## Rate Constants

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Kennedi2J
Posts: 101
Joined: Wed Sep 18, 2019 12:20 am

### Rate Constants

What does it mean for a reaction to be pseudo-first-order?

Daniel Yu 1E
Posts: 100
Joined: Sat Aug 24, 2019 12:15 am

### Re: Rate Constants

A Pseudo First Order Reaction is reaction which is not first-order reaction naturally but made first order by increasing or decreasing the concentration of one or the other reactant is known as Pseudo first order reaction. Pseudo means 'fake'.

Posts: 50
Joined: Fri Sep 27, 2019 12:29 am

### Re: Rate Constants

To be pseudo-first-order, the concentration of one of the reactants in the rate law must be increased to be much larger than the concentration of the other reactant. For example, look at #15 from ENDGAME. The concentration of A has been increased so much that even when A is used up by the reaction, there is essentially no change in the concentration of A. Because of that, we can say that the rate law of that reaction is only dependent on [B] and is, therefore, pseudo-first-order even though the actual rate law = k[A][B] which is 2nd order.

Philomena 4F
Posts: 27
Joined: Wed Feb 27, 2019 12:16 am

### Re: Rate Constants

Anuradha S 1F wrote:To be pseudo-first-order, the concentration of one of the reactants in the rate law must be increased to be much larger than the concentration of the other reactant. For example, look at #15 from ENDGAME. The concentration of A has been increased so much that even when A is used up by the reaction, there is essentially no change in the concentration of A. Because of that, we can say that the rate law of that reaction is only dependent on [B] and is, therefore, pseudo-first-order even though the actual rate law = k[A][B] which is 2nd order.

Is there a guide on how large one [reactant] has to be compared to the other [reactant] for this concept to be applied?