# Effect of catalyst on kinetics

Watch
Announcements
#1
Hello

I'm reading through my lecture slides and here's what it says :

When reactant concentration <<<< catalyst concentration
• the reaction is first order. The catalyst affects absolute rate, not the relationship between rate and [reactant]

When reactant concentration >>>>> catalyst concentration
• the reaction is Zero order. The amount of catalyst controls reaction rate totally. The catalyst is 'maxed out'.

I don't understand the bit which talks about 'absolute rate'. What could this mean?? Also what do they mean by the relationship between the rate and [reactant] ?

The second bold bit I think I understand as more of the reactant is present, hence only the enzyme is the limiting factor and also the enzyme catalyst is maxed out.
0
5 years ago
#2
(Original post by James A)
Hello

I'm reading through my lecture slides and here's what it says :

When reactant concentration <<<< catalyst concentration
• the reaction is first order. The catalyst affects absolute rate, not the relationship between rate and [reactant]

When reactant concentration >>>>> catalyst concentration
• the reaction is Zero order. The amount of catalyst controls reaction rate totally. The catalyst is 'maxed out'.

I don't understand the bit which talks about 'absolute rate'. What could this mean?? Also what do they mean by the relationship between the rate and [reactant] ?

The second bold bit I think I understand as more of the reactant is present, hence only the enzyme is the limiting factor and also the enzyme catalyst is maxed out.
Hmm, I think it's quite possibly an example of some poorly worded slides. My take on it is that whilst the catalyst is greatly in excess it doesn't effect the order with respect to each reactant. I.e. It's pseudo-first order.

For instance think about it without the catalyst present where you have a system going from A --> B where v=k[A]^x
By adding a catalyst, you're not actually changing x but you are changing the absolute rate i.e. what you actually observe.
0
#3
(Original post by haydyb123)
Hmm, I think it's quite possibly an example of some poorly worded slides. My take on it is that whilst the catalyst is greatly in excess it doesn't effect the order with respect to each reactant. I.e. It's pseudo-first order.

For instance think about it without the catalyst present where you have a system going from A --> B where v=k[A]^x
By adding a catalyst, you're not actually changing x but you are changing the absolute rate i.e. what you actually observe.

In a nutshell, the catalyst has no effect on the Rate equation (except k of course as that varies with catalyst concentration) but it speeds up the reaction and what we observe?
0
5 years ago
#4
(Original post by James A)

In a nutshell, the catalyst has no effect on the Rate equation (except k of course as that varies with catalyst concentration) but it speeds up the reaction and what we observe?
Exactly; I see you study pharmacy so i'm not sure how much kinetics you've done but the fundamental equation here is the Arrhenius which links k with the activation energy.

k=Ae^-(Ea/RT) where Ea is the activation energy, so the catalyst just lowers this hence an increase in rate.
0
#5
(Original post by haydyb123)
Exactly; I see you study pharmacy so i'm not sure how much kinetics you've done but the fundamental equation here is the Arrhenius which links k with the activation energy.

k=Ae^-(Ea/RT) where Ea is the activation energy, so the catalyst just lowers this hence an increase in rate.

Ah I get you, so we can handily link k with Ea

Is there a particular reason for why this might come useful in general?

Does it allow us to calculate the k values of different temperatures?

And also there's the other form of the arrhenius equation which was basically multiplied by ln throughout?
0
5 years ago
#6
(Original post by James A)
Hello

I'm reading through my lecture slides and here's what it says :

When reactant concentration <<<< catalyst concentration
• the reaction is first order. The catalyst affects absolute rate, not the relationship between rate and [reactant]

When reactant concentration >>>>> catalyst concentration
• the reaction is Zero order. The amount of catalyst controls reaction rate totally. The catalyst is 'maxed out'.

I don't understand the bit which talks about 'absolute rate'. What could this mean?? Also what do they mean by the relationship between the rate and [reactant] ?

The second bold bit I think I understand as more of the reactant is present, hence only the enzyme is the limiting factor and also the enzyme catalyst is maxed out.
Right some editing of your question is required... Strange eh

When reactant concentration >>>>> catalyst concentration
• the reaction is Zero order. The amount of catalyst controls reaction rate totally. The catalyst is 'maxed out'.

should be

When reactant concentration >>>>> catalyst concentration
• the reaction is Zero order with respect to [reactant]. The amount of catalyst controls reaction rate totally. The catalyst is 'maxed out'.

You can consider the system to be saturated by the catalyst. In terms of enzymes, there are more active sites than molecules of reactant so has no increase in rate by increasing [enzyme]

When reactant concentration >>>>> catalyst concentration
• the reaction is Zero order. The amount of catalyst controls reaction rate totally. The catalyst is 'maxed out'.

This is fine.... In this situation you can consider it the reverse. The catalyst/enzyme active site is saturated by substrate so increasing the [reactant] doesn't increase the rate. (this assumes the catalysed reaction is much faster than the uncatalysed)
1
5 years ago
#7
(Original post by James A)

Ah I get you, so we can handily link k with Ea

Is there a particular reason for why this might come useful in general?

Does it allow us to calculate the k values of different temperatures?

And also there's the other form of the arrhenius equation which was basically multiplied by ln throughout?
Basically it's useful as it allows you to actually calculate the activation energy given the rate constant. It also allows you to thread together a lot of other thermodynamic values i.e. equilibrium constant, so it really is quite useful!

The form you're describing is just where you've taken the natural logarithm of both sides i.e. lnK= lnA - EA/RT which means that if you plot ln K against 1/RT you can obtain the negative of the activation energy as the gradient.

So pretty damn useful!
1
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (138)
59.23%
No (52)
22.32%
Not sure (43)
18.45%