Quantcast

Kaplan FL1 Q20

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

pine138

New Member
Joined
Dec 11, 2014
Messages
5
Reaction score
1

Members don't see this ad.
The reaction
2A + B <--> C + D
is found to take place in two steps, the first, which follows second order kinetics is slow and the second is fast. Which of the following CANNOT be the mechanism for the rate determining step of this reaction?
A) A collides with B
B) A collides with A
C) B collides with B
D) 2 A collides with 1 B


D) 2A collides with 1 B

The rate determining step follows second order kinetics. So, it must only involve two reactant molecules. Although it is often said the kinetics of a reaction cannot be derived from its stoichiometry, this is only true of the overall reaction. When a complex reaction is broken down to a series of elementary reactions, we can derive the rate law form slowest of elementary reactions.
D) has 3 reactant molecules so its wrong

I do not understand where they are getting the relationship between reaction order and the number of reactants. Can anyone shed some light?
 

aldol16

Full Member
5+ Year Member
Joined
Nov 1, 2015
Messages
5,435
Reaction score
4,225
For an elementary step, reaction order for that step is directly related to stoichiometry. This only applies for elementary steps - if you don't remember what those are, you should review kinetics. Basically, at the elementary step, reactant molecules react directly (i.e. collision) to form a product molecule with only one transition state. So literally, A and A colliding would characterize an elementary step. Since you know that the RDS involves a reaction order of two, you know that that elementary step involves two particles colliding.
 

aldol16

Full Member
5+ Year Member
Joined
Nov 1, 2015
Messages
5,435
Reaction score
4,225
What do you mean two particles colliding? Isn't 2A + B three particles?

There are multiple steps occurring here. Two particles are colliding, sticking together, and then the third particle hits them. The three particles don't come together simultaneously - the chances of that happening are quite low.
 

nostra_damus

Full Member
2+ Year Member
Joined
Dec 25, 2015
Messages
198
Reaction score
168
So because it says that the second order kinetics is slow, and you know this is the elementary step because it's the "slow step" you know that only two molecules are participating?
 

aldol16

Full Member
5+ Year Member
Joined
Nov 1, 2015
Messages
5,435
Reaction score
4,225
So because it says that the second order kinetics is slow, and you know this is the elementary step because it's the "slow step" you know that only two molecules are participating?

Here's the reasoning: It tells you explicitly that this reaction occurs in two steps. Step 1: it tells you that this is slow and follows second-order kinetics. Step 2: it tells you that this is fast. Because step 1 is the 'slow' step, it must be the rate determining step. So step 1 is what the question is asking about. Ignore step 2 from this point forward. Now, it also told you that step 1 follows second-order kinetics. That means that it depends on the concentrations of two reagents, i.e. those two reagents collide to form an intermediate that then reacts via step 2.
 
Top