Asked by omar
Consider the first order reaction A --> products where 25% of A disappears in 24 seconds. What is the half life of this reaction?
...i thought it would just be doubled but it isn't. i'd appreciate an explanation.
...i thought it would just be doubled but it isn't. i'd appreciate an explanation.
Answers
Answered by
DrBob222
Probably I work these the long way but I do that because then I know I will get it right.
ln(No/N) = kt
I assume a value of 100 for No, and if 25% disappears, that means 75% is left so starting with 100 gives me 75 at the end of 24 seconds.
ln(100/75) = 24*k
solve for k and I get 0.01199 (I know that's too many places but I just leave those numbers in the calculator. Then for the half life
ln(100/50) = 0.01199t<sub>1/2</sub>
solve for t<sub>1/2</sub>
Check my math but I obtained 57.8 seconds which would be rounded to 58 sec if the 24 value has just 2 s.f.
ln(No/N) = kt
I assume a value of 100 for No, and if 25% disappears, that means 75% is left so starting with 100 gives me 75 at the end of 24 seconds.
ln(100/75) = 24*k
solve for k and I get 0.01199 (I know that's too many places but I just leave those numbers in the calculator. Then for the half life
ln(100/50) = 0.01199t<sub>1/2</sub>
solve for t<sub>1/2</sub>
Check my math but I obtained 57.8 seconds which would be rounded to 58 sec if the 24 value has just 2 s.f.
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