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Chemical kinetics question involving rate constant for first order reaction? 10 points!?
It took 629.9 s for 52.36% of a substance to react according to first-order kinetics. What is the value of the rate constant?
I can't seem to get the correct answer using the formula ln[A]t/[A]0=-kt, can someone please show me how to do this? thanks!
Oh my gosh, I figured out what i did wrong...I was plugging in the ratio as 47.64 instead of 0.4764...such a dumb mistake
3 Answers
- 9 years agoFavorite Answer
You're probably plugging in .5236 as the ratio.
Since 52.36% is how much has reacted, 47.64% remains.
So the ratio would be 0.4764. Try it this way, and you'll get
k=0.001177
- 9 years ago
Hello,
Using ln([A]initial/[A]remaining) = k x time b/c this is a first order reaction (meaning that half life is "independent" of concentration AND the half life is "constant")
Where [ ] is the concentration of the substance in Molarity (moles/liter), A is the substance, the rate constant in 1/seconds b/c time is in seconds.
**Since 52.36% "reacts," (or decomposes) we must find how percent is much is "remaining."
100% - 52.36% = 47.64% of substance A remains after 629.9 seconds thus...
ln(100% initial / 47.64% remaining) = k x (629.9 seconds)
.001177 1/seconds or 1.18E-3 1/seconds = k
Source(s): U of H Brain - 9 years ago
Using the same formula you mentioned, i obtained the equation,
k=2.303log(a/a-x)/t
=2.303log(100/47.64)/629.9
=2.303x0.3220/629.9
=0.0001169
Guess the mistake u did was by taking 52.36 instead of 47.64...While solving such questions,always take concentration of reactants.
