## How can a reaction be zero order?

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

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Vivian Chen 3A
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### How can a reaction be zero order?

I still don't quite understand why a reaction can be zero order. Does this mean that the reaction depends only on the activation energy or the catalyst?

Neil DSilva 1L
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: How can a reaction be zero order?

I'm not sure if it means that the reaction depends only on the activation energy or the catalyst, but the way I picture zero order reactions is like this:

Picture a hemoglobin molecule (this is going to be a very simplified picture of the body). Each hemoglobin molecule can only carry four oxygen molecules to the rest of the body. That means that if you have one hemoglobin molecule, no matter what the concentration of oxygen is, the rate of oxygenation of the body doesn't change (it is always dependent on that one hemoglobin molecule). So you can increase the concentration of the oxygen indefinitely, but the rate of oxygenation will always be the same.

So the rate is zero order with respect to oxygen and some other order with respect to hemoglobin. I believe it's a similar concept with zero order reactions: there's some type of active site where all the reactions can occur and when that site is completely saturated, the reaction can't occur any faster, so no matter how much reactant you add, the rate doesn't change.

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### Re: How can a reaction be zero order?

No reaction can really be zero order forever, because once the reactant is depleted there cannot be any more reaction (but the zero-order equation predicts a constant rate forever!)

Instead, zero-order mechanics (or rather "pseudo-zero-order") is an approximation to when a reactant is in high excess. Consider the enzymatic binding E+S-->ES (E means enzyme and S means substrate). If both reactants are first order then we have Rate=k[E][S]. But in enzymatic experiments, we keep the amount of enzyme at a fixed constant, so the rate is now Rate=k'[S]. Finally, in living cells, [S] is much much greater than [E], so [S] will barely change during the course of reaction and remains about constant. So we end up with just Rate=k''. Now, under the conditions of study, we have a zero-order model.

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