AM4EVA

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Hi guys
Im going crazy trying to figure out the explanation:

The question is:
Decomposition of N2O5 yields NO2 + O2 and has a rate constant of 3.4 *10'-5sec-1 at 20 degrees celcius
Assuming first order kinetics how long will it take to decompose 50% of the N2O5?

His explantion went into half life formula that equals 0.693????? Where does that figure even come from? Please guide me

Thank you

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Jan 10, 2013
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Hello,
There's a lot of extraneous information in that question that you have to filter out. All you need to know is that it assumes first order kinetics, and that it's asking for the half life.

You should be familiar with the half lives for the rate reactions for the Chem portion of the DAT.

0th: t(1/2) = [Ao]/2k
1st: ln2/k = .693/k
2nd: 1/(k[Ao])

Hope this helps
 

Ari Rezaei

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What the guy said above ^^

Also, the 0.693 comes from the ln(2). It's a good number to know considering you won't have a calculator on the chem portion of the DAT. The explanation just wrote another version of the first order half life equation.
 
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A

AM4EVA

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Wow you guys rock!!!!
Thx a billion!!!

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PocketRocket

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0th: t(1/2) = [Ao]/2k
1st: ln2/k = .693/k
2nd: 1/(k[Ao])
srry to bring up an old thread but im doing this one and a bit confused on the formulas that were listed ^^
Where did the ln2 come from? Whats [Ao]? and where is time(t) in the formulas for 1st and 2nd order?

Are you guys understanding this conceptually or just plugging stuff into the formula?

Also, what if the question asked for anything other than a 50% decomposition, ie 40%?
 

FeralisExtremum

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srry to bring up an old thread but im doing this one and a bit confused on the formulas that were listed ^^
Where did the ln2 come from? Whats [Ao]? and where is time(t) in the formulas for 1st and 2nd order?

Are you guys understanding this conceptually or just plugging stuff into the formula?

Also, what if the question asked for anything other than a 50% decomposition, ie 40%?
Ln 2 comes from the half-life formula for a first-order reaction. Mathematically, Ln 2 = .693 so the formula for half-life of a first order reaction is often abbreviated to .693/k rather than writing out (Ln 2)/k.

[Ao] represents initial concentration. The "t½" (half-life) time is what you are solving for with these equations. For half-life times I think knowing the formulas is sufficient:
Zero order: t½ = [Ao] / 2k
First order: t½ = 0.693 / k
Second order: t½ = 1 / k [Ao]

In other words this formula is not used to give you the amount of a substance at a given half-life, it is used to determine what the length of the half-life is. Notice that for zero and second order reactions, the half-life time changes as the concentration remaining changes (since radioactive decay is first order which means the half-life is constant, we usually don't have to worry about it). I don't think they would ask you for anything other than 50% decomposition if they only gave you half-life information. In the event they did you'd have to estimate it between the amounts known at certain half-lives.
 

orgoman22

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Hi guys
Im going crazy trying to figure out the explanation:

The question is:
Decomposition of N2O5 yields NO2 + O2 and has a rate constant of 3.4 *10'-5sec-1 at 20 degrees celcius
Assuming first order kinetics how long will it take to decompose 50% of the N2O5?

His explantion went into half life formula that equals 0.693????? Where does that figure even come from? Please guide me

Thank you

Sent from my SGH-T989 using SDN Mobile
The half life is the time for 50% of a sample to decompose,,,,,,,,the needed formual for the DAT exam is Kt1/2 = .693.......where t1/2 simply is the half life time......... No need to derive it......but like any kinetic equation...it comes from integral calculus......which is not needed for the DAT. If you want me to forward you a link on the derivation message me.

Hope this helps

Dr. Jim Romano