Rate equations inhibitor order Watch

Toasticide
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so as i was doing some question on determining the order in respect to a reactant from experimental data i got curious as to what order an inhibitor would be. from my knowledge in AS biology the more inhibitor there is the lower the rate, therefore the order of an inhibitor wont be zero. however since the order is what power is the increase of a reactant raised to to get the increase of rate, and when an inhibitor is added the rate decreases, does this mean it will have fractional order?

i may be over thinking this and im sure it wont be asked (inhibitors i mean, i know fractional ones are with stuff like rate determining step and what not)
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Toasticide
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bump... kinda dissapointed, tried to put up a picture of mr. bump for the lols but it wouldnt work :/
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MexicanKeith
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(Original post by Toasticide)
so as i was doing some question on determining the order in respect to a reactant from experimental data i got curious as to what order an inhibitor would be. from my knowledge in AS biology the more inhibitor there is the lower the rate, therefore the order of an inhibitor wont be zero. however since the order is what power is the increase of a reactant raised to to get the increase of rate, and when an inhibitor is added the rate decreases, does this mean it will have fractional order?

i may be over thinking this and im sure it wont be asked (inhibitors i mean, i know fractional ones are with stuff like rate determining step and what not)
The kinetics of inhibitor reactions are quite complicated. Not least because different inhibitors act in different ways on different steps of enzyme mechanism. The standard enzyme mechanism (without inhibition) is that described by Michaelis-Menten kinetics (https://en.wikipedia.org/wiki/Michae...enten_kinetics), this alone is more complicated than what you're used to at A level because it involves more than one equilibrium step, then use of the steady state approximation! Adding in an inhibitor makes the kinetics even more complicated!

More of a qualitative description would simply say, an inhibitor, which slows reaction with high concentrations should have a negative order, (ie if you write the rate equation as a fraction, it appears on the bottom) so a greater concentration increases the denominator and hence the rate decreases. But in reality the rate laws for these multistep processes are always more complicated than simple
rate= k [A]^a [B]^b [C]^c rate laws that are covered at A-level. So, good question, but pretty complicated answer!
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Toasticide
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(Original post by MexicanKeith)
The kinetics of inhibitor reactions are quite complicated. Not least because different inhibitors act in different ways on different steps of enzyme mechanism. The standard enzyme mechanism (without inhibition) is that described by Michaelis-Menten kinetics (https://en.wikipedia.org/wiki/Michae...enten_kinetics), this alone is more complicated than what you're used to at A level because it involves more than one equilibrium step, then use of the steady state approximation! Adding in an inhibitor makes the kinetics even more complicated!

More of a qualitative description would simply say, an inhibitor, which slows reaction with high concentrations should have a negative order, (ie if you write the rate equation as a fraction, it appears on the bottom) so a greater concentration increases the denominator and hence the rate decreases. But in reality the rate laws for these multistep processes are always more complicated than simple
rate= k [A]^a [B]^b [C]^c rate laws that are covered at A-level. So, good question, but pretty complicated answer!
ah thanks for the answer
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