pseudo first order [ENDORSED]

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Ashley McClearnen 1B
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pseudo first order

In pseudo 1st order rate laws, why are the reactants in large excess left out?

Chem_Mod
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Re: pseudo first order

If one of the concentration is extremely large, then its concentration essentially remains unchanged. Thus the concentration for this item could be considered constant and be incorporated into the rate constant. As a result, it will look as if it were only a first order reaction.

Marina Gollas 1A
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Re: pseudo first order

What is a pseudo first order rate law again? My TA tried to explain it, but I still didn't really understand it. Sorry if this is a dumb question.

Megan_Ervin_1F
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Joined: Fri Sep 28, 2018 12:18 am

Re: pseudo first order  [ENDORSED]

You make the other concentrations so big that they are essentially viewed as constant so the rate law becomes "fake" first order

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Re: pseudo first order

So is it technically a second order reaction, but because one of the concentrations is so big its basically constant?

paytonm1H
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Joined: Fri Sep 28, 2018 12:18 am

Re: pseudo first order

Madison Hurst wrote:So is it technically a second order reaction, but because one of the concentrations is so big its basically constant?

it is pseudo first order with respect to the reactant you are focusing on, I think