?

Answer 13

you would take the natural log of both sides to get rid of the \(e\) function, but that will let you solve for the rate constant, which is not what you need to do for this problem

Answer 14

you'd take the \(k\) and plug it into the second equation that @aaronq gave