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Can someone please explain to me why half-life is independent of concentration for all first order reaction? What about 2nd order?
Thanks.
Thanks.
Half-life question
- Thread starter Lilith
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I had to bust out my Chem 1 book but it ain't gonna be the answer you want to hear (or maybe it is).
First, the derivation from Rate = k[A] to the equation of half life requires calculus. So the good news it that part is out of scope. Now, using calculus, for first order reactions, you get:
ln([A(t)]/[A(o)] = -kt where [A(t)] is conc at time t and [A(o)] is orig conc
Then half life occurs when [A(t)] = (1/2)[A(o)] because we are looking for half-life afterall
Plug and solve, you get:
ln(1/2) = -kt
ln(2) = kt
t = ln(2)/k
So, for first order rxns the half life is t = ln(2)/k which is independent of concentration [A]!!!
FYI, Second order half lives are out of scope (because it is much harder). It is also derived using calculus. Should I go over it too? Well, start with Rate = k[A]^2 and you can derive the following using calculus:
1/[A(t)] - 1/[A(o)] = kt
Solve for half life where [A(t)] = (1/2)[A(o)], you get:
t = 1/{k[A(o)]}
So for 2nd order half life, time is dependent on concentration. Again, 2nd order half life is OUT OF SCOPE for the DAT.
Hope this helps. Sorry if I lost you...it is a bit hard to explain. I'd recommend reading up on a Chem text for more context if this didn't make sense.
First, the derivation from Rate = k[A] to the equation of half life requires calculus. So the good news it that part is out of scope. Now, using calculus, for first order reactions, you get:
ln([A(t)]/[A(o)] = -kt where [A(t)] is conc at time t and [A(o)] is orig conc
Then half life occurs when [A(t)] = (1/2)[A(o)] because we are looking for half-life afterall
Plug and solve, you get:
ln(1/2) = -kt
ln(2) = kt
t = ln(2)/k
So, for first order rxns the half life is t = ln(2)/k which is independent of concentration [A]!!!
FYI, Second order half lives are out of scope (because it is much harder). It is also derived using calculus. Should I go over it too? Well, start with Rate = k[A]^2 and you can derive the following using calculus:
1/[A(t)] - 1/[A(o)] = kt
Solve for half life where [A(t)] = (1/2)[A(o)], you get:
t = 1/{k[A(o)]}
So for 2nd order half life, time is dependent on concentration. Again, 2nd order half life is OUT OF SCOPE for the DAT.
Hope this helps. Sorry if I lost you...it is a bit hard to explain. I'd recommend reading up on a Chem text for more context if this didn't make sense.
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