Asked by Jessica

If compound X has a first-order half-life of 24 seconds, how long would you have to wait for only 33% of the original material to be left? Explain why not knowing original Molarity of X does not effect this problem.
Answered by DrBob222
k = 0.693/t<sub>1/2</sub>

Substitute k into the below equation.

ln(No/N) = kt
No = 100%
N = 33%
k from above
Solve for t = time in seconds.
Answered by Jessica
0.693/In(.100/.33)= -.5804
Answered by DrBob222
No. You never calculated k.
Answered by DrBob222
Also you know that a negative time makes no sense.
Answered by Jessica
-0.693/
In(.100/.33) = 17.23 seconds

I know I did my math off but I'm not sure if I did it correct this time would it be 17.23 sec
Answered by DrBob222
Still no. You still did not calculate k.
k = 0.693/t<sub>1/2</sub>
t<sub>1/2</sub> from the problem is 24 seconds.
k = 0.693/24 = ?
Then substitute k into the ln(No/N) = kt formula and solve for t in seconds.
Answered by Jessica
K=0.693/24 = .028875

In(.100/.33)/-.028875= 41.34 sec

Answered by DrBob222
Almost but not quite.
You changed by ln 100/33 to ln 0.1/0.33 and that's the error. You divided by 100 so ln 100/33 or ln 1.00/0.33 would be right. Again, that ln 0.100/0.33 gives you a negative number and that gives a negative time which obviously can't be right.
ln(1.00/0.33) = 0.0288t so
t = about 38 seconds.
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