## true statement?

$V_{0} = \frac{V_{max}[S]}{K_{M} + [S]}; K_{M} = \frac{[E][S]}{[ES]}$

inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

### true statement?

"the lower the temperature, the higher the rate enhancement will be when a catalyst is added"

is this a correct statement? this question relates to question 7E.3 in the 7th edition- the whole process of it is very confusing to me but id like to understand the conceptual part first.

Kobe_Wright
Posts: 83
Joined: Fri Sep 28, 2018 12:16 am

### Re: true statement?

Yes. When you look at the Arrhenius equation we can see that increasing temperature makes the rate larger, as does lowering the activation energy, which is what a catalyst does, so doing these together should theoretically increase the enhancement of the rate.

Destiny Diaz 4D
Posts: 51
Joined: Fri Sep 28, 2018 12:28 am

### Re: true statement?

the way I think of this, just to be quick is the higher the temp. the more collisions within the reaction meaning it will react at a faster rate. so the statement is true in this context

Jeannine 1I
Posts: 73
Joined: Fri Sep 28, 2018 12:27 am

### Re: true statement?

Kobe_Wright wrote:Yes. When you look at the Arrhenius equation we can see that increasing temperature makes the rate larger, as does lowering the activation energy, which is what a catalyst does, so doing these together should theoretically increase the enhancement of the rate.

But isn't the statement saying that the lower the temperature, the greater the catalyst will work?

Kobe_Wright
Posts: 83
Joined: Fri Sep 28, 2018 12:16 am

### Re: true statement?

Jeannine 1I wrote:
Kobe_Wright wrote:Yes. When you look at the Arrhenius equation we can see that increasing temperature makes the rate larger, as does lowering the activation energy, which is what a catalyst does, so doing these together should theoretically increase the enhancement of the rate.

But isn't the statement saying that the lower the temperature, the greater the catalyst will work?

Yeah my bad when i looked at the equation i explained it right but answered wrong the statement is incorrect.